CBSE Class 12 Mathematics Vector Algebra Assignment Set 02

Topic 9: Vector Algebra

Schematic Diagram

Some Important Results/Concepts

• Position vector of point \( A(x, y, z) = \vec{OA} = x\hat{i} + y\hat{j} + z\hat{k} \)
• If \( A(x_1, y_1, z_1) \) and point \( B(x_2, y_2, z_2) \) then \( \vec{AB} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k} \)
• If \( \vec{a} = x\hat{i} + y\hat{j} + z\hat{k} \) ; \( |\vec{a}| = \sqrt{x^2 + y^2 + z^2} \)
• Unit vector parallel to \( \vec{a} = \frac{\vec{a}}{|\vec{a}|} \)
• Scalar Product (dot product) between two vectors: \( \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta \) ; \( \theta \) is angle between the vectors
• \( \cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \)
• If \( \vec{a} = a_1\hat{i} + b_1\hat{j} + c_1\hat{k} \) and \( \vec{b} = a_2\hat{i} + b_2\hat{j} + c_2\hat{k} \) then \( \vec{a} \cdot \vec{b} = a_1 a_2 + b_1 b_2 + c_1 c_2 \)
• If \( \vec{a} \) is perpendicular to \( \vec{b} \) then \( \vec{a} \cdot \vec{b} = 0 \)
• \( \vec{a} \cdot \vec{a} = |\vec{a}|^2 \)
• Projection of \( \vec{a} \) on \( \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \)
• Vector product between two vectors: \( \vec{a} \times \vec{b} = |\vec{a}||\vec{b}|\sin\theta\,\hat{n} \) ; \( \hat{n} \) is the normal unit vector which is perpendicular to both \( \vec{a} \) & \( \vec{b} \)
• \( \hat{n} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} \)
• If \( \vec{a} \) is parallel to \( \vec{b} \) then \( \vec{a} \times \vec{b} = \vec{0} \)
• Area of triangle (whose sides are given by \( \vec{a} \) and \( \vec{b} \)) = \( \frac{1}{2}|\vec{a} \times \vec{b}| \)
• Area of parallelogram (whose adjacent sides are given by \( \vec{a} \) and \( \vec{b} \)) = \( |\vec{a} \times \vec{b}| \)
• Area of parallelogram (whose diagonals are given by \( \vec{a} \) and \( \vec{b} \)) = \( \frac{1}{2}|\vec{a} \times \vec{b}| \)

Assignments

(i) Vector and scalars, Direction ratio and direction cosines&Unit vector

Level I

Question 1. If \( \vec{a} = \hat{i} + \hat{j} - 5\hat{k} \) and \( \vec{b} = \hat{i} - 4\hat{j} + 3\hat{k} \) find a unit vector parallel to \( \vec{a} + \vec{b} \)
Answer: First compute the resultant vector by taking the sum of \( \vec{a} \) and \( \vec{b} \):
\( \vec{c} = \vec{a} + \vec{b} = (\hat{i} + \hat{j} - 5\hat{k}) + (\hat{i} - 4\hat{j} + 3\hat{k}) = 2\hat{i} - 3\hat{j} - 2\hat{k} \).
Next, evaluate the magnitude of this sum:
\( |\vec{c}| = \sqrt{2^2 + (-3)^2 + (-2)^2} = \sqrt{4 + 9 + 4} = \sqrt{17} \).
A unit vector aligned with \( \vec{a} + \vec{b} \) is obtained by dividing the vector by its length:
\( \hat{c} = \pm \frac{\vec{c}}{|\vec{c}|} = \pm \frac{2\hat{i} - 3\hat{j} - 2\hat{k}}{\sqrt{17}} = \pm \left( \frac{2}{\sqrt{17}}\hat{i} - \frac{3}{\sqrt{17}}\hat{j} - \frac{2}{\sqrt{17}}\hat{k} \right) \).
In simple words: Add the two vectors together by components, work out their total length, and divide each coordinate by that length.

Exam Tip: Remember to include the \( \pm \) sign whenever a question asks for a vector "parallel to" a given direction.

Question 2. Write a vector of magnitude 15 units in the direction of vector \( \hat{i} - 2\hat{j} + 2\hat{k} \)
Answer: Let \( \vec{a} = \hat{i} - 2\hat{j} + 2\hat{k} \).
Find the magnitude of \( \vec{a} \):
\( |\vec{a}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \).
The unit vector pointing along \( \vec{a} \) is:
\( \hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{\hat{i} - 2\hat{j} + 2\hat{k}}{3} \).
Now scale this unit vector by the desired magnitude of 15 units:
\( \vec{v} = 15\,\hat{a} = 15 \left( \frac{\hat{i} - 2\hat{j} + 2\hat{k}}{3} \right) = 5(\hat{i} - 2\hat{j} + 2\hat{k}) = 5\hat{i} - 10\hat{j} + 10\hat{k} \).
In simple words: Find the 1-unit direction vector first, then multiply it by 15.

Exam Tip: Never multiply the unnormalized vector directly by the requested magnitude; always normalize to a unit vector first.

Question 3. If \( \vec{a} = \hat{i} + \hat{j} - \hat{k} \), \( \vec{b} = \hat{i} - \hat{j} + \hat{k} \), \( \vec{c} = -\hat{i} + \hat{j} + \hat{k} \) find a unit vector in the direction of \( \vec{a} + \vec{b} + \vec{c} \)
Answer: Let \( \vec{r} = \vec{a} + \vec{b} + \vec{c} \). Adding their respective components gives:
\( \vec{r} = (1 + 1 - 1)\hat{i} + (1 - 1 + 1)\hat{j} + (-1 + 1 + 1)\hat{k} = \hat{i} + \hat{j} + \hat{k} \).
Calculate the magnitude of \( \vec{r} \):
\( |\vec{r}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \).
The required unit vector along this direction is:
\( \hat{r} = \frac{\vec{r}}{|\vec{r}|} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k} \).
In simple words: Combine all three vectors by adding like parts, then divide by the square root of 3.

Exam Tip: Add corresponding \( \hat{i}, \hat{j}, \hat{k} \) coefficients carefully to avoid simple sign slips.

Question 4. Find a unit vector in the direction of the vector \( \vec{a} = 2\hat{i} + \hat{j} + 2\hat{k} \) [ CBSE 2011]
Answer: Compute the absolute length of vector \( \vec{a} \):
\( |\vec{a}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \).
The unit vector in the same direction is:
\( \hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{2\hat{i} + \hat{j} + 2\hat{k}}{3} = \frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} + \frac{2}{3}\hat{k} \).
In simple words: Divide each coordinate of the vector by its total length of 3.

Exam Tip: Writing the unit vector in separated component form \( \frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} + \frac{2}{3}\hat{k} \) earns full credit under board marking schemes.

Question 5. Find a vector in the direction of vector \( \vec{a} = \hat{i} - 2\hat{j} \), whose magnitude is 7
Answer: First determine the norm of \( \vec{a} \):
\( |\vec{a}| = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \).
The unit vector pointing in the direction of \( \vec{a} \) is:
\( \hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{\hat{i} - 2\hat{j}}{\sqrt{5}} \).
Multiplying by the required magnitude 7 produces:
\( \vec{v} = 7\,\hat{a} = \frac{7}{\sqrt{5}}(\hat{i} - 2\hat{j}) = \frac{7}{\sqrt{5}}\hat{i} - \frac{14}{\sqrt{5}}\hat{j} \).
In simple words: Find the unit vector by dividing by the square root of 5, then scale it up by multiplying with 7.

Exam Tip: Keep radicals in the denominator or rationalize neatly; leaving the fraction factored out as \( \frac{7}{\sqrt{5}}(\hat{i} - 2\hat{j}) \) is entirely acceptable.

Level II

Question 1. Find a vector of magnitude 5 units, perpendicular to each of the vectors \( (\vec{a} + \vec{b}) \), \( (\vec{a} - \vec{b}) \) where \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} + 3\hat{k} \).
Answer: First determine the vectors \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \):
\( \vec{a} + \vec{b} = (\hat{i} + \hat{j} + \hat{k}) + (\hat{i} + 2\hat{j} + 3\hat{k}) = 2\hat{i} + 3\hat{j} + 4\hat{k} \)
\( \vec{a} - \vec{b} = (\hat{i} + \hat{j} + \hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) = 0\hat{i} - \hat{j} - 2\hat{k} \).
A vector normal to both is given by their cross product:
\( \vec{n} = (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 0 & -1 & -2 \end{vmatrix} \)
\( = \hat{i}(-6 - (-4)) - \hat{j}(-4 - 0) + \hat{k}(-2 - 0) = -2\hat{i} + 4\hat{j} - 2\hat{k} \).
Evaluate the magnitude of \( \vec{n} \):
\( |\vec{n}| = \sqrt{(-2)^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt{6} \).
The unit normal vector is:
\( \hat{n} = \pm \frac{\vec{n}}{|\vec{n}|} = \pm \frac{-2\hat{i} + 4\hat{j} - 2\hat{k}}{2\sqrt{6}} = \pm \frac{-\hat{i} + 2\hat{j} - \hat{k}}{\sqrt{6}} \).
Multiplying by the requested magnitude of 5 units gives:
\( \vec{v} = \pm \frac{5}{\sqrt{6}}(-\hat{i} + 2\hat{j} - \hat{k}) \).
In simple words: Take the cross product of the sum and difference vectors to find a perpendicular line, turn it into a unit vector, and multiply by 5.

Exam Tip: Remember that two opposite directions are both perpendicular to a given pair, so always write \( \pm \).

Question 2. If the sum of two unit vectors is a unit vector, show that the magnitude of their difference is \( \sqrt{3} \).
Answer: Let \( \hat{a} \) and \( \hat{b} \) be two unit vectors such that \( |\hat{a}| = 1 \), \( |\hat{b}| = 1 \), and their sum is also a unit vector, so \( |\hat{a} + \hat{b}| = 1 \).
Expanding the square of the magnitude of their sum:
\( |\hat{a} + \hat{b}|^2 = |\hat{a}|^2 + |\hat{b}|^2 + 2(\hat{a} \cdot \hat{b}) \)
\( \implies 1^2 = 1^2 + 1^2 + 2(\hat{a} \cdot \hat{b}) \)
\( \implies 1 = 2 + 2(\hat{a} \cdot \hat{b}) \)
\( \implies 2(\hat{a} \cdot \hat{b}) = -1 \).
Now consider the square of the magnitude of their difference:
\( |\hat{a} - \hat{b}|^2 = |\hat{a}|^2 + |\hat{b}|^2 - 2(\hat{a} \cdot \hat{b}) \)
Substitute \( 2(\hat{a} \cdot \hat{b}) = -1 \):
\( |\hat{a} - \hat{b}|^2 = 1 + 1 - (-1) = 2 + 1 = 3 \).
Taking the positive square root gives:
\( |\hat{a} - \hat{b}| = \sqrt{3} \).
Hence proved.
In simple words: Use the length formula on the sum to find the dot product, then put that dot product into the difference formula.

Exam Tip: Avoid evaluating the angle \( \theta \) explicitly; solving directly for \( 2(\hat{a} \cdot \hat{b}) \) makes the proof faster and eliminates trigonometric errors.

Question 3. If \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \), \( \vec{b} = 4\hat{i} - 2\hat{j} + 3\hat{k} \) and \( \vec{c} = \hat{i} - 2\hat{j} + \hat{k} \), find a vector of magnitude 6 units which is parallel to the vector \( 2\vec{a} - \vec{b} + 3\vec{c} \)
Answer: Let \( \vec{r} = 2\vec{a} - \vec{b} + 3\vec{c} \).
Evaluate the linear combination:
\( \vec{r} = 2(\hat{i} + \hat{j} + \hat{k}) - (4\hat{i} - 2\hat{j} + 3\hat{k}) + 3(\hat{i} - 2\hat{j} + \hat{k}) \)
\( = (2 - 4 + 3)\hat{i} + (2 + 2 - 6)\hat{j} + (2 - 3 + 3)\hat{k} \)
\( = \hat{i} - 2\hat{j} + 2\hat{k} \).
Calculate the magnitude of \( \vec{r} \):
\( |\vec{r}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \).
The unit vector parallel to \( \vec{r} \) is:
\( \hat{r} = \pm \frac{\vec{r}}{|\vec{r}|} = \pm \frac{\hat{i} - 2\hat{j} + 2\hat{k}}{3} \).
Multiplying by the desired magnitude of 6 units:
\( \vec{v} = \pm 6\,\hat{r} = \pm 6 \left( \frac{\hat{i} - 2\hat{j} + 2\hat{k}}{3} \right) = \pm 2(\hat{i} - 2\hat{j} + 2\hat{k}) = \pm (2\hat{i} - 4\hat{j} + 4\hat{k}) \).
In simple words: Calculate the combined vector, divide by its length 3 to get a 1-unit vector, and scale up by 6.

Exam Tip: Be sure to include both signs (\( \pm \)) because vectors parallel to a given line can point in either direction.

Level - III

Question 1. If a line make α,β, γ with the X - axis , Y– axis and Z – axis respectively, then find the value of \( \sin^2\alpha + \sin^2\beta + \sin^2\gamma \)
Answer: Let \( l = \cos\alpha \), \( m = \cos\beta \), and \( n = \cos\gamma \) be the direction cosines of the given line.
A fundamental identity of direction cosines states that:
\( \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \).
Using the trigonometric relation \( \sin^2\theta = 1 - \cos^2\theta \) for each angle:
\( \sin^2\alpha + \sin^2\beta + \sin^2\gamma = (1 - \cos^2\alpha) + (1 - \cos^2\beta) + (1 - \cos^2\gamma) \)
\( = 3 - (\cos^2\alpha + \cos^2\beta + \cos^2\gamma) \)
\( = 3 - 1 = 2 \).
Therefore, the value is 2.
In simple words: Change every sine squared into 1 minus cosine squared, then use the fact that the sum of squared direction cosines is 1.

Exam Tip: State the direction cosine property \( l^2 + m^2 + n^2 = 1 \) explicitly to establish the starting step clearly.

Question 2. For what value of p, is \( (\hat{i} + \hat{j} + \hat{k})p \) a unit vector?
Answer: Let \( \vec{v} = p(\hat{i} + \hat{j} + \hat{k}) \).
For \( \vec{v} \) to be a unit vector, its length must equal 1:
\( |\vec{v}| = 1 \)
\( \implies |p|\sqrt{1^2 + 1^2 + 1^2} = 1 \)
\( \implies |p|\sqrt{3} = 1 \)
\( \implies |p| = \frac{1}{\sqrt{3}} \)
\( \implies p = \pm \frac{1}{\sqrt{3}} \).
In simple words: The magnitude of the sum of unit vectors along the axes is root 3, so p must be plus or minus 1 divided by root 3.

Exam Tip: Always remember that \( p \) can be either positive or negative because the magnitude involves an absolute value.

Question 3. What is the cosine of the angle which the vector \( \sqrt{2}\hat{i} + \hat{j} + \hat{k} \) makes with Y-axis
Answer: Let \( \vec{a} = \sqrt{2}\hat{i} + \hat{j} + \hat{k} \).
Compute the norm of \( \vec{a} \):
\( |\vec{a}| = \sqrt{(\sqrt{2})^2 + 1^2 + 1^2} = \sqrt{2 + 1 + 1} = \sqrt{4} = 2 \).
The cosine of the angle \( \beta \) that a vector makes with the Y-axis is its direction cosine along the y-coordinate:
\( \cos\beta = \frac{a_y}{|\vec{a}|} = \frac{1}{2} \).
Hence, the cosine of the angle with the Y-axis is \( \frac{1}{2} \).
In simple words: Take the y-component of the vector and divide it by the vector's total length.

Exam Tip: If the question asks for the angle itself, state \( \beta = \frac{\pi}{3} \) or \( 60^\circ \); here, provide the cosine value \( \frac{1}{2} \) as asked.

Question 4. Write the value of p for which \( \vec{a} = 3\hat{i} + 2\hat{j} + 9\hat{k} \) and \( \vec{b} = \hat{i} + p\hat{j} + 3\hat{k} \) are parallel vectors.
Answer: Two non-zero vectors \( \vec{a} \) and \( \vec{b} \) are parallel if their corresponding scalar components are proportional:
\( \frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} \).
Substituting the given values:
\( \frac{3}{1} = \frac{2}{p} = \frac{9}{3} \).
Since \( \frac{3}{1} = 3 \) and \( \frac{9}{3} = 3 \), equating with the middle term gives:
\( 3 = \frac{2}{p} 
\implies p = \frac{2}{3} \).
In simple words: Set the ratios of matching components equal to each other and solve for p.

Exam Tip: Verify that both outer ratios match before equating to the middle ratio to be sure the vectors can indeed be parallel.

(ii)Position vector of a point and collinear vectors

Level - I

Question 1. Find the position vector of the midpoint of the line segment joining the points \( A(5\hat{i} + 3\hat{j}) \) and \( B(3\hat{i} – \hat{j}) \).
Answer: Let \( \vec{a} = 5\hat{i} + 3\hat{j} \) and \( \vec{b} = 3\hat{i} - \hat{j} \) represent the position vectors of points A and B respectively.
The position vector \( \vec{m} \) of the midpoint of line segment AB is given by the average of their position vectors:
\( \vec{m} = \frac{\vec{a} + \vec{b}}{2} = \frac{(5\hat{i} + 3\hat{j}) + (3\hat{i} - \hat{j})}{2} = \frac{8\hat{i} + 2\hat{j}}{2} = 4\hat{i} + \hat{j} \).
Thus, the midpoint's position vector is \( 4\hat{i} + \hat{j} \).
In simple words: Add the coordinates of the two endpoints together and divide by 2.

Exam Tip: State the midpoint formula \( \vec{r} = \frac{\vec{a} + \vec{b}}{2} \) before carrying out the addition.

Question 2. In a triangle ABC, the sides AB and BC are represents by vectors \( 2\hat{i} - \hat{j} + 2\hat{k} \) , \( \hat{i} + 3\hat{j} + 5\hat{k} \) respectively. Find the vector representing CA.
Answer: In triangle ABC, by the polygon law of vector addition, the sum of vectors forming a closed loop is zero:
\( \vec{AB} + \vec{BC} + \vec{CA} = \vec{0} \)
\( \implies \vec{CA} = -(\vec{AB} + \vec{BC}) \).
Summing the two given vectors:
\( \vec{AB} + \vec{BC} = (2\hat{i} - \hat{j} + 2\hat{k}) + (\hat{i} + 3\hat{j} + 5\hat{k}) = 3\hat{i} + 2\hat{j} + 7\hat{k} \).
Taking the negative gives \( \vec{CA} \):
\( \vec{CA} = -(3\hat{i} + 2\hat{j} + 7\hat{k}) = -3\hat{i} - 2\hat{j} - 7\hat{k} \).
In simple words: Add vector AB and vector BC, then reverse all the signs to get vector CA.

Exam Tip: Pay attention to vector orientation: \( \vec{AC} = \vec{AB} + \vec{BC} \), while \( \vec{CA} = -(\vec{AB} + \vec{BC}) \).

Question 3. Show that the points (1,0), (6,0) ,(0,0) are collinear.
Answer: Let the points be \( O(0, 0) \), \( P(1, 0) \), and \( Q(6, 0) \).
Their position vectors with respect to the origin are:
\( \vec{OP} = 1\hat{i} + 0\hat{j} = \hat{i} \)
\( \vec{OQ} = 6\hat{i} + 0\hat{j} = 6\hat{i} \).
Notice that \( \vec{OQ} = 6\,\vec{OP} \).
Because \( \vec{OQ} \) is a scalar multiple of \( \vec{OP} \) and both vectors share the common point O, the points O, P, and Q lie along the exact same straight line (the X-axis).
Hence, the three points are collinear.
In simple words: Since all three points have a y-value of 0, they all lie on the x-axis, making them collinear.

Exam Tip: Showing that two vectors sharing a common point are scalar multiples of each other (\( \vec{AB} = \lambda\vec{AC} \)) is the standard vector method to establish collinearity.

Level - II

Question 1. Write the position vector of a point R which divides the line joining the points P and Q whose position vectors are \( \hat{i} + 2\hat{j} - \hat{k} \) and \( -\hat{i} + \hat{j} + \hat{k} \) respectively in the ratio 2 : 1 externally.
Answer: Let \( \vec{p} = \hat{i} + 2\hat{j} - \hat{k} \) and \( \vec{q} = -\hat{i} + \hat{j} + \hat{k} \).
The section formula for external division in the ratio \( m : n = 2 : 1 \) is:
\( \vec{r} = \frac{m\vec{q} - n\vec{p}}{m - n} \).
Substitute the given vectors and values:
\( \vec{r} = \frac{2(-\hat{i} + \hat{j} + \hat{k}) - 1(\hat{i} + 2\hat{j} - \hat{k})}{2 - 1} \)
\( = \frac{(-2\hat{i} + 2\hat{j} + 2\hat{k}) - (\hat{i} + 2\hat{j} - \hat{k})}{1} \)
\( = (-2 - 1)\hat{i} + (2 - 2)\hat{j} + (2 + 1)\hat{k} = -3\hat{i} + 0\hat{j} + 3\hat{k} = -3\hat{i} + 3\hat{k} \).
Therefore, the position vector of point R is \( -3\hat{i} + 3\hat{k} \).
In simple words: Apply the external division formula by multiplying the second point by 2, subtracting the first point, and dividing by 1.

Exam Tip: External division uses a minus sign in both the numerator and the denominator; internal division uses an addition sign.

Question 2. Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are \( (2\vec{a} + \vec{b}) \) and \( (\vec{a} - 3\vec{b}) \) respectively, externally in the ratio 1:2. Also, show that P is the mid-point of the line segment RQ
Answer: Let the position vector of P be \( \vec{p} = 2\vec{a} + \vec{b} \) and that of Q be \( \vec{q} = \vec{a} - 3\vec{b} \).
For external division in the ratio \( m : n = 1 : 2 \), the section formula gives:
\( \vec{r} = \frac{m\vec{q} - n\vec{p}}{m - n} = \frac{1(\vec{a} - 3\vec{b}) - 2(2\vec{a} + \vec{b})}{1 - 2} \)
\( = \frac{\vec{a} - 3\vec{b} - 4\vec{a} - 2\vec{b}}{-1} = \frac{-3\vec{a} - 5\vec{b}}{-1} = 3\vec{a} + 5\vec{b} \).
Now compute the midpoint of line segment RQ:
\( \vec{m}_{\text{RQ}} = \frac{\vec{r} + \vec{q}}{2} = \frac{(3\vec{a} + 5\vec{b}) + (\vec{a} - 3\vec{b})}{2} \)
\( = \frac{4\vec{a} + 2\vec{b}}{2} = 2\vec{a} + \vec{b} = \vec{p} \).
Since the midpoint of RQ matches the position vector of P, point P is indeed the mid-point of segment RQ.
In simple words: Use the external ratio formula to find R, then average R and Q to verify that you get back point P.

Exam Tip: Be mindful of the negative sign in the denominator \( (1 - 2 = -1) \) when simplifying external division with \( m < n \).

(iii) Dot product of two vectors

Level - I

Question 1. Find \( \vec{a} \cdot \vec{b} \) if \( \vec{a} = 3\hat{i} - \hat{j} + 2\hat{k} \) and \( \vec{b} = 2\hat{i} + 3\hat{j} + 3\hat{k} \) .
Answer: Calculate the scalar product by summing the products of matching components:
\( \vec{a} \cdot \vec{b} = (3)(2) + (-1)(3) + (2)(3) = 6 - 3 + 6 = 9 \).
In simple words: Multiply matching coordinates together and add them up.

Exam Tip: Remember that dot product produces a scalar number, not a vector with \( \hat{i}, \hat{j}, \hat{k} \).

Question 2. If \( |\vec{a}| = \sqrt{3} \) , \( |\vec{b}| = 2 \) and \( \vec{a} \cdot \vec{b} = \sqrt{6} \) . Then find the angle between \( \vec{a} \) and \( \vec{b} \) .
Answer: The angle \( \theta \) between vectors \( \vec{a} \) and \( \vec{b} \) satisfies:
\( \cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \).
Substitute the given magnitudes and scalar product:
\( \cos\theta = \frac{\sqrt{6}}{\sqrt{3} \times 2} = \frac{\sqrt{3} \times \sqrt{2}}{2\sqrt{3}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \).
Since \( \cos\theta = \frac{1}{\sqrt{2}} \), the angle is \( \theta = \frac{\pi}{4} \) (or \( 45^\circ \)).
In simple words: Divide the dot product by the product of the two lengths to get \( 1/\sqrt{2} \), which corresponds to 45 degrees.

Exam Tip: Always state both radian and degree measures when giving standard angle answers.

Question 3. Write the angle between two vectors \( \vec{a} \) and \( \vec{b} \) with magnitudes \( \sqrt{3} \) and 2 respectively having \( \vec{a} \cdot \vec{b} = \sqrt{6} \) [ CBSE 2011]
Answer: Using the cosine formula for angle \( \theta \):
\( \cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{\sqrt{6}}{\sqrt{3} \times 2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \).
This gives \( \theta = \frac{\pi}{4} \) (or \( 45^\circ \)).
In simple words: The cosine ratio works out to 1 over root 2, so the angle is 45 degrees.

Exam Tip: Simplify square roots carefully: write \( \sqrt{6} \) as \( \sqrt{3}\sqrt{2} \) to cancel \( \sqrt{3} \) directly.

Level - II

Question 1. The dot products of a vector with the vectors \( \hat{i} - 3\hat{j} \), \( \hat{i} - 2\hat{j} \) and \( \hat{i} + \hat{j} + 4\hat{k} \) are 0 , 5 and 8 respectively. Find the vectors.
Answer: Let the required vector be \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).
From the given dot products, write three linear equations:
1. \( \vec{r} \cdot (\hat{i} - 3\hat{j}) = x - 3y = 0 \implies x = 3y \)
2. \( \vec{r} \cdot (\hat{i} - 2\hat{j}) = x - 2y = 5 \)
Substitute \( x = 3y \) into the second equation:
\( 3y - 2y = 5 \implies y = 5 \).
Then \( x = 3(5) = 15 \).
3. \( \vec{r} \cdot (\hat{i} + \hat{j} + 4\hat{k}) = x + y + 4z = 8 \)
Substitute \( x = 15 \) and \( y = 5 \):
\( 15 + 5 + 4z = 8 \implies 20 + 4z = 8 \implies 4z = -12 \implies z = -3 \).
Thus, the required vector is \( \vec{r} = 15\hat{i} + 5\hat{j} - 3\hat{k} \).
In simple words: Write out three simple algebra equations from the dot products and solve for x, y, and z one by one.

Exam Tip: Substituting \( x = 3y \) immediately simplifies the system without needing matrix methods.

Question 2. If \( \vec{a} \) and \( \vec{b} \) are two vectors such that \( |\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}| \), then what is the angle between \( \vec{a} \) and \( \vec{b} \) .
Answer: By definition, for vectors \( \vec{a} \) and \( \vec{b} \) with angle \( \theta \):
\( |\vec{a} \cdot \vec{b}| = |\vec{a}||\vec{b}|\cos\theta \) and \( |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta \).
Equating the two magnitudes:
\( |\vec{a}||\vec{b}|\cos\theta = |\vec{a}||\vec{b}|\sin\theta \).
Assuming non-zero vectors, divide both sides by \( |\vec{a}||\vec{b}|\cos\theta \):
\( \tan\theta = 1 \).
In the range \( 0 \le \theta \le \pi \), this yields \( \theta = \frac{\pi}{4} \) (or \( 45^\circ \)).
In simple words: The sine and cosine of the angle must be equal, which occurs at 45 degrees.

Exam Tip: Remember that the angle between two vectors always lies between \( 0 \) and \( \pi \).

Question 3. If \( \vec{a} = 2\hat{i} + 2\hat{j} + 3\hat{k} \) , \( \vec{b} = -\hat{i} + 2\hat{j} + \hat{k} \) and \( \vec{c} = 3\hat{i} + \hat{j} \) are such that \( \vec{a} + \lambda\vec{b} \) is perpendicular to \( \vec{c} \), find the value of λ.
Answer: Express \( \vec{a} + \lambda\vec{b} \) in terms of components:
\( \vec{a} + \lambda\vec{b} = (2\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(-\hat{i} + 2\hat{j} + \hat{k}) = (2 - \lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (3 + \lambda)\hat{k} \).
Because \( \vec{a} + \lambda\vec{b} \) is perpendicular to \( \vec{c} = 3\hat{i} + \hat{j} + 0\hat{k} \), their scalar product must be zero:
\( (\vec{a} + \lambda\vec{b}) \cdot \vec{c} = 0 \)
\( \implies (2 - \lambda)(3) + (2 + 2\lambda)(1) + (3 + \lambda)(0) = 0 \)
\( \implies 6 - 3\lambda + 2 + 2\lambda = 0 \)
\( \implies 8 - \lambda = 0 \implies \lambda = 8 \).
Thus, the required value of \( \lambda \) is 8.
In simple words: Combine the vectors with lambda, take the dot product with vector c, set it to zero, and solve.

Exam Tip: Notice that \( \vec{c} \) has a \( 0\hat{k} \) component; multiplying the third term by zero avoids unnecessary work.

Level - III

Question 1. If \( \vec{a} \) & \( \vec{b} \) are unit vectors inclined at an angle θ , prove that \( \sin\frac{\theta}{2} = \frac{1}{2}|\vec{a} - \vec{b}| \) .
Answer: Since \( \vec{a} \) and \( \vec{b} \) are unit vectors, \( |\vec{a}| = |\vec{b}| = 1 \).
Consider the squared magnitude of their difference:
\( |\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b}) \)
\( = 1 + 1 - 2(|\vec{a}||\vec{b}|\cos\theta) = 2 - 2\cos\theta = 2(1 - \cos\theta) \).
Using the half-angle formula \( 1 - \cos\theta = 2\sin^2\frac{\theta}{2} \):
\( |\vec{a} - \vec{b}|^2 = 2\left(2\sin^2\frac{\theta}{2}\right) = 4\sin^2\frac{\theta}{2} \).
Taking the square root on both sides (noting \( \sin\frac{\theta}{2} \ge 0 \) for \( 0 \le \theta \le \pi \)):
\( |\vec{a} - \vec{b}| = 2\sin\frac{\theta}{2} \)
\( \implies \sin\frac{\theta}{2} = \frac{1}{2}|\vec{a} - \vec{b}| \).
Hence proved.
In simple words: Square the difference of the two unit vectors to get \( 2 - 2\cos\theta \), then replace it with the half-angle sine identity.

Exam Tip: Always cite the half-angle trigonometric identity \( 1 - \cos\theta = 2\sin^2\frac{\theta}{2} \) clearly in your derivation.

Question 2. If |\( \vec{a} + \vec{b} \)| = |\( \vec{a} - \vec{b} \)|, then find the angle between \( \vec{a} \) and \( \vec{b} \) .
Answer: Given that \( |\vec{a} + \vec{b}| = |\vec{a} - \vec{b}| \), square both sides:
\( |\vec{a} + \vec{b}|^2 = |\vec{a} - \vec{b}|^2 \)
\( \implies |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b}) \).
Canceling \( |\vec{a}|^2 + |\vec{b}|^2 \) from both sides:
\( 2(\vec{a} \cdot \vec{b}) = -2(\vec{a} \cdot \vec{b}) \)
\( \implies 4(\vec{a} \cdot \vec{b}) = 0 \)
\( \implies \vec{a} \cdot \vec{b} = 0 \).
Since their scalar product is zero, the vectors are perpendicular.
Therefore, the angle between \( \vec{a} \) and \( \vec{b} \) is \( \frac{\pi}{2} \) (or \( 90^\circ \)).
In simple words: When the sum and difference have the same length, the vectors are at right angles to each other.

Exam Tip: Geometrically, this means that the diagonals of a parallelogram are equal, which proves that the figure is a rectangle.

Question 3. For what values of λ, vectors \( \vec{a} = 3\hat{i} - 2\hat{j} + 4\hat{k} \) and \( \vec{b} = \lambda\hat{i} - 4\hat{j} + 8\hat{k} \) are (i) Orthogonal (ii) Parallel
Answer:
(i) For the vectors to be orthogonal, their dot product must vanish:
\( \vec{a} \cdot \vec{b} = 0 \)
\( \implies 3(\lambda) + (-2)(-4) + 4(8) = 0 \)
\( \implies 3\lambda + 8 + 32 = 0 \)
\( \implies 3\lambda + 40 = 0 \implies \lambda = -\frac{40}{3} \).

(ii) For the vectors to be parallel, their scalar components must be in proportion:
\( \frac{3}{\lambda} = \frac{-2}{-4} = \frac{4}{8} \).
Since \( \frac{-2}{-4} = \frac{1}{2} \) and \( \frac{4}{8} = \frac{1}{2} \):
\( \frac{3}{\lambda} = \frac{1}{2} \implies \lambda = 6 \).
In simple words: Set the dot product to zero for perpendicular vectors, and set component ratios equal for parallel vectors.

Exam Tip: Address both parts separately with clear headings (i) and (ii) to match marking criteria.

Question 4. Find|\( \vec{x} \)|, if for a unit vector \( \vec{a} \) , (\( \vec{x} \) - \( \vec{a} \)).(\( \vec{x} \) + \( \vec{a} \))=15.
Answer: Expanding the dot product using the identity \( (\vec{u} - \vec{v}) \cdot (\vec{u} + \vec{v}) = |\vec{u}|^2 - |\vec{v}|^2 \):
\( (\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = |\vec{x}|^2 - |\vec{a}|^2 = 15 \).
Since \( \vec{a} \) is a unit vector, \( |\vec{a}| = 1 \), so \( |\vec{a}|^2 = 1 \).
Substituting this gives:
\( |\vec{x}|^2 - 1 = 15 \)
\( \implies |\vec{x}|^2 = 16 \)
\( \implies |\vec{x}| = 4 \).
In simple words: The difference of squares equals 15, and since the unit vector length squared is 1, the magnitude squared is 16.

Exam Tip: Magnitude represents length and is strictly non-negative; do not write \( \pm 4 \).

Question 5. If \( \vec{a} = 5\hat{i} - \hat{j} + 7\hat{k} \) and \( \vec{b} = \hat{i} - \hat{j} + \mu\hat{k} \), find μ, such that \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \) are orthogonal.
Answer: Two vectors are orthogonal when their dot product is zero:
\( (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 0 \).
Using the identity \( (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = |\vec{a}|^2 - |\vec{b}|^2 \):
\( |\vec{a}|^2 - |\vec{b}|^2 = 0 \implies |\vec{a}|^2 = |\vec{b}|^2 \).
Compute \( |\vec{a}|^2 \) and \( |\vec{b}|^2 \):
\( |\vec{a}|^2 = 5^2 + (-1)^2 + 7^2 = 25 + 1 + 49 = 75 \)
\( |\vec{b}|^2 = 1^2 + (-1)^2 + \mu^2 = 1 + 1 + \mu^2 = 2 + \mu^2 \).
Equating the two magnitudes:
\( 2 + \mu^2 = 75 \)
\( \implies \mu^2 = 73 \)
\( \implies \mu = \pm\sqrt{73} \).
In simple words: Sum and difference are orthogonal only when both original vectors have the exact same length.

Exam Tip: Remember to write \( \pm\sqrt{73} \) because squaring removes the sign of \( \mu \).

Question 6. Show that the vector \( 2\hat{i} - \hat{j} + \hat{k} \), \( \hat{i} - 3\hat{j} - 5\hat{k} \) and \( 3\hat{i} - 4\hat{j} - 4\hat{k} \) form sides of a right angled triangle.
Answer: Let \( \vec{a} = 2\hat{i} - \hat{j} + \hat{k} \), \( \vec{b} = \hat{i} - 3\hat{j} - 5\hat{k} \), and \( \vec{c} = 3\hat{i} - 4\hat{j} - 4\hat{k} \).
First verify whether these vectors form a triangle:
\( \vec{a} + \vec{b} = (2 + 1)\hat{i} + (-1 - 3)\hat{j} + (1 - 5)\hat{k} = 3\hat{i} - 4\hat{j} - 4\hat{k} = \vec{c} \).
Since \( \vec{a} + \vec{b} = \vec{c} \), the vectors form the sides of a triangle.
Now calculate the square of the magnitude of each vector:
\( |\vec{a}|^2 = 2^2 + (-1)^2 + 1^2 = 4 + 1 + 1 = 6 \)
\( |\vec{b}|^2 = 1^2 + (-3)^2 + (-5)^2 = 1 + 9 + 25 = 35 \)
\( |\vec{c}|^2 = 3^2 + (-4)^2 + (-4)^2 = 9 + 16 + 16 = 41 \).
Notice that:
\( |\vec{a}|^2 + |\vec{b}|^2 = 6 + 35 = 41 = |\vec{c}|^2 \).
Since the sum of the squares of two sides equals the square of the third side, Pythagoras' theorem is satisfied.
Hence, the vectors form the sides of a right-angled triangle.
In simple words: Two vectors add up to give the third, and the sum of their squared lengths equals the third squared length.

Exam Tip: You must show two things: that the vectors form a triangle (\( \vec{a} + \vec{b} = \vec{c} \)), and that Pythagoras' theorem holds.

Question 7. Let \( \vec{a} = \hat{i} + 4\hat{j} + 2\hat{k} \) , \( \vec{b} = 3\hat{i} - 2\hat{j} + 7\hat{k} \) and \( \vec{c} = 2\hat{i} - \hat{j} + 4\hat{k} \). Find a vector \( \vec{d} \) which is perpendicular to both \( \vec{a} \) and \( \vec{b} \) and \( \vec{c} \cdot \vec{d} = 18 \).
Answer: Since \( \vec{d} \) is perpendicular to both \( \vec{a} \) and \( \vec{b} \), it must be parallel to their cross product \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 2 \\ 3 & -2 & 7 \end{vmatrix} \)
\( = \hat{i}(28 - (-4)) - \hat{j}(7 - 6) + \hat{k}(-2 - 12) = 32\hat{i} - \hat{j} - 14\hat{k} \).
Therefore, \( \vec{d} = k(32\hat{i} - \hat{j} - 14\hat{k}) \) for some scalar \( k \).
Given \( \vec{c} \cdot \vec{d} = 18 \):
\( (2\hat{i} - \hat{j} + 4\hat{k}) \cdot k(32\hat{i} - \hat{j} - 14\hat{k}) = 18 \)
\( \implies k [2(32) + (-1)(-1) + 4(-14)] = 18 \)
\( \implies k [64 + 1 - 56] = 18 \)
\( \implies 9k = 18 \implies k = 2 \).
Substituting \( k = 2 \) into \( \vec{d} \):
\( \vec{d} = 2(32\hat{i} - \hat{j} - 14\hat{k}) = 64\hat{i} - 2\hat{j} - 28\hat{k} \).
In simple words: Find the cross product to get the perpendicular direction, then use the dot product with c to find the scaling factor 2.

Exam Tip: Using \( \vec{d} = k(\vec{a} \times \vec{b}) \) avoids having to set up and solve a system of three linear equations.

Question 8. If \( \vec{a} \) , \( \vec{b} \) , \( \vec{c} \) are three mutually perpendicular vectors of equal magnitudes, prove that \( \vec{a} + \vec{b} + \vec{c} \) is equally inclined with the vectors \( \vec{a} \) , \( \vec{b} \) , \( \vec{c} \) .
Answer: Let \( |\vec{a}| = |\vec{b}| = |\vec{c}| = \lambda \).
Since the vectors are mutually perpendicular, we have:
\( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0 \).
Let \( \vec{r} = \vec{a} + \vec{b} + \vec{c} \). The square of its magnitude is:
\( |\vec{r}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \)
\( = \lambda^2 + \lambda^2 + \lambda^2 + 0 = 3\lambda^2 \implies |\vec{r}| = \sqrt{3}\lambda \).
Let \( \alpha, \beta, \gamma \) be the angles that \( \vec{r} \) makes with \( \vec{a}, \vec{b}, \vec{c} \) respectively:
\( \cos\alpha = \frac{\vec{r} \cdot \vec{a}}{|\vec{r}||\vec{a}|} = \frac{(\vec{a} + \vec{b} + \vec{c}) \cdot \vec{a}}{\sqrt{3}\lambda \cdot \lambda} = \frac{|\vec{a}|^2 + 0 + 0}{\sqrt{3}\lambda^2} = \frac{\lambda^2}{\sqrt{3}\lambda^2} = \frac{1}{\sqrt{3}} \).
Similarly:
\( \cos\beta = \frac{\vec{r} \cdot \vec{b}}{|\vec{r}||\vec{b}|} = \frac{\lambda^2}{\sqrt{3}\lambda^2} = \frac{1}{\sqrt{3}} \)
\( \cos\gamma = \frac{\vec{r} \cdot \vec{c}}{|\vec{r}||\vec{c}|} = \frac{\lambda^2}{\sqrt{3}\lambda^2} = \frac{1}{\sqrt{3}} \).
Since \( \cos\alpha = \cos\beta = \cos\gamma = \frac{1}{\sqrt{3}} \), it follows that \( \alpha = \beta = \gamma = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \).
Thus, \( \vec{a} + \vec{b} + \vec{c} \) is equally inclined to \( \vec{a}, \vec{b}, \vec{c} \).
In simple words: The dot product of the sum with each vector is the same, so the angle with all three vectors is identical.

Exam Tip: Show the calculation of at least one angle in full, then state the other two by symmetry to save time while earning full marks.

Question 9. Let \( \vec{a} \) , \( \vec{b} \) , \( \vec{c} \) be three vectors such that \( |\vec{a}| = 3, |\vec{b}| = 4, |\vec{c}| = 5 \) and each of them being perpendicular to the sum of the other two, find \( |\vec{a} + \vec{b} + \vec{c}| \).
Answer: Given that each vector is orthogonal to the sum of the other two:
1. \( \vec{a} \cdot (\vec{b} + \vec{c}) = 0 \implies \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 0 \)
2. \( \vec{b} \cdot (\vec{c} + \vec{a}) = 0 \implies \vec{b} \cdot \vec{c} + \vec{b} \cdot \vec{a} = 0 \)
3. \( \vec{c} \cdot (\vec{a} + \vec{b}) = 0 \implies \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} = 0 \).
Adding all three equations:
\( 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \).
Now evaluate the squared length of the sum vector:
\( |\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \)
\( = 3^2 + 4^2 + 5^2 + 0 = 9 + 16 + 25 = 50 \).
Taking the square root:
\( |\vec{a} + \vec{b} + \vec{c}| = \sqrt{50} = 5\sqrt{2} \).
In simple words: The cross-terms in the squared length formula add up to zero, leaving just the sum of the squared individual lengths.

Exam Tip: Adding the three orthogonality relations directly proves that the sum of all pairwise dot products is zero.

(iv) Projection of a vector

Level - I

Question 1. Find the projection of \( \vec{a} \) on \( \vec{b} \) if \( \vec{a} \cdot \vec{b} = 8 \) and \( \vec{b} = 2\hat{i} + 6\hat{j} + 3\hat{k} \) .
Answer: The projection of \( \vec{a} \) onto \( \vec{b} \) is given by the formula:
\( \text{Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \).
Compute the length of \( \vec{b} \):
\( |\vec{b}| = \sqrt{2^2 + 6^2 + 3^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7 \).
Substitute the values into the formula:
\( \text{Projection} = \frac{8}{7} \).
In simple words: Divide the dot product of the vectors by the magnitude of the base vector.

Exam Tip: Projection of \( \vec{a} \) on \( \vec{b} \) divides by \( |\vec{b}| \), whereas projection of \( \vec{b} \) on \( \vec{a} \) divides by \( |\vec{a}| \).

Question 2. Write the projection of the vector \( \hat{i} - \hat{j} \) on the vector \( \hat{i} + \hat{j} \) [ CBSE 2011]
Answer: Let \( \vec{a} = \hat{i} - \hat{j} \) and \( \vec{b} = \hat{i} + \hat{j} \).
The projection of \( \vec{a} \) on \( \vec{b} \) is:
\( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{(1)(1) + (-1)(1)}{\sqrt{1^2 + 1^2}} = \frac{1 - 1}{\sqrt{2}} = \frac{0}{\sqrt{2}} = 0 \).
Therefore, the projection is 0.
In simple words: Because the two vectors are perpendicular to each other, the projection is zero.

Exam Tip: A projection of 0 means the vectors are orthogonal; stating this adds clarity to your answer.

Question 3. Find the angle between the vectors \( \hat{i} - 2\hat{j} + 3\hat{k} \) and \( 3\hat{i} - 2\hat{j} + \hat{k} \)
Answer: Let \( \vec{a} = \hat{i} - 2\hat{j} + 3\hat{k} \) and \( \vec{b} = 3\hat{i} - 2\hat{j} + \hat{k} \).
Calculate their dot product:
\( \vec{a} \cdot \vec{b} = (1)(3) + (-2)(-2) + (3)(1) = 3 + 4 + 3 = 10 \).
Calculate their magnitudes:
\( |\vec{a}| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \)
\( |\vec{b}| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14} \).
The cosine of the angle \( \theta \) between them is:
\( \cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{10}{\sqrt{14}\sqrt{14}} = \frac{10}{14} = \frac{5}{7} \).
Thus, the angle is \( \theta = \cos^{-1}\left(\frac{5}{7}\right) \).
In simple words: Find the dot product and divide it by 14, then express the answer as inverse cosine.

Exam Tip: Simplify the fraction \( \frac{10}{14} \) to \( \frac{5}{7} \) before writing the inverse cosine.

Question 4. Find the projection of the vector \( \hat{i} + 3\hat{j} + 7\hat{k} \) on the vector \( 7\hat{i} - \hat{j} + 8\hat{k} \)
Answer: Let \( \vec{a} = \hat{i} + 3\hat{j} + 7\hat{k} \) and \( \vec{b} = 7\hat{i} - \hat{j} + 8\hat{k} \).
Evaluate the scalar product \( \vec{a} \cdot \vec{b} \):
\( \vec{a} \cdot \vec{b} = (1)(7) + (3)(-1) + (7)(8) = 7 - 3 + 56 = 60 \).
Evaluate the length of \( \vec{b} \):
\( |\vec{b}| = \sqrt{7^2 + (-1)^2 + 8^2} = \sqrt{49 + 1 + 64} = \sqrt{114} \).
The projection of \( \vec{a} \) on \( \vec{b} \) is:
\( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{60}{\sqrt{114}} \).
In simple words: Work out the dot product on top and the length of the target vector on the bottom.

Exam Tip: Make sure you divide by the magnitude of the vector after "on", not the first vector.

Level - II

Question 1. Three vertices of a triangle are A(0, -1, -2), B(3,1,4) and C(5,7,1). Show that it is a right angled triangle. Also find the other two angles.
Answer: Find the vectors representing the sides of triangle ABC:
\( \vec{AB} = (3 - 0)\hat{i} + (1 - (-1))\hat{j} + (4 - (-2))\hat{k} = 3\hat{i} + 2\hat{j} + 6\hat{k} \)
\( \vec{BC} = (5 - 3)\hat{i} + (7 - 1)\hat{j} + (1 - 4)\hat{k} = 2\hat{i} + 6\hat{j} - 3\hat{k} \)
\( \vec{CA} = (0 - 5)\hat{i} + (-1 - 7)\hat{j} + (-2 - 1)\hat{k} = -5\hat{i} - 8\hat{j} - 3\hat{k} \).
Compute their magnitudes:
\( |\vec{AB}| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7 \)
\( |\vec{BC}| = \sqrt{2^2 + 6^2 + (-3)^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7 \)
\( |\vec{CA}| = \sqrt{(-5)^2 + (-8)^2 + (-3)^2} = \sqrt{25 + 64 + 9} = \sqrt{98} = 7\sqrt{2} \).
Evaluate the dot product of \( \vec{AB} \) and \( \vec{BC} \):
\( \vec{AB} \cdot \vec{BC} = (3)(2) + (2)(6) + (6)(-3) = 6 + 12 - 18 = 0 \).
Since \( \vec{AB} \cdot \vec{BC} = 0 \), the sides AB and BC are perpendicular, meaning \( \angle B = 90^\circ \) (or \( \frac{\pi}{2} \)).
Because \( |\vec{AB}| = |\vec{BC}| = 7 \), triangle ABC is an isosceles right-angled triangle.
Therefore, the remaining two angles are equal:
\( \angle A = \angle C = 45^\circ \) (or \( \frac{\pi}{4} \)).
In simple words: The dot product of AB and BC is zero so angle B is 90 degrees, and because both arms have length 7, the other angles are 45 degrees.

Exam Tip: Showing that \( \vec{AB} \cdot \vec{BC} = 0 \) is much quicker for proving a right angle than using side lengths in the Pythagorean theorem.

Question 2. Show that the angle between any two diagonals of a cube is \( \cos^{-1}\left(\frac{1}{3}\right) \) .
Answer: Let a cube of side length \( a \) have one corner at the origin \( O(0,0,0) \) and its edges along the positive X, Y, and Z axes.
The coordinates of its eight vertices are:
\( O(0,0,0), A(a,0,0), B(0,a,0), C(0,0,a), D(a,a,0), E(0,a,a), F(a,0,a), G(a,a,a) \).
The four main diagonals are lines joining opposite vertices, such as OG and AF.
The vector for diagonal OG is:
\( \vec{d}_1 = a\hat{i} + a\hat{j} + a\hat{k} \).
The vector for diagonal AF is (from \( A(a,0,0) \) to \( E(0,a,a) \)):
\( \vec{d}_2 = -a\hat{i} + a\hat{j} + a\hat{k} \).
Compute their magnitudes:
\( |\vec{d}_1| = \sqrt{a^2 + a^2 + a^2} = a\sqrt{3} \)
\( |\vec{d}_2| = \sqrt{(-a)^2 + a^2 + a^2} = a\sqrt{3} \).
Compute their dot product:
\( \vec{d}_1 \cdot \vec{d}_2 = a(-a) + a(a) + a(a) = -a^2 + a^2 + a^2 = a^2 \).
The angle \( \theta \) between the two diagonals satisfies:
\( \cos\theta = \frac{\vec{d}_1 \cdot \vec{d}_2}{|\vec{d}_1||\vec{d}_2|} = \frac{a^2}{(a\sqrt{3})(a\sqrt{3})} = \frac{a^2}{3a^2} = \frac{1}{3} \).
Thus, \( \theta = \cos^{-1}\left(\frac{1}{3}\right) \).
Hence proved.
In simple words: Express the diagonals using coordinates, take their dot product to get \( a^2 \), and divide by the product of lengths \( 3a^2 \).

Exam Tip: State the coordinates of the chosen opposite vertices clearly before writing down the diagonal vectors.

Question 3. If \( \vec{a} \) , \( \vec{b} \) , \( \vec{c} \) are non - zero and non – coplanar vectors, prove that \( \vec{a} – 2\vec{b} +3\vec{c} \), \( \vec{a} – 3\vec{b} + 5\vec{c} \) and \( – 2\vec{a} + 3\vec{b} – 4\vec{c} \) are also coplanar
Answer: Three vectors are coplanar if and only if their scalar triple product equals zero.
Let:
\( \vec{u} = \vec{a} - 2\vec{b} + 3\vec{c} \)
\( \vec{v} = \vec{a} - 3\vec{b} + 5\vec{c} \)
\( \vec{w} = -2\vec{a} + 3\vec{b} - 4\vec{c} \).
The scalar triple product \( [\vec{u}, \vec{v}, \vec{w}] \) can be expressed in determinant form as:
\( [\vec{u}, \vec{v}, \vec{w}] = \begin{vmatrix} 1 & -2 & 3 \\ 1 & -3 & 5 \\ -2 & 3 & -4 \end{vmatrix} [\vec{a}, \vec{b}, \vec{c}] \).
Evaluate the determinant of coefficients:
\( \begin{vmatrix} 1 & -2 & 3 \\ 1 & -3 & 5 \\ -2 & 3 & -4 \end{vmatrix} = 1((-3)(-4) - 3(5)) - (-2)(1(-4) - (-2)(5)) + 3(1(3) - (-2)(-3)) \)
\( = 1(12 - 15) + 2(-4 + 10) + 3(3 - 6) \)
\( = 1(-3) + 2(6) + 3(-3) = -3 + 12 - 9 = 0 \).
Therefore:
\( [\vec{u}, \vec{v}, \vec{w}] = 0 \cdot [\vec{a}, \vec{b}, \vec{c}] = 0 \).
Since their scalar triple product is zero, the three vectors are coplanar.
Hence proved.
In simple words: Put the coefficients into a 3 by 3 determinant; since the determinant works out to zero, the three vectors lie in the same plane.

Exam Tip: Use the property \( [\vec{u}, \vec{v}, \vec{w}] = \det(C)\,[\vec{a}, \vec{b}, \vec{c}] \) to quickly evaluate linear combinations in coplanarity proofs.

Level - III

Question 1. If a unit vector \( \vec{a} \) makes angles π /4, with \( \hat{i} \) , π/3 with \( \hat{j} \) and an acute angle θ with \( \hat{k} \) , then find the component of \( \vec{a} \) and angle θ .
Answer: Let the direction cosines of unit vector \( \vec{a} \) be \( l = \cos\frac{\pi}{4} \), \( m = \cos\frac{\pi}{3} \), and \( n = \cos\theta \).
We have:
\( l = \frac{1}{\sqrt{2}} \), \( m = \frac{1}{2} \).
Since \( l^2 + m^2 + n^2 = 1 \):
\( \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{2}\right)^2 + \cos^2\theta = 1 \)
\( \implies \frac{1}{2} + \frac{1}{4} + \cos^2\theta = 1 \)
\( \implies \frac{3}{4} + \cos^2\theta = 1 \)
\( \implies \cos^2\theta = \frac{1}{4} \).
Given that \( \theta \) is an acute angle, \( \cos\theta > 0 \):
\( \cos\theta = \frac{1}{2} \implies \theta = \frac{\pi}{3} \) (or \( 60^\circ \)).
The components of the unit vector \( \vec{a} \) are its direction cosines:
\( \vec{a} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{2}\hat{j} + \frac{1}{2}\hat{k} \).
The scalar components are \( \frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2} \), and the angle is \( \theta = \frac{\pi}{3} \).
In simple words: Use the fact that direction cosines squared sum to 1 to find that cosine theta is 1/2, giving an angle of 60 degrees.

Exam Tip: Specify that \( \cos\theta = +\frac{1}{2} \) because the question states that \( \theta \) is acute.

Question 2. If \( \vec{a} \) , \( \vec{b} \) , \( \vec{c} \) are three mutually perpendicular vectors of equal magnitudes, prove that \( \vec{a} + \vec{b} + \vec{c} \) is equally inclined with the vectors \( \vec{a} \) , \( \vec{b} \) , \( \vec{c} \) .
Answer: Let \( |\vec{a}| = |\vec{b}| = |\vec{c}| = k \).
Because the vectors are mutually perpendicular:
\( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0 \).
Let \( \vec{s} = \vec{a} + \vec{b} + \vec{c} \). Then:
\( |\vec{s}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \)
\( = k^2 + k^2 + k^2 + 0 = 3k^2 \implies |\vec{s}| = \sqrt{3}k \).
The cosine of the angle \( \theta_1 \) between \( \vec{s} \) and \( \vec{a} \) is:
\( \cos\theta_1 = \frac{\vec{s} \cdot \vec{a}}{|\vec{s}||\vec{a}|} = \frac{\vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{a} + \vec{c} \cdot \vec{a}}{\sqrt{3}k \cdot k} = \frac{k^2 + 0 + 0}{\sqrt{3}k^2} = \frac{1}{\sqrt{3}} \).
Similarly, for the angles \( \theta_2 \) and \( \theta_3 \) with \( \vec{b} \) and \( \vec{c} \):
\( \cos\theta_2 = \frac{1}{\sqrt{3}} \) and \( \cos\theta_3 = \frac{1}{\sqrt{3}} \).
Since \( \theta_1 = \theta_2 = \theta_3 = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \), the sum vector is equally inclined to all three vectors.
Hence proved.
In simple words: Each dot product yields the same cosine value of 1 divided by root 3, so all three angles match.

Exam Tip: Clearly define a variable such as \( k \) for the equal magnitudes to keep intermediate steps tidy.

Question 3. If with reference to the right handed system of mutually perpendicular unit vectors \( \hat{i} , \hat{j} ,and \hat{k} \), \( \vec{\alpha} = 3\hat{i} – \hat{j} \), \( \vec{\beta} = 2\hat{i} + \hat{j} – 3\hat{k} \) then express \( \vec{\beta} \) in the form of \( \vec{\beta}_1 + \vec{\beta}_2 \), where \( \vec{\beta}_1 \) is parallel to \( \vec{\alpha} \) and \( \vec{\beta}_2 \) is perpendicular to \( \vec{\alpha} \) .
Answer: Since \( \vec{\beta}_1 \) is parallel to \( \vec{\alpha} \), we can write \( \vec{\beta}_1 = c\vec{\alpha} = c(3\hat{i} - \hat{j}) \) for some scalar \( c \).
Because \( \vec{\beta} = \vec{\beta}_1 + \vec{\beta}_2 \), the perpendicular component is:
\( \vec{\beta}_2 = \vec{\beta} - \vec{\beta}_1 = (2\hat{i} + \hat{j} - 3\hat{k}) - c(3\hat{i} - \hat{j}) = (2 - 3c)\hat{i} + (1 + c)\hat{j} - 3\hat{k} \).
Since \( \vec{\beta}_2 \) is perpendicular to \( \vec{\alpha} \), their scalar product is zero:
\( \vec{\beta}_2 \cdot \vec{\alpha} = 0 \)
\( \implies [(2 - 3c)\hat{i} + (1 + c)\hat{j} - 3\hat{k}] \cdot (3\hat{i} - \hat{j}) = 0 \)
\( \implies 3(2 - 3c) - 1(1 + c) = 0 \)
\( \implies 6 - 9c - 1 - c = 0 \)
\( \implies 5 - 10c = 0 \implies c = \frac{1}{2} \).
Substitute \( c = \frac{1}{2} \) back into both components:
\( \vec{\beta}_1 = \frac{1}{2}(3\hat{i} - \hat{j}) = \frac{3}{2}\hat{i} - \frac{1}{2}\hat{j} \)
\( \vec{\beta}_2 = \left(2 - \frac{3}{2}\right)\hat{i} + \left(1 + \frac{1}{2}\right)\hat{j} - 3\hat{k} = \frac{1}{2}\hat{i} + \frac{3}{2}\hat{j} - 3\hat{k} \).
Thus, \( \vec{\beta} = \left(\frac{3}{2}\hat{i} - \frac{1}{2}\hat{j}\right) + \left(\frac{1}{2}\hat{i} + \frac{3}{2}\hat{j} - 3\hat{k}\right) \).
In simple words: Write the parallel part as a scalar multiple of alpha, set the dot product of the remaining part with alpha to zero to solve for the scalar, and compute both vectors.

Exam Tip: Check your final answer by verifying that \( \vec{\beta}_1 + \vec{\beta}_2 = \vec{\beta} \) and \( \vec{\beta}_1 \cdot \vec{\beta}_2 = 0 \).

Question 4. Show that the points A, B, C with position vectors \( \vec{a} = 3\hat{i} – 4\hat{j} – 4\hat{k} \), \( \vec{b} = 2\hat{i} – \hat{j} + \hat{k} \) and \( \vec{c} = \hat{i} – 3\hat{j} – 5\hat{k} \) respectively form the vertices of a right angled triangle.
Answer: Determine the side vectors of triangle ABC from the given position vectors:
\( \vec{AB} = \vec{b} - \vec{a} = (2 - 3)\hat{i} + (-1 - (-4))\hat{j} + (1 - (-4))\hat{k} = -\hat{i} + 3\hat{j} + 5\hat{k} \)
\( \vec{BC} = \vec{c} - \vec{b} = (1 - 2)\hat{i} + (-3 - (-1))\hat{j} + (-5 - 1)\hat{k} = -\hat{i} - 2\hat{j} - 6\hat{k} \)
\( \vec{CA} = \vec{a} - \vec{c} = (3 - 1)\hat{i} + (-4 - (-3))\hat{j} + (-4 - (-5))\hat{k} = 2\hat{i} - \hat{j} + \hat{k} \).
Compute the squared lengths of each side:
\( |\vec{AB}|^2 = (-1)^2 + 3^2 + 5^2 = 1 + 9 + 25 = 35 \)
\( |\vec{BC}|^2 = (-1)^2 + (-2)^2 + (-6)^2 = 1 + 4 + 36 = 41 \)
\( |\vec{CA}|^2 = 2^2 + (-1)^2 + 1^2 = 4 + 1 + 1 = 6 \).
Notice that:
\( |\vec{AB}|^2 + |\vec{CA}|^2 = 35 + 6 = 41 = |\vec{BC}|^2 \).
Also evaluate the dot product between \( \vec{AB} \) and \( \vec{CA} \):
\( \vec{AB} \cdot \vec{CA} = (-1)(2) + (3)(-1) + (5)(1) = -2 - 3 + 5 = 0 \).
This confirms that side AB is perpendicular to side CA at vertex A.
Hence, points A, B, and C form the vertices of a right-angled triangle.
In simple words: Find the three side vectors, verify their lengths satisfy Pythagoras' theorem, and confirm the dot product of perpendicular sides is zero.

Exam Tip: Showing \( \vec{AB} \cdot \vec{CA} = 0 \) gives undeniable proof that angle A is \( 90^\circ \).

Question 5. If \( \vec{a} \) & \( \vec{b} \) are unit vectors inclined at an angle θ , prove that (i) \( \sin\frac{\theta}{2} = \frac{1}{2}|\vec{a} - \vec{b}| \) (ii) \( \tan\frac{\theta}{2} = \frac{|\vec{a} - \vec{b}|}{|\vec{a} + \vec{b}|} \)
Answer:
(i) Since \( \vec{a} \) and \( \vec{b} \) are unit vectors, \( |\vec{a}| = |\vec{b}| = 1 \).
Evaluate \( |\vec{a} - \vec{b}|^2 \):
\( |\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b}) = 1 + 1 - 2\cos\theta = 2(1 - \cos\theta) \).
Using \( 1 - \cos\theta = 2\sin^2\frac{\theta}{2} \):
\( |\vec{a} - \vec{b}|^2 = 4\sin^2\frac{\theta}{2} \implies |\vec{a} - \vec{b}| = 2\sin\frac{\theta}{2} \implies \sin\frac{\theta}{2} = \frac{1}{2}|\vec{a} - \vec{b}| \).

(ii) Similarly evaluate \( |\vec{a} + \vec{b}|^2 \):
\( |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = 1 + 1 + 2\cos\theta = 2(1 + \cos\theta) \).
Using \( 1 + \cos\theta = 2\cos^2\frac{\theta}{2} \):
\( |\vec{a} + \vec{b}|^2 = 4\cos^2\frac{\theta}{2} \implies |\vec{a} + \vec{b}| = 2\cos\frac{\theta}{2} \implies \cos\frac{\theta}{2} = \frac{1}{2}|\vec{a} + \vec{b}| \).
Dividing the two results:
\( \tan\frac{\theta}{2} = \frac{\sin(\theta/2)}{\cos(\theta/2)} = \frac{\frac{1}{2}|\vec{a} - \vec{b}|}{\frac{1}{2}|\vec{a} + \vec{b}|} = \frac{|\vec{a} - \vec{b}|}{|\vec{a} + \vec{b}|} \).
Hence proved.
In simple words: Express the length of the difference as sine of half the angle and the sum as cosine of half the angle, then divide them to get tangent.

Exam Tip: Deriving part (ii) directly from part (i) by calculating \( \cos\frac{\theta}{2} \) is the cleanest method to earn full marks.

(vii)Cross product of two vectors

Level - I

Question 1. If |\( \vec{a} \)| = 3 , |\( \vec{b} \)| = 5 and \( \vec{a} \cdot \vec{b} = 9 \) . Find |\( \vec{a} \times \vec{b} \)|
Answer: Using Lagrange's identity relating dot product and cross product:
\( |\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 \).
Substitute the known values:
\( |\vec{a} \times \vec{b}|^2 + 9^2 = (3^2)(5^2) \)
\( \implies |\vec{a} \times \vec{b}|^2 + 81 = (9)(25) = 225 \)
\( \implies |\vec{a} \times \vec{b}|^2 = 225 - 81 = 144 \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{144} = 12 \).
Therefore, the value of \( |\vec{a} \times \vec{b}| \) is 12.
In simple words: Lagrange's identity connects dot product and cross product like Pythagoras' theorem, allowing you to solve for 12.

Exam Tip: Stating Lagrange's identity \( |\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 \) saves you from calculating the intermediate angle.

Question 2. Find|\( \vec{a} \times \vec{b} \)| , if \( \vec{a} = \hat{i} - 7\hat{j} + 7\hat{k} \) and \( \vec{b} = 3\hat{i} + 2\hat{j} + 2\hat{k} \)
Answer: Compute the vector cross product using a determinant:
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -7 & 7 \\ 3 & 2 & 2 \end{vmatrix} \)
\( = \hat{i}((-7)(2) - (7)(2)) - \hat{j}((1)(2) - (7)(3)) + \hat{k}((1)(2) - (-7)(3)) \)
\( = \hat{i}(-14 - 14) - \hat{j}(2 - 21) + \hat{k}(2 + 21) \)
\( = -28\hat{i} + 19\hat{j} + 23\hat{k} \).
Now evaluate its magnitude:
\( |\vec{a} \times \vec{b}| = \sqrt{(-28)^2 + 19^2 + 23^2} = \sqrt{784 + 361 + 529} = \sqrt{1674} = 3\sqrt{186} \).
In simple words: Expand the cross-product determinant row by row, then find the square root of the sum of squared numbers.

Exam Tip: Be careful with signs when calculating the middle \( \hat{j} \) term of the cross product determinant.

Question 3. Find |\( \vec{x} \)|, if \( \vec{p} \) is a unit vector and , (\( \vec{x} \) - \( \vec{p} \)).(\( \vec{x} \) + \( \vec{p} \))= 80.
Answer: Expand the dot product using difference of squares:
\( (\vec{x} - \vec{p}) \cdot (\vec{x} + \vec{p}) = |\vec{x}|^2 - |\vec{p}|^2 = 80 \).
Since \( \vec{p} \) is a unit vector, \( |\vec{p}| = 1 \), so \( |\vec{p}|^2 = 1 \).
This gives:
\( |\vec{x}|^2 - 1 = 80 \)
\( \implies |\vec{x}|^2 = 81 \)
\( \implies |\vec{x}| = 9 \).
In simple words: The equation simplifies to length squared minus 1 equals 80, which means length squared is 81 and the length is 9.

Exam Tip: Magnitude is never negative, so reject \( -9 \) immediately.

Question 4. Find p, if \( (2\hat{i} + 6\hat{j} + 27\hat{k}) \times (\hat{i} + 3\hat{j} + p\hat{k}) = \vec{0} \) .
Answer: Two vectors have a zero cross product if and only if they are collinear (parallel).
For parallel vectors, their corresponding scalar components must be in equal ratio:
\( \frac{2}{1} = \frac{6}{3} = \frac{27}{p} \).
Since \( \frac{2}{1} = 2 \) and \( \frac{6}{3} = 2 \):
\( 2 = \frac{27}{p} 
\implies p = \frac{27}{2} \).
In simple words: A zero cross product means the vectors are parallel, so compare the matching coordinate ratios to find p.

Exam Tip: Equating component ratios is much faster than setting up and expanding the full \( 3 \times 3 \) cross-product determinant.

Level - II

Question 1. Find λ, if \( (2\hat{i} + 6\hat{j} + 14\hat{k}) \times (\hat{i} - \lambda\hat{j} + 7\hat{k}) = \vec{0} \) .
Answer: Since the vector product of the two vectors is the zero vector, the vectors are parallel.
Their components must therefore be proportional:
\( \frac{2}{1} = \frac{6}{-\lambda} = \frac{14}{7} \).
Since \( \frac{2}{1} = 2 \) and \( \frac{14}{7} = 2 \):
\( 2 = \frac{6}{-\lambda} \implies -2\lambda = 6 \implies \lambda = -3 \).
In simple words: Cross product is zero for parallel vectors; equating ratios gives lambda equals -3.

Exam Tip: Do not overlook the negative sign attached to \( \lambda \) in the middle component.

Question 2. Show that (\( \vec{a} \) - \( \vec{b} \)) x (\( \vec{a} \) + \( \vec{b} \)) = 2(\( \vec{a} \) x \( \vec{b} \))
Answer: Expand the cross product using distributivity:
\( \text{LHS} = (\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = \vec{a} \times \vec{a} + \vec{a} \times \vec{b} - \vec{b} \times \vec{a} - \vec{b} \times \vec{b} \).
Since the cross product of any vector with itself is zero, \( \vec{a} \times \vec{a} = \vec{0} \) and \( \vec{b} \times \vec{b} = \vec{0} \).
Furthermore, cross product is anti-commutative, so \( -\vec{b} \times \vec{a} = \vec{a} \times \vec{b} \).
Substituting these properties in:
\( \text{LHS} = \vec{0} + \vec{a} \times \vec{b} + \vec{a} \times \vec{b} - \vec{0} = 2(\vec{a} \times \vec{b}) = \text{RHS} \).
Hence proved.
In simple words: Self-cross-products vanish, and flipping the order of the remaining cross product doubles the original term.

Exam Tip: State the property \( \vec{b} \times \vec{a} = -\vec{a} \times \vec{b} \) explicitly to justify why the terms add together.

Question 3. Find the angle between two vectors \( \vec{a} \) and \( \vec{b} \) if |\( \vec{a} \)| = 3, |\( \vec{b} \)| =4 and |\( \vec{a} \times \vec{b} \)| = 6.
Answer: The magnitude of the cross product relates to angle \( \theta \) by:
\( |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta \).
Substitute the given magnitudes:
\( 6 = (3)(4)\sin\theta \)
\( \implies 6 = 12\sin\theta \)
\( \implies \sin\theta = \frac{6}{12} = \frac{1}{2} \).
For \( 0 \le \theta \le \pi \), \( \sin\theta = \frac{1}{2} \) gives two solutions:
\( \theta = \frac{\pi}{6} \) (or \( 30^\circ \)) and \( \theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \) (or \( 150^\circ \)).
In simple words: Divide the cross product length by 12 to find sine theta is 1/2, which gives either 30 or 150 degrees.

Exam Tip: Always list both supplementary angles \( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \); providing only the acute angle loses half credit.

Question 4. Let \( \vec{a} \) , \( \vec{b} \) , \( \vec{c} \) be unit vectors such that \( \vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} = 0 \) and the angle between \( \vec{b} \) and \( \vec{c} \) is π /6, prove that \( \vec{a} = \pm 2(\vec{b} \times \vec{c}) \).
Answer: Since \( \vec{a} \cdot \vec{b} = 0 \) and \( \vec{a} \cdot \vec{c} = 0 \), vector \( \vec{a} \) is perpendicular to both \( \vec{b} \) and \( \vec{c} \).
Consequently, \( \vec{a} \) is parallel to \( \vec{b} \times \vec{c} \), so we can write:
\( \vec{a} = \pm \frac{\vec{b} \times \vec{c}}{|\vec{b} \times \vec{c}|} \).
Compute the magnitude \( |\vec{b} \times \vec{c}| \):
\( |\vec{b} \times \vec{c}| = |\vec{b}||\vec{c}|\sin\frac{\pi}{6} \).
Since \( \vec{b} \) and \( \vec{c} \) are unit vectors, \( |\vec{b}| = 1 \) and \( |\vec{c}| = 1 \):
\( |\vec{b} \times \vec{c}| = (1)(1)\left(\frac{1}{2}\right) = \frac{1}{2} \).
Substituting this magnitude back into the unit vector expression:
\( \vec{a} = \pm \frac{\vec{b} \times \vec{c}}{\frac{1}{2}} = \pm 2(\vec{b} \times \vec{c}) \).
Hence proved.
In simple words: Being perpendicular to both vectors means vector a points along their cross product, and dividing by the cross product's length of 1/2 multiplies it by 2.

Exam Tip: State clearly that \( \vec{a} \) being a unit vector normal to the plane of \( \vec{b} \) and \( \vec{c} \) makes \( \vec{a} = \pm \hat{n} \).

Level - III

Question 1. Find the value of the following: \( \hat{i}.(\hat{j} \times \hat{k}) + \hat{j}.(\hat{i} \times \hat{k}) + \hat{k}.(\hat{i} \times \hat{j}) \)
Answer: Using standard unit vector cross product relations:
1. \( \hat{j} \times \hat{k} = \hat{i} \implies \hat{i} \cdot (\hat{j} \times \hat{k}) = \hat{i} \cdot \hat{i} = 1 \)
2. \( \hat{i} \times \hat{k} = -\hat{j} \implies \hat{j} \cdot (\hat{i} \times \hat{k}) = \hat{j} \cdot (-\hat{j}) = -1 \)
3. \( \hat{i} \times \hat{j} = \hat{k} \implies \hat{k} \cdot (\hat{i} \times \hat{j}) = \hat{k} \cdot \hat{k} = 1 \).
Adding these three terms together:
\( 1 + (-1) + 1 = 1 \).
Therefore, the value is 1.
In simple words: The cyclic products evaluate to 1, -1, and 1, adding up to 1.

Exam Tip: Note the anti-clockwise orientation: \( \hat{i} \times \hat{k} = -\hat{j} \), which introduces the negative sign.

Question 2. Vectors \( \vec{a} \) and \( \vec{b} \) are such that \( |\vec{a}| = \sqrt{3} \) , \( |\vec{b}| = \frac{2}{3} \) , and \( \vec{a} \times \vec{b} \) is a unit vector. Write the angle between \( \vec{a} \) and \( \vec{b} \)
Answer: Given that \( \vec{a} \times \vec{b} \) is a unit vector, its length is 1:
\( |\vec{a} \times \vec{b}| = 1 \).
By the definition of the cross product magnitude:
\( |\vec{a}||\vec{b}|\sin\theta = 1 \).
Substitute the known values:
\( \sqrt{3} \left(\frac{2}{3}\right)\sin\theta = 1 \)
\( \implies \frac{2}{\sqrt{3}}\sin\theta = 1 \)
\( \implies \sin\theta = \frac{\sqrt{3}}{2} \).
For \( 0 \le \theta \le \pi \), this gives:
\( \theta = \frac{\pi}{3} \) (or \( 60^\circ \)) and \( \theta = \frac{2\pi}{3} \) (or \( 120^\circ \)).
In simple words: Substitute the lengths into the cross-product equation to find sine theta is root 3 over 2, giving 60 or 120 degrees.

Exam Tip: When \( \sin\theta = \frac{\sqrt{3}}{2} \), always mention both possible values: \( \frac{\pi}{3} \) and \( \frac{2\pi}{3} \).

Question 3. If \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \) and \( \vec{b} = \hat{j} - \hat{k} \) , find a vector \( \vec{c} \) such that \( \vec{a} \times \vec{c} = \vec{b} \) and \( \vec{a} \cdot \vec{c} = 3 \).
Answer: Take the cross product of \( \vec{a} \) with both sides of \( \vec{a} \times \vec{c} = \vec{b} \):
\( \vec{a} \times (\vec{a} \times \vec{c}) = \vec{a} \times \vec{b} \).
Using the vector triple product expansion identity:
\( (\vec{a} \cdot \vec{c})\vec{a} - (\vec{a} \cdot \vec{a})\vec{c} = \vec{a} \times \vec{b} \).
We are given \( \vec{a} \cdot \vec{c} = 3 \), and \( \vec{a} \cdot \vec{a} = |\vec{a}|^2 = 1^2 + 1^2 + 1^2 = 3 \).
Calculate \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 0 & 1 & -1 \end{vmatrix} = \hat{i}(-1 - 1) - \hat{j}(-1 - 0) + \hat{k}(1 - 0) = -2\hat{i} + \hat{j} + \hat{k} \).
Substitute into the identity:
\( 3(\hat{i} + \hat{j} + \hat{k}) - 3\vec{c} = -2\hat{i} + \hat{j} + \hat{k} \)
\( \implies 3\vec{c} = (3\hat{i} + 3\hat{j} + 3\hat{k}) - (-2\hat{i} + \hat{j} + \hat{k}) \)
\( \implies 3\vec{c} = 5\hat{i} + 2\hat{j} + 2\hat{k} \)
\( \implies \vec{c} = \frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k} \).
In simple words: Apply vector triple product to unwrap the cross product, plug in the given values, and solve for c.

Exam Tip: The vector triple product formula \( \vec{a} \times (\vec{a} \times \vec{c}) = (\vec{a}\cdot\vec{c})\vec{a} - |\vec{a}|^2\vec{c} \) is the most efficient method for solving this problem.

Question 4. If \( \vec{a} \times \vec{b} = \vec{c} \times \vec{d} \) and \( \vec{a} \times \vec{c} = \vec{b} \times \vec{d} \) show that (\( \vec{a} \) - \( \vec{d} \)) is parallel to \( \vec{b} \) - \( \vec{c} \) , where \( \vec{a} \neq \vec{d} \) and \( \vec{b} \neq \vec{c} \) .
Answer: We are given two equations:
1. \( \vec{a} \times \vec{b} = \vec{c} \times \vec{d} \)
2. \( \vec{a} \times \vec{c} = \vec{b} \times \vec{d} \).
Subtract equation (2) from equation (1):
\( (\vec{a} \times \vec{b}) - (\vec{a} \times \vec{c}) = (\vec{c} \times \vec{d}) - (\vec{b} \times \vec{d}) \).
Using the distributive property of cross products on the left side:
\( \vec{a} \times (\vec{b} - \vec{c}) \).
Rearranging the right side using the anti-commutative property:
\( (\vec{c} \times \vec{d}) - (\vec{b} \times \vec{d}) = -(\vec{d} \times \vec{c}) + (\vec{d} \times \vec{b}) = \vec{d} \times (\vec{b} - \vec{c}) \).
Equating both sides:
\( \vec{a} \times (\vec{b} - \vec{c}) = \vec{d} \times (\vec{b} - \vec{c}) \)
\( \implies \vec{a} \times (\vec{b} - \vec{c}) - \vec{d} \times (\vec{b} - \vec{c}) = \vec{0} \)
\( \implies (\vec{a} - \vec{d}) \times (\vec{b} - \vec{c}) = \vec{0} \).
Since \( \vec{a} \neq \vec{d} \) and \( \vec{b} \neq \vec{c} \), neither vector is the zero vector.
A cross product of two non-zero vectors equals zero if and only if they are parallel.
Therefore, \( (\vec{a} - \vec{d}) \) is parallel to \( (\vec{b} - \vec{c}) \).
Hence proved.
In simple words: Subtract the two equations and factor out the common vector to show that their cross product is zero.

Exam Tip: Be mindful of factor ordering: \( \vec{c} \times \vec{d} = -\vec{d} \times \vec{c} \) is essential for grouping terms correctly.

Question 5. Express \( 2\hat{i} - \hat{j} + 3\hat{k} \) as the sum of a vector parallel and perpendicular to \( 2\hat{i} + 4\hat{j} - 2\hat{k} \).
Answer: Let \( \vec{r} = 2\hat{i} - \hat{j} + 3\hat{k} \) and \( \vec{m} = 2\hat{i} + 4\hat{j} - 2\hat{k} \).
We write \( \vec{r} = \vec{r}_1 + \vec{r}_2 \), where \( \vec{r}_1 \parallel \vec{m} \) and \( \vec{r}_2 \perp \vec{m} \).
The vector projection of \( \vec{r} \) onto \( \vec{m} \) gives the parallel component:
\( \vec{r}_1 = \left( \frac{\vec{r} \cdot \vec{m}}{|\vec{m}|^2} \right)\vec{m} \).
Compute \( \vec{r} \cdot \vec{m} \):
\( \vec{r} \cdot \vec{m} = 2(2) + (-1)(4) + 3(-2) = 4 - 4 - 6 = -6 \).
Compute \( |\vec{m}|^2 \):
\( |\vec{m}|^2 = 2^2 + 4^2 + (-2)^2 = 4 + 16 + 4 = 24 \).
Thus:
\( \vec{r}_1 = \frac{-6}{24}\vec{m} = -\frac{1}{4}(2\hat{i} + 4\hat{j} - 2\hat{k}) = -\frac{1}{2}\hat{i} - \hat{j} + \frac{1}{2}\hat{k} \).
Now determine the perpendicular component \( \vec{r}_2 \):
\( \vec{r}_2 = \vec{r} - \vec{r}_1 = (2\hat{i} - \hat{j} + 3\hat{k}) - \left(-\frac{1}{2}\hat{i} - \hat{j} + \frac{1}{2}\hat{k}\right) \)
\( = \left(2 + \frac{1}{2}\right)\hat{i} + (-1 + 1)\hat{j} + \left(3 - \frac{1}{2}\right)\hat{k} = \frac{5}{2}\hat{i} + \frac{5}{2}\hat{k} \).
Therefore, the decomposition is:
\( 2\hat{i} - \hat{j} + 3\hat{k} = \left(-\frac{1}{2}\hat{i} - \hat{j} + \frac{1}{2}\hat{k}\right) + \left(\frac{5}{2}\hat{i} + \frac{5}{2}\hat{k}\right) \).
In simple words: Project the vector onto m to find the parallel part, then subtract that part from the original to get the perpendicular part.

Exam Tip: Quickly verify that \( \vec{r}_2 \cdot \vec{m} = \frac{5}{2}(2) + 0(4) + \frac{5}{2}(-2) = 5 - 5 = 0 \) to make sure the perpendicular component is correct.

(viii)Area of a triangle & Area of a parallelogram

Level - I

Question 1. Find the area of Parallelogram whose adjacent sides are represented by the vectors \( \vec{a} = 3\hat{i} + \hat{j} - 2\hat{k} \) and \( \vec{b} = \hat{i} - 3\hat{j} + 4\hat{k} \).
Answer: The area of a parallelogram with adjacent sides \( \vec{a} \) and \( \vec{b} \) is given by \( |\vec{a} \times \vec{b}| \).
Evaluate the cross product:
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & -3 & 4 \end{vmatrix} \)
\( = \hat{i}(4 - 6) - \hat{j}(12 - (-2)) + \hat{k}(-9 - 1) \)
\( = -2\hat{i} - 14\hat{j} - 10\hat{k} \).
Find the magnitude of \( \vec{a} \times \vec{b} \):
\( \text{Area} = \sqrt{(-2)^2 + (-14)^2 + (-10)^2} = \sqrt{4 + 196 + 100} = \sqrt{300} = 10\sqrt{3} \text{ sq units} \).
In simple words: Find the cross product of the two sides and calculate its length to get the area.

Exam Tip: Always state the unit "sq units" at the end of an area problem.

Question 2. If \( \vec{a} \) and \( \vec{b} \) represent the two adjacent sides of a Parallelogram, then write the area of parallelogram in terms of \( \vec{a} \) and \( \vec{b} \) .
Answer: The area of a parallelogram having adjacent side vectors \( \vec{a} \) and \( \vec{b} \) is given by:
\( \text{Area} = |\vec{a} \times \vec{b}| \).
In simple words: The magnitude of the cross product of the two adjacent side vectors gives the area.

Exam Tip: Be sure to distinguish between adjacent sides (\( |\vec{a} \times \vec{b}| \)) and diagonals (\( \frac{1}{2}|\vec{d}_1 \times \vec{d}_2| \)).

Question 3. Find the area of triangle having the points A(1,1,1) , B(1,2,3) and C(2,3,1) as its vertices.
Answer: Find two co-initial edge vectors of triangle ABC:
\( \vec{AB} = (1 - 1)\hat{i} + (2 - 1)\hat{j} + (3 - 1)\hat{k} = 0\hat{i} + \hat{j} + 2\hat{k} \)
\( \vec{AC} = (2 - 1)\hat{i} + (3 - 1)\hat{j} + (1 - 1)\hat{k} = \hat{i} + 2\hat{j} + 0\hat{k} \).
Compute their cross product:
\( \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 2 & 0 \end{vmatrix} \)
\( = \hat{i}(0 - 4) - \hat{j}(0 - 2) + \hat{k}(0 - 1) = -4\hat{i} + 2\hat{j} - \hat{k} \).
The area of triangle ABC is half the magnitude of this cross product:
\( \text{Area} = \frac{1}{2}|\vec{AB} \times \vec{AC}| = \frac{1}{2}\sqrt{(-4)^2 + 2^2 + (-1)^2} = \frac{1}{2}\sqrt{16 + 4 + 1} = \frac{\sqrt{21}}{2} \text{ sq units} \).
In simple words: Form two side vectors from vertex A, compute their cross product, and halve the resulting length.

Exam Tip: Remember the factor of \( \frac{1}{2} \) for triangle areas, which is easy to overlook under exam pressure.

Level - II

Question 1. Show that the area of the Parallelogram having diagonals ( \( 3\hat{i} + \hat{j} - 2\hat{k} \) ) and ( \( \hat{i} - 3\hat{j} + 4\hat{k} \)) is 5√3 Sq units.
Answer: Let \( \vec{d}_1 = 3\hat{i} + \hat{j} - 2\hat{k} \) and \( \vec{d}_2 = \hat{i} - 3\hat{j} + 4\hat{k} \) be the diagonal vectors of the parallelogram.
The area of a parallelogram in terms of its diagonals is given by:
\( \text{Area} = \frac{1}{2}|\vec{d}_1 \times \vec{d}_2| \).
Compute the vector product \( \vec{d}_1 \times \vec{d}_2 \):
\( \vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & -3 & 4 \end{vmatrix} \)
\( = \hat{i}(4 - 6) - \hat{j}(12 - (-2)) + \hat{k}(-9 - 1) \)
\( = -2\hat{i} - 14\hat{j} - 10\hat{k} \).
Find its magnitude:
\( |\vec{d}_1 \times \vec{d}_2| = \sqrt{(-2)^2 + (-14)^2 + (-10)^2} = \sqrt{4 + 196 + 100} = \sqrt{300} = 10\sqrt{3} \).
Therefore, the area is:
\( \text{Area} = \frac{1}{2}(10\sqrt{3}) = 5\sqrt{3} \text{ sq units} \).
Hence proved.
In simple words: For diagonals, take their cross product, find its length, and multiply by one-half.

Exam Tip: Always remember that the diagonal formula requires a factor of \( \frac{1}{2} \), unlike the adjacent sides formula.

Question 2. If \( \vec{a} \) , \( \vec{b} \) , \( \vec{c} \) are the position vectors of the vertices of a Δ ABC, show that the area of the Δ ABC is \( \frac{1}{2}|\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}| \) .
Answer: Let points A, B, and C have position vectors \( \vec{a}, \vec{b}, \vec{c} \) respectively.
The side vectors \( \vec{AB} \) and \( \vec{AC} \) are:
\( \vec{AB} = \vec{b} - \vec{a} \)
\( \vec{AC} = \vec{c} - \vec{a} \).
The area of triangle ABC is given by:
\( \text{Area} = \frac{1}{2}|\vec{AB} \times \vec{AC}| = \frac{1}{2}|(\vec{b} - \vec{a}) \times (\vec{c} - \vec{a})| \).
Expanding the cross product using distributivity:
\( (\vec{b} - \vec{a}) \times (\vec{c} - \vec{a}) = \vec{b} \times \vec{c} - \vec{b} \times \vec{a} - \vec{a} \times \vec{c} + \vec{a} \times \vec{a} \).
Using the identity \( \vec{a} \times \vec{a} = \vec{0} \) and the anti-commutative property \( -\vec{b} \times \vec{a} = \vec{a} \times \vec{b} \) and \( -\vec{a} \times \vec{c} = \vec{c} \times \vec{a} \):
\( (\vec{b} - \vec{a}) \times (\vec{c} - \vec{a}) = \vec{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{c} \times \vec{a} \)
\( = \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} \).
Substituting this back into the area formula:
\( \text{Area} = \frac{1}{2}|\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}| \).
Hence proved.
In simple words: Write the two sides using vertex position vectors and expand the cross product using anti-symmetry.

Exam Tip: A corollary to remember: if points A, B, and C are collinear, then \( \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = \vec{0} \).

Question 3. Using Vectors, find the area of the triangle with vertices A(1,1,2), B(2,3,5) and C(1,5,5) [ CBSE 2011]
Answer: Find the vectors representing sides AB and AC:
\( \vec{AB} = (2 - 1)\hat{i} + (3 - 1)\hat{j} + (5 - 2)\hat{k} = \hat{i} + 2\hat{j} + 3\hat{k} \)
\( \vec{AC} = (1 - 1)\hat{i} + (5 - 1)\hat{j} + (5 - 2)\hat{k} = 0\hat{i} + 4\hat{j} + 3\hat{k} \).
Compute their cross product:
\( \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 4 & 3 \end{vmatrix} \)
\( = \hat{i}(6 - 12) - \hat{j}(3 - 0) + \hat{k}(4 - 0) \)
\( = -6\hat{i} - 3\hat{j} + 4\hat{k} \).
Calculate the magnitude of the cross product:
\( |\vec{AB} \times \vec{AC}| = \sqrt{(-6)^2 + (-3)^2 + 4^2} = \sqrt{36 + 9 + 16} = \sqrt{61} \).
The area of triangle ABC is:
\( \text{Area} = \frac{1}{2}|\vec{AB} \times \vec{AC}| = \frac{\sqrt{61}}{2} \text{ sq units} \).
In simple words: Form side vectors AB and AC, calculate their cross product, and take half the length of that vector.

Exam Tip: Double-check the \( \hat{j} \) term: \( -(3 - 0) = -3 \); dropping the minus sign is a frequent source of lost marks.

Questions for Self Evaluation

Question 1. The scalar product of the vector \( \hat{i} + \hat{j} + \hat{k} \) with the unit vector along the sum of vectors \( 2\hat{i} + 4\hat{j} - 5\hat{k} \) and \( \lambda\hat{i} + 2\hat{j} + 3\hat{k} \) is equal to one. Find the value of λ.
Answer: Let \( \vec{s} \) be the sum of the two vectors:
\( \vec{s} = (2\hat{i} + 4\hat{j} - 5\hat{k}) + (\lambda\hat{i} + 2\hat{j} + 3\hat{k}) = (2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k} \).
The magnitude of \( \vec{s} \) is:
\( |\vec{s}| = \sqrt{(2 + \lambda)^2 + 6^2 + (-2)^2} = \sqrt{\lambda^2 + 4\lambda + 4 + 36 + 4} = \sqrt{\lambda^2 + 4\lambda + 44} \).
The unit vector along \( \vec{s} \) is \( \hat{s} = \frac{\vec{s}}{|\vec{s}|} \).
The scalar product of \( \hat{i} + \hat{j} + \hat{k} \) with \( \hat{s} \) is equal to 1:
\( (\hat{i} + \hat{j} + \hat{k}) \cdot \frac{(2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{\lambda^2 + 4\lambda + 44}} = 1 \)
\( \implies \frac{(2 + \lambda) + 6 - 2}{\sqrt{\lambda^2 + 4\lambda + 44}} = 1 \)
\( \implies \frac{\lambda + 6}{\sqrt{\lambda^2 + 4\lambda + 44}} = 1 \)
\( \implies \lambda + 6 = \sqrt{\lambda^2 + 4\lambda + 44} \).
Squaring both sides:
\( (\lambda + 6)^2 = \lambda^2 + 4\lambda + 44 \)
\( \implies \lambda^2 + 12\lambda + 36 = \lambda^2 + 4\lambda + 44 \)
\( \implies 12\lambda - 4\lambda = 44 - 36 \)
\( \implies 8\lambda = 8 \implies \lambda = 1 \).
In simple words: Add the vectors, divide by their length, dot with the given vector, and square both sides to find lambda equals 1.

Exam Tip: Check for extraneous roots when squaring equations: \( 1 + 6 = 7 \) and \( \sqrt{1 + 4 + 44} = \sqrt{49} = 7 \), confirming \( \lambda = 1 \).

Question 2. If \( \vec{a} \) , \( \vec{b} \) and \( \vec{c} \) be three vectors such that |\( \vec{a} \)| = 3, |\( \vec{b} \)| =4 , |\( \vec{c} \)| =5 and each one of them being perpendicular to the sum of the other two, find |\( \vec{a} + \vec{b} + \vec{c} \)|.
Answer: Given that each vector is perpendicular to the sum of the remaining two:
\( \vec{a} \cdot (\vec{b} + \vec{c}) = 0 \implies \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 0 \)
\( \vec{b} \cdot (\vec{c} + \vec{a}) = 0 \implies \vec{b} \cdot \vec{c} + \vec{b} \cdot \vec{a} = 0 \)
\( \vec{c} \cdot (\vec{a} + \vec{b}) = 0 \implies \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} = 0 \).
Adding these three equations together yields:
\( 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \).
Now evaluate the magnitude of the sum vector:
\( |\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \)
\( = 3^2 + 4^2 + 5^2 + 0 = 9 + 16 + 25 = 50 \).
Taking the square root:
\( |\vec{a} + \vec{b} + \vec{c}| = \sqrt{50} = 5\sqrt{2} \).
In simple words: The pairwise dot product sum is 0, so the squared total length is simply 9 + 16 + 25 = 50.

Exam Tip: Simplify \( \sqrt{50} \) into \( 5\sqrt{2} \) to present the final answer in simplest radical form.

Question 3. If |\( \vec{a} + \vec{b} \)| = |\( \vec{a} - \vec{b} \)|, then find the angle between \( \vec{a} \) and \( \vec{b} \) .
Answer: Squaring both sides of the equation:
\( |\vec{a} + \vec{b}|^2 = |\vec{a} - \vec{b}|^2 \)
\( \implies |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b}) \).
Subtracting \( |\vec{a}|^2 + |\vec{b}|^2 \) from both sides:
\( 4(\vec{a} \cdot \vec{b}) = 0 \implies \vec{a} \cdot \vec{b} = 0 \).
Since their scalar product is zero, the vectors are perpendicular.
Therefore, the angle between \( \vec{a} \) and \( \vec{b} \) is \( \frac{\pi}{2} \) (or \( 90^\circ \)).
In simple words: When the sum of two vectors has the same length as their difference, they must be at right angles.

Exam Tip: Clearly show the cancellation of \( |\vec{a}|^2 \) and \( |\vec{b}|^2 \) to demonstrate full algebraic working.

Question 4. Dot product of a vector with \( \hat{i} + \hat{j} – 3\hat{k} \), \( \hat{i} + 3\hat{j} – 2\hat{k} \), and \( 2\hat{i} + \hat{j} + 4\hat{k} \) are 0, 5, 8 respectively. Find the vector.
Answer: Let the unknown vector be \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).
Using the given dot products, write three equations:
1. \( x + y - 3z = 0 \)
2. \( x + 3y - 2z = 5 \)
3. \( 2x + y + 4z = 8 \).
Subtract equation (1) from equation (2):
\( (x + 3y - 2z) - (x + y - 3z) = 5 - 0 \)
\( \implies 2y + z = 5 \implies z = 5 - 2y \).
From equation (1), express \( x \) in terms of \( y \) and \( z \):
\( x = 3z - y = 3(5 - 2y) - y = 15 - 6y - y = 15 - 7y \).
Substitute \( x = 15 - 7y \) and \( z = 5 - 2y \) into equation (3):
\( 2(15 - 7y) + y + 4(5 - 2y) = 8 \)
\( \implies 30 - 14y + y + 20 - 8y = 8 \)
\( \implies 50 - 21y = 8 \)
\( \implies -21y = 8 - 50 = -42 \implies y = 2 \).
Now solve for \( z \) and \( x \):
\( z = 5 - 2(2) = 1 \)
\( x = 15 - 7(2) = 1 \).
Hence, the required vector is \( \vec{r} = \hat{i} + 2\hat{j} + \hat{k} \).
In simple words: Write out the three dot product equations, eliminate variables to find x = 1, y = 2, and z = 1.

Exam Tip: Substitute your final coordinates back into all three original equations to guarantee full accuracy.

Question 5. Find the components of a vector which is perpendicular to the vectors \( \hat{i} + 2\hat{j} – \hat{k} \) and \( 3\hat{i} – \hat{j} + 2\hat{k} \).
Answer: A vector perpendicular to both \( \vec{a} = \hat{i} + 2\hat{j} - \hat{k} \) and \( \vec{b} = 3\hat{i} - \hat{j} + 2\hat{k} \) is given by their cross product:
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{vmatrix} \)
\( = \hat{i}(4 - 1) - \hat{j}(2 - (-3)) + \hat{k}(-1 - 6) \)
\( = 3\hat{i} - 5\hat{j} - 7\hat{k} \).
Any vector perpendicular to both given vectors is of the form \( k(3\hat{i} - 5\hat{j} - 7\hat{k}) \), where \( k \) is a non-zero scalar.
Its scalar components are proportional to \( 3, -5, -7 \) (or for \( k = 1 \), the components are \( 3, -5, -7 \)).
In simple words: Take the cross product of the two vectors to find the perpendicular direction components 3, -5, and -7.

Exam Tip: State that any scalar multiple \( 3k, -5k, -7k \) (with \( k \neq 0 \)) is a valid set of components.