OP Malhotra Class 12 Maths Solutions Chapter 21 Vectors Chapter Test

Question 1. Find the magnitude of the vector \( \vec{a} = 2 \hat{i} - 6 \hat{j} - 3 \hat{k} \).
Answer:
Given vector is \( \vec{a} = 2 \hat{i} - 6 \hat{j} - 3 \hat{k} \).
To find the magnitude of this vector, we take the square root of the sum of the squares of its components.
\( |\vec{a}| = \sqrt{(2)^2 + (-6)^2 + (-3)^2} \)
\( = \sqrt{4 + 36 + 9} \)
\( = \sqrt{49} \)
\( = 7 \)
So, the magnitude of the vector \( \vec{a} \) is 7 units.
In simple words: To find how long a vector is, you square each of its number parts, add them up, and then take the square root of the total. This gives you the vector's length or magnitude.

🎯 Exam Tip: Remember to include the square of the negative signs, as \((-6)^2\) and \((-3)^2\) both result in positive values, which is a common mistake for students.

Question 2. Find the direction ratios and the direction cosines of the vector \( \vec{r} = 2 \hat{i} + 3 \hat{j} + \hat{k} \).
Answer:
Given vector is \( \vec{r} = 2 \hat{i} + 3 \hat{j} + \hat{k} \).
The coefficients of \( \hat{i}, \hat{j}, \hat{k} \) are the direction ratios of the vector.
So, the direction ratios of vector \( \vec{r} \) are \( \langle 2, 3, 1 \rangle \).
Next, we find the magnitude of the vector.
\( |\vec{r}| = \sqrt{(2)^2 + (3)^2 + (1)^2} \)
\( = \sqrt{4 + 9 + 1} \)
\( = \sqrt{14} \)
The direction cosines are found by dividing each direction ratio by the magnitude of the vector.
Thus, the direction cosines of \( \vec{r} \) are \( \langle \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{1}{\sqrt{14}} \rangle \). These values help define the orientation of the vector in space.
In simple words: The numbers in front of \( \hat{i}, \hat{j}, \hat{k} \) are the "direction ratios". To get "direction cosines", first find the vector's total length, then divide each direction ratio by that length.

🎯 Exam Tip: Direction ratios can be any set of numbers proportional to the actual direction, but direction cosines are unique and represent the cosines of the angles the vector makes with the coordinate axes.

Question 3.
(i) Find the position vector of a point R which divides the line joining the points P(\hat{i} + 2 \hat{j} - \hat{k}) and Q(-\hat{i} + \hat{j} + \hat{k}) in the ratio 2 : 1 internally.
(ii) externally.
(iii) Find the position vector of a point which divides the join of points with position vector \( \vec{a} - 2 \vec{b} \) and \( 2 \vec{a} + \vec{b} \) externally in the ratio 2 : 1.
Answer:
Let the position vectors of points P and Q be \( \vec{p} \) and \( \vec{q} \) respectively.
\( \vec{p} = \hat{i} + 2 \hat{j} - \hat{k} \)
\( \vec{q} = -\hat{i} + \hat{j} + \hat{k} \)
The ratio is \( m : n = 2 : 1 \).

(i) For internal division, the position vector \( \vec{r} \) is given by the formula \( \vec{r} = \frac{m\vec{q} + n\vec{p}}{m+n} \).
\( \vec{r} = \frac{2(-\hat{i} + \hat{j} + \hat{k}) + 1(\hat{i} + 2 \hat{j} - \hat{k})}{2+1} \)
\( = \frac{(-2\hat{i} + 2\hat{j} + 2\hat{k}) + (\hat{i} + 2\hat{j} - \hat{k})}{3} \)
\( = \frac{(-2+1)\hat{i} + (2+2)\hat{j} + (2-1)\hat{k}}{3} \)
\( = \frac{-\hat{i} + 4\hat{j} + \hat{k}}{3} \)
\( = -\frac{1}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{1}{3}\hat{k} \)

(ii) For external division, the position vector \( \vec{r}' \) is given by the formula \( \vec{r}' = \frac{m\vec{q} - n\vec{p}}{m-n} \).
\( \vec{r}' = \frac{2(-\hat{i} + \hat{j} + \hat{k}) - 1(\hat{i} + 2 \hat{j} - \hat{k})}{2-1} \)
\( = \frac{(-2\hat{i} + 2\hat{j} + 2\hat{k}) - (\hat{i} + 2\hat{j} - \hat{k})}{1} \)
\( = (-2-1)\hat{i} + (2-2)\hat{j} + (2-(-1))\hat{k} \)
\( = -3\hat{i} + 0\hat{j} + 3\hat{k} \)
\( = -3\hat{i} + 3\hat{k} \)

(iii) Let the two position vectors be \( \vec{A} = \vec{a} - 2 \vec{b} \) and \( \vec{B} = 2 \vec{a} + \vec{b} \). The ratio is \( m : n = 2 : 1 \) (externally).
The required position vector \( \vec{r}'' \) is given by \( \vec{r}'' = \frac{m\vec{B} - n\vec{A}}{m-n} \).
\( \vec{r}'' = \frac{2(2\vec{a} + \vec{b}) - 1(\vec{a} - 2\vec{b})}{2-1} \)
\( = \frac{(4\vec{a} + 2\vec{b}) - (\vec{a} - 2\vec{b})}{1} \)
\( = 4\vec{a} + 2\vec{b} - \vec{a} + 2\vec{b} \)
\( = 3\vec{a} + 4\vec{b} \)
These calculations are based on the section formula for vectors, which is similar to how we find a point dividing a line segment in coordinate geometry.
In simple words: To find a point that splits a line in a certain ratio, use a special formula. For "internal" division, you add the parts, and for "external" division, you subtract them. Just swap the numbers around carefully for each case.

🎯 Exam Tip: Always be careful with the signs in the external division formula \( \frac{m\vec{q} - n\vec{p}}{m-n} \), as a single sign error can lead to a completely different result.

Question 4. Find the value of p for which \( p(\hat{i} + \hat{j} + \hat{k}) \) is a unit vector.
Answer:
Let the given vector be \( \vec{v} = p(\hat{i} + \hat{j} + \hat{k}) \).
This can be written as \( \vec{v} = p\hat{i} + p\hat{j} + p\hat{k} \).
For a vector to be a unit vector, its magnitude must be 1.
First, let's find the magnitude of \( \vec{v} \):
\( |\vec{v}| = \sqrt{(p)^2 + (p)^2 + (p)^2} \)
\( = \sqrt{p^2 + p^2 + p^2} \)
\( = \sqrt{3p^2} \)
\( = |p|\sqrt{3} \)
Since \( \vec{v} \) is a unit vector, \( |\vec{v}| = 1 \).
So, \( |p|\sqrt{3} = 1 \).
This implies \( p\sqrt{3} = \pm 1 \).
Dividing by \( \sqrt{3} \), we get:
\( p = \pm \frac{1}{\sqrt{3}} \)
The parameter 'p' scales the direction of the vector to make its length exactly one unit.
In simple words: A "unit vector" has a length of 1. So, if you have a vector with a 'p' in it, calculate its total length. Then, set that length equal to 1 and solve for 'p'. Remember 'p' can be positive or negative.

🎯 Exam Tip: When finding the value of a scalar like 'p' that makes a vector a unit vector, always consider both positive and negative solutions since squaring removes the sign information.

Question 5. 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.
Answer:
Given vectors are \( \vec{a} = 3 \hat{i} + 2 \hat{j} + 9 \hat{k} \) and \( \vec{b} = \hat{i} + p \hat{j} + 3 \hat{k} \).
If two vectors \( \vec{a} \) and \( \vec{b} \) are parallel, then one vector must be a scalar multiple of the other. This means \( \vec{a} = \lambda \vec{b} \) for some non-zero scalar \( \lambda \).
Substitute the given vectors into this equation:
\( 3 \hat{i} + 2 \hat{j} + 9 \hat{k} = \lambda(\hat{i} + p \hat{j} + 3 \hat{k}) \)
\( \implies 3 \hat{i} + 2 \hat{j} + 9 \hat{k} = \lambda \hat{i} + \lambda p \hat{j} + 3\lambda \hat{k} \)
Now, we compare the coefficients of \( \hat{i}, \hat{j}, \) and \( \hat{k} \) on both sides of the equation.
For \( \hat{i} \): \( 3 = \lambda \)
For \( \hat{j} \): \( 2 = \lambda p \)
For \( \hat{k} \): \( 9 = 3\lambda \)
From the first equation, we find \( \lambda = 3 \).
Let's check this with the third equation: \( 9 = 3(3) \), which is \( 9 = 9 \). This confirms our value for \( \lambda \).
Now, substitute \( \lambda = 3 \) into the second equation:
\( 2 = (3)p \)
\( \implies 3p = 2 \)
\( \implies p = \frac{2}{3} \)
When vectors are parallel, their corresponding components are proportional. This fundamental property is crucial for solving such problems.
In simple words: If two vectors point in the same (or opposite) direction, they are parallel. This means all their matching number parts must be in the same proportion. Find the 'proportion number' first, then use it to find the unknown 'p'.

🎯 Exam Tip: Always compare all corresponding coefficients to ensure consistency. If the values of \( \lambda \) obtained from different components are not the same, the vectors are not parallel.

Question 6. Find the position vector of the mid-point of the line segment A B, where A is the point (3, 4, -2) and B is the point (1, 2, 4).
Answer:
Let the position vector of point A be \( \vec{A} \) and point B be \( \vec{B} \).
Given point A is (3, 4, -2), so its position vector is \( \vec{A} = 3 \hat{i} + 4 \hat{j} - 2 \hat{k} \).
Given point B is (1, 2, 4), so its position vector is \( \vec{B} = 1 \hat{i} + 2 \hat{j} + 4 \hat{k} \).
The position vector of the mid-point of a line segment connecting two points is found by averaging their position vectors.
The position vector of the mid-point (M) is given by the formula:
\( \vec{M} = \frac{\vec{A} + \vec{B}}{2} \)
Substitute the position vectors of A and B:
\( \vec{M} = \frac{(3 \hat{i} + 4 \hat{j} - 2 \hat{k}) + (\hat{i} + 2 \hat{j} + 4 \hat{k})}{2} \)
Combine the corresponding components:
\( \vec{M} = \frac{(3+1)\hat{i} + (4+2)\hat{j} + (-2+4)\hat{k}}{2} \)
\( \vec{M} = \frac{4\hat{i} + 6\hat{j} + 2\hat{k}}{2} \)
Divide each component by 2:
\( \vec{M} = 2\hat{i} + 3\hat{j} + \hat{k} \)
This mid-point vector represents the exact center of the line segment AB.
In simple words: To find the middle point of a line, just add the two starting position vectors together and then divide the whole thing by 2. This gives you the position vector for the midpoint.

🎯 Exam Tip: The midpoint formula is a specific case of the section formula where the ratio is 1:1. It's often easier to remember as simply averaging the coordinates or vectors.

Question 7.
(i) Write a vector of magnitude 15 units in the direction of vector \( \hat{i} - 2 \hat{j} + 2 \hat{k} \).
(ii) 7 units in the direction \( \vec{a} = \hat{i} - 2 \hat{j} \).
Answer:
To find a vector with a specific magnitude in a given direction, we first find the unit vector in that direction and then multiply it by the desired magnitude.

(i) Let the given vector be \( \vec{v} = \hat{i} - 2 \hat{j} + 2 \hat{k} \).
First, calculate the magnitude of \( \vec{v} \):
\( |\vec{v}| = \sqrt{(1)^2 + (-2)^2 + (2)^2} \)
\( = \sqrt{1 + 4 + 4} \)
\( = \sqrt{9} = 3 \)
Next, find the unit vector in the direction of \( \vec{v} \):
\( \hat{v} = \frac{\vec{v}}{|\vec{v}|} = \frac{\hat{i} - 2 \hat{j} + 2 \hat{k}}{3} \)
Now, to find a vector with magnitude 15 units in this direction, multiply the unit vector by 15:
Required vector \( = 15 \hat{v} \)
\( = 15 \left( \frac{\hat{i} - 2 \hat{j} + 2 \hat{k}}{3} \right) \)
\( = 5(\hat{i} - 2 \hat{j} + 2 \hat{k}) \)
\( = 5\hat{i} - 10\hat{j} + 10\hat{k} \)

(ii) Let the given vector be \( \vec{a} = \hat{i} - 2 \hat{j} \).
First, calculate the magnitude of \( \vec{a} \):
\( |\vec{a}| = \sqrt{(1)^2 + (-2)^2} \)
\( = \sqrt{1 + 4} \)
\( = \sqrt{5} \)
Next, find the unit vector in the direction of \( \vec{a} \):
\( \hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{\hat{i} - 2 \hat{j}}{\sqrt{5}} \)
Now, to find a vector with magnitude 7 units in this direction, multiply the unit vector by 7:
Required vector \( = 7 \hat{a} \)
\( = 7 \left( \frac{\hat{i} - 2 \hat{j}}{\sqrt{5}} \right) \)
\( = \frac{7}{\sqrt{5}}\hat{i} - \frac{14}{\sqrt{5}}\hat{j} \)
To rationalize the denominator, multiply the numerator and denominator by \( \sqrt{5} \):
\( = \frac{7\sqrt{5}}{5}\hat{i} - \frac{14\sqrt{5}}{5}\hat{j} \)
This method ensures the resulting vector has the correct length while maintaining its original orientation.
In simple words: First, find a "unit vector" for the given direction by dividing the original vector by its length. Then, multiply this unit vector by the new length you want the vector to have.

🎯 Exam Tip: Always calculate the unit vector correctly before scaling it to the desired magnitude. Rationalize denominators if instructed or for cleaner final answers, especially for direction cosines or vector components.

Question 8. 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:
First, we need to find the resultant vector \( \vec{V} = 2 \vec{a} - \vec{b} + 3 \vec{c} \).
Substitute the given vectors:
\( \vec{V} = 2(\hat{i} + \hat{j} + \hat{k}) - (4 \hat{i} - 2 \hat{j} + 3 \hat{k}) + 3(\hat{i} - 2 \hat{j} + \hat{k}) \)
Distribute the scalar multiples:
\( \vec{V} = (2\hat{i} + 2\hat{j} + 2\hat{k}) - (4\hat{i} - 2\hat{j} + 3\hat{k}) + (3\hat{i} - 6\hat{j} + 3\hat{k}) \)
Combine the \( \hat{i} \) components, \( \hat{j} \) components, and \( \hat{k} \) components separately:
\( \vec{V} = (2 - 4 + 3)\hat{i} + (2 - (-2) - 6)\hat{j} + (2 - 3 + 3)\hat{k} \)
\( \vec{V} = (2 - 4 + 3)\hat{i} + (2 + 2 - 6)\hat{j} + (2 - 3 + 3)\hat{k} \)
\( \vec{V} = (1)\hat{i} + (-2)\hat{j} + (2)\hat{k} \)
\( \vec{V} = \hat{i} - 2\hat{j} + 2\hat{k} \)

Next, we find the magnitude of this resultant vector \( \vec{V} \):
\( |\vec{V}| = \sqrt{(1)^2 + (-2)^2 + (2)^2} \)
\( = \sqrt{1 + 4 + 4} \)
\( = \sqrt{9} = 3 \)

Now, we need to find the unit vector in the direction of \( \vec{V} \):
\( \hat{V} = \frac{\vec{V}}{|\vec{V}|} = \frac{\hat{i} - 2\hat{j} + 2\hat{k}}{3} \)

Finally, to find a vector of magnitude 6 units parallel to \( \vec{V} \), we multiply the unit vector \( \hat{V} \) by 6:
Required vector \( = 6 \hat{V} \)
\( = 6 \left( \frac{\hat{i} - 2\hat{j} + 2\hat{k}}{3} \right) \)
\( = 2(\hat{i} - 2\hat{j} + 2\hat{k}) \)
\( = 2\hat{i} - 4\hat{j} + 4\hat{k} \)
This process allows us to create a new vector that has the desired length and orientation.
In simple words: First, combine the given vectors according to the math problem to get a new vector. Then, find the length of this new vector. Divide the new vector by its length to get a "unit vector" (length 1). Finally, multiply this unit vector by the desired length (6 in this case) to get your final answer.

🎯 Exam Tip: Pay close attention to the signs when performing vector addition and subtraction, especially when a negative sign precedes a parenthesis, as this is a frequent source of errors.

Question 9. Show that the points A(-2 \hat{i} + 3 \hat{j} + 5 \hat{k}), B(\hat{i} + 2 \hat{j} + 3 \hat{k}) and C(7 \hat{i} - \hat{k}) are collinear.
Answer:
To show that three points A, B, and C are collinear, we need to prove that the vector \( \vec{AB} \) is parallel to the vector \( \vec{BC} \) (or \( \vec{AC} \)) and that they share a common point.
First, let's write down the position vectors of the points:
Position vector of A (\(\vec{A}\)) \( = -2 \hat{i} + 3 \hat{j} + 5 \hat{k} \)
Position vector of B (\(\vec{B}\)) \( = \hat{i} + 2 \hat{j} + 3 \hat{k} \)
Position vector of C (\(\vec{C}\)) \( = 7 \hat{i} + 0 \hat{j} - \hat{k} \) (we can write \(0 \hat{j}\) for clarity)

Next, calculate the vector \( \vec{AB} \):
\( \vec{AB} = \vec{B} - \vec{A} \)
\( = (\hat{i} + 2 \hat{j} + 3 \hat{k}) - (-2 \hat{i} + 3 \hat{j} + 5 \hat{k}) \)
\( = (1 - (-2))\hat{i} + (2 - 3)\hat{j} + (3 - 5)\hat{k} \)
\( = (1 + 2)\hat{i} - \hat{j} - 2\hat{k} \)
\( = 3\hat{i} - \hat{j} - 2\hat{k} \)

Now, calculate the vector \( \vec{BC} \):
\( \vec{BC} = \vec{C} - \vec{B} \)
\( = (7 \hat{i} - \hat{k}) - (\hat{i} + 2 \hat{j} + 3 \hat{k}) \)
\( = (7 - 1)\hat{i} + (0 - 2)\hat{j} + (-1 - 3)\hat{k} \)
\( = 6\hat{i} - 2\hat{j} - 4\hat{k} \)

Now, compare \( \vec{AB} \) and \( \vec{BC} \). We can see a relationship:
\( \vec{BC} = 6\hat{i} - 2\hat{j} - 4\hat{k} \)
We can factor out a 2 from \( \vec{BC} \):
\( \vec{BC} = 2(3\hat{i} - \hat{j} - 2\hat{k}) \)
Notice that the expression in the parenthesis is exactly \( \vec{AB} \).
\( \implies \vec{BC} = 2 \vec{AB} \)
Since \( \vec{BC} \) is a scalar multiple of \( \vec{AB} \), this means \( \vec{AB} \) and \( \vec{BC} \) are parallel vectors. Also, point B is common to both vectors \( \vec{AB} \) and \( \vec{BC} \).
Therefore, points A, B, and C lie on the same straight line, meaning they are collinear.
This demonstrates that when vectors between points are parallel and share a point, the points must be aligned.
In simple words: To show three points are in a straight line, calculate the vector from the first point to the second (like \( \vec{AB} \)), and then the vector from the second to the third (like \( \vec{BC} \)). If one vector is just a scaled-up version of the other, and they share a point, then all three points are on the same line.

🎯 Exam Tip: When proving collinearity, always ensure that in addition to showing the vectors are parallel, you explicitly state that a common point exists between them to complete the proof.

Question 10. Show that the points A, B and 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 \( \vec{c} = \hat{i} - 3 \hat{j} - 5 \hat{k} \) are vertices of a right angled triangle.
Answer:
To prove that the points form a right-angled triangle, we need to show that the sum of the squares of the magnitudes of two sides equals the square of the magnitude of the third side (Pythagoras theorem).
Given position vectors:
\( \vec{A} = 3 \hat{i} - 4 \hat{j} - 4 \hat{k} \)
\( \vec{B} = 2 \hat{i} - \hat{j} + \hat{k} \)
\( \vec{C} = \hat{i} - 3 \hat{j} - 5 \hat{k} \)

First, calculate the vectors representing the sides of the triangle:
Vector \( \vec{AB} = \vec{B} - \vec{A} \)
\( = (2 \hat{i} - \hat{j} + \hat{k}) - (3 \hat{i} - 4 \hat{j} - 4 \hat{k}) \)
\( = (2-3)\hat{i} + (-1-(-4))\hat{j} + (1-(-4))\hat{k} \)
\( = -\hat{i} + 3\hat{j} + 5\hat{k} \)

Vector \( \vec{BC} = \vec{C} - \vec{B} \)
\( = (\hat{i} - 3 \hat{j} - 5 \hat{k}) - (2 \hat{i} - \hat{j} + \hat{k}) \)
\( = (1-2)\hat{i} + (-3-(-1))\hat{j} + (-5-1)\hat{k} \)
\( = -\hat{i} - 2\hat{j} - 6\hat{k} \)

Vector \( \vec{CA} = \vec{A} - \vec{C} \)
\( = (3 \hat{i} - 4 \hat{j} - 4 \hat{k}) - (\hat{i} - 3 \hat{j} - 5 \hat{k}) \)
\( = (3-1)\hat{i} + (-4-(-3))\hat{j} + (-4-(-5))\hat{k} \)
\( = 2\hat{i} - \hat{j} + \hat{k} \)

Next, calculate the square of the magnitude of each side:
\( |\vec{AB}|^2 = (-1)^2 + (3)^2 + (5)^2 = 1 + 9 + 25 = 35 \)
\( |\vec{BC}|^2 = (-1)^2 + (-2)^2 + (-6)^2 = 1 + 4 + 36 = 41 \)
\( |\vec{CA}|^2 = (2)^2 + (-1)^2 + (1)^2 = 4 + 1 + 1 = 6 \)

Now, check if Pythagoras theorem holds for any combination of sides:
Consider \( |\vec{AB}|^2 + |\vec{CA}|^2 = 35 + 6 = 41 \)
We see that \( |\vec{AB}|^2 + |\vec{CA}|^2 = |\vec{BC}|^2 \).
Since \( AB^2 + CA^2 = BC^2 \), the triangle formed by points A, B, and C is a right-angled triangle. The right angle is at the vertex A, which is opposite to the longest side BC.
The Pythagorean theorem works for vectors just as it does for side lengths, confirming the triangle's shape.
In simple words: To check if a triangle is right-angled using vectors, first find the vectors for each of its three sides. Then, calculate the square of the length (magnitude) of each side. If the square of one side's length is equal to the sum of the squares of the other two sides' lengths, then it's a right-angled triangle.

🎯 Exam Tip: After finding the square of the magnitudes, identify the largest value. The sum of the other two squared magnitudes should equal this largest value for a right-angled triangle. The vertex opposite the longest side is where the right angle is located.