GSEB Class 12 Maths Solutions Chapter 10 Vector Algebra Exercise 10.2

Question 1. Compute the magnitudes of the following vectors:
\( \vec{a} = \hat {i} + \hat {j} + \hat {k} \)
\( \vec{b} = 2\hat {i} - 7\hat {j} - 3\hat {k} \)
\( \vec{c} = \frac{1}{\sqrt{3}}\hat {i} + \frac{1}{\sqrt{3}}\hat {j} - \frac{1}{\sqrt{3}}\hat {k} \)
Answer:
(i) For \( \vec{a} = \hat {i} + \hat {j} + \hat {k} \), the magnitude is calculated as:
\( |\vec{a}| = \sqrt{1^{2} + 1^{2} + 1^{2}} = \sqrt{1 + 1 + 1} = \sqrt{3} \).
(ii) For \( \vec{b} = 2\hat {i} - 7\hat {j} - 3\hat {k} \), the magnitude is found by:
\( |\vec{b}| = \sqrt{2^{2} + (-7)^{2} + (-3)^{2}} = \sqrt{4 + 49 + 9} = \sqrt{62} \).
(iii) For \( \vec{c} = \frac{1}{\sqrt{3}}\hat {i} + \frac{1}{\sqrt{3}}\hat {j} - \frac{1}{\sqrt{3}}\hat {k} \), its magnitude is:
\( |\vec{c}| = \sqrt{ (\frac{1}{\sqrt{3}})^{2} + (\frac{1}{\sqrt{3}})^{2} + (-\frac{1}{\sqrt{3}})^{2} } = \sqrt{ \frac{1}{3} + \frac{1}{3} + \frac{1}{3} } = \sqrt{ \frac{3}{3} } = \sqrt{1} = 1 \).
In simple words: To find a vector's magnitude, square each of its component numbers, add them up, and then take the square root of the total.

Exam Tip: Remember that the magnitude of a vector \( x\hat {i} + y\hat {j} + z\hat {k} \) is always given by \( \sqrt{x^2 + y^2 + z^2} \). Pay careful attention to negative signs when squaring components, as the square of a negative number is positive.

Question 2. Write two different vectors having the same magnitude.
Answer: We can consider the vectors \( \vec{a} = \hat {i} + 2\hat {j} + \hat {k} \) and \( \vec{b} = \hat {i} + \hat {j} + 2\hat {k} \).
The magnitude of \( \vec{a} \) is \( |\vec{a}| = \sqrt{1^{2} + 2^{2} + 1^{2}} = \sqrt{1 + 4 + 1} = \sqrt{6} \).
The magnitude of \( \vec{b} \) is \( |\vec{b}| = \sqrt{1^{2} + 1^{2} + 2^{2}} = \sqrt{1 + 1 + 4} = \sqrt{6} \).
Therefore, these two different vectors, \( \vec{a} \) and \( \vec{b} \), have the same magnitude.
In simple words: We can create different vectors that have the same length. For example, if you swap the numbers in a vector like \( (1, 2, 1) \) to get \( (1, 1, 2) \), their lengths will stay the same because you are just adding the squares of the same numbers.

Exam Tip: To create vectors with the same magnitude, simply rearrange the absolute values of the components. The order of the components changes the direction, but the magnitude remains the same if the squares of the numbers are identical.

Question 3. Write two different vectors having same direction.
Answer: Let the two vectors be \( \vec{a} = \hat {i} + \hat {j} + \hat {k} \) and \( \vec{b} = 3\hat {i} + 3\hat {j} + 3\hat {k} \).
The direction cosines of \( \vec{a} \) are \( (\frac{1}{\sqrt{1^2+1^2+1^2}}, \frac{1}{\sqrt{1^2+1^2+1^2}}, \frac{1}{\sqrt{1^2+1^2+1^2}}) = (\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}) \).
The direction cosines of \( \vec{b} \) are \( (\frac{3}{\sqrt{3^2+3^2+3^2}}, \frac{3}{\sqrt{3^2+3^2+3^2}}, \frac{3}{\sqrt{3^2+3^2+3^2}}) = (\frac{3}{\sqrt{27}}, \frac{3}{\sqrt{27}}, \frac{3}{\sqrt{27}}) \), which simplifies to \( (\frac{3}{3\sqrt{3}}, \frac{3}{3\sqrt{3}}, \frac{3}{3\sqrt{3}}) = (\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}) \).
Thus, the vectors \( \vec{a} \) and \( \vec{b} \) possess the same direction but different magnitudes.
In simple words: Vectors that point in the exact same direction are just scaled versions of each other. If you multiply a vector by a positive number, it gets longer but still points the same way.

Exam Tip: Two vectors have the same direction if their corresponding components are proportional and the scalar multiple is positive. Calculating direction cosines is a reliable method to confirm if two vectors are indeed pointing in the same direction.

Question 4. Find the values of x and y so that the vectors \( 2\hat {i} + 3\hat {j} \) and \( x\hat {i} + y\hat {j} \) are equal.
Answer: If two vectors are equal, their corresponding components must also be equal.
Given that \( 2\hat {i} + 3\hat {j} = x\hat {i} + y\hat {j} \).
By equating the coefficients of \( \hat {i} \) and \( \hat {j} \), we get:
For \( \hat {i} \): \( x = 2 \)
For \( \hat {j} \): \( y = 3 \)
So, the values are \( x = 2 \) and \( y = 3 \).
In simple words: When two vectors are identical, the numbers in front of \( \hat {i} \) must match, and the numbers in front of \( \hat {j} \) must also match.

Exam Tip: Remember the fundamental principle that for two vectors to be equal, their respective components along each axis must be identical. This concept is crucial for solving many vector-related problems.

Question 5. Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (- 5, 7).
Answer: Let the initial point be \( A(2, 1) \) and the terminal point be \( B(-5, 7) \).
The vector \( \overrightarrow{\mathrm{AB}} \) is calculated using the formula \( (x_2 - x_1)\hat {i} + (y_2 - y_1)\hat {j} \).
So, \( \overrightarrow{\mathrm{AB}} = (-5 - 2)\hat {i} + (7 - 1)\hat {j} = -7\hat {i} + 6\hat {j} \).
The vector components are \( -7\hat {i} \) and \( 6\hat {j} \).
The scalar components are \( -7 \) and \( 6 \).
In simple words: To find a vector between two points, subtract the starting x-coordinate from the ending x-coordinate for the \( \hat {i} \) part, and do the same for the y-coordinates for the \( \hat {j} \) part. The numbers by themselves are the scalar parts, and the numbers with \( \hat {i} \) or \( \hat {j} \) are the vector parts.

Exam Tip: Be careful with the order of subtraction; always subtract the initial point's coordinates from the terminal point's coordinates. The scalar components are just the numbers, while the vector components include the unit vectors \( \hat {i} \) and \( \hat {j} \).

Question 6. Find the sum of the vectors \( \vec{a} = \hat {i} – 2\hat {j} + \hat {k} \), \( \vec{b} = – 2\hat {i} + 4\hat {j} + 5\hat {k} \) and \( \vec{c} = \hat {i} – 6\hat {j} – 7\hat {k} \).
Answer: To find the sum of vectors, we add their corresponding components.
Sum of vectors \( = \vec{a} + \vec{b} + \vec{c} \)
\( = (\hat {i} – 2\hat {j} + \hat {k}) + (-2\hat {i} + 4\hat {j} + 5\hat {k}) + (\hat {i} – 6\hat {j} – 7\hat {k}) \)
\( = (1 - 2 + 1)\hat {i} + (-2 + 4 - 6)\hat {j} + (1 + 5 - 7)\hat {k} \)
\( = 0\hat {i} - 4\hat {j} - 1\hat {k} \)
\( = -4\hat {j} - \hat {k} \).
In simple words: To add vectors, you just add up all the \( \hat {i} \) parts, all the \( \hat {j} \) parts, and all the \( \hat {k} \) parts separately.

Exam Tip: When adding vectors, always group and sum the coefficients of \( \hat {i} \), \( \hat {j} \), and \( \hat {k} \) independently. Be careful with positive and negative signs during addition.

Question 7. Find the unit vector in the direction of the vector \( \vec{a} = \hat {i} + \hat {j} + 2\hat {k} \).
Answer: First, we need to determine the magnitude of the given vector \( \vec{a} \).
The vector is \( \vec{a} = \hat {i} + \hat {j} + 2\hat {k} \).
Its magnitude is \( |\vec{a}| = \sqrt{1^{2} + 1^{2} + 2^{2}} = \sqrt{1 + 1 + 4} = \sqrt{6} \).
The unit vector in the direction of \( \vec{a} \) is given by \( \hat{a} = \frac{\vec{a}}{|\vec{a}|} \).
\( \hat{a} = \frac{1}{\sqrt{6}}(\hat {i} + \hat {j} + 2\hat {k}) \)
\( = \frac{1}{\sqrt{6}}\hat {i} + \frac{1}{\sqrt{6}}\hat {j} + \frac{2}{\sqrt{6}}\hat {k} \).
In simple words: A unit vector is a special kind of vector that has a length of exactly 1, but it points in the same direction as the original vector. You find it by dividing each part of the original vector by its total length.

Exam Tip: Always remember that a unit vector is obtained by dividing the vector by its magnitude. Ensure the magnitude is calculated correctly to avoid errors in the unit vector components.

Question 8. Find the unit vector in the direction of vector \( \overrightarrow{\mathrm{PQ}} \), where P and Q are the points (1, 2, 3) and (4, 5, 6) respectively.
Answer: The points are given as \( P(1, 2, 3) \) and \( Q(4, 5, 6) \).
First, we calculate the vector \( \overrightarrow{\mathrm{PQ}} \). The formula for a vector from point P to Q is \( (x_2 - x_1)\hat {i} + (y_2 - y_1)\hat {j} + (z_2 - z_1)\hat {k} \).
\( \overrightarrow{\mathrm{PQ}} = (4 - 1)\hat {i} + (5 - 2)\hat {j} + (6 - 3)\hat {k} \)
\( = 3\hat {i} + 3\hat {j} + 3\hat {k} \).
Next, we find the magnitude of \( \overrightarrow{\mathrm{PQ}} \).
\( |\overrightarrow{\mathrm{PQ}}| = \sqrt{3^{2} + 3^{2} + 3^{2}} = \sqrt{9 + 9 + 9} = \sqrt{27} = 3\sqrt{3} \).
Finally, the unit vector in the direction of \( \overrightarrow{\mathrm{PQ}} \) is \( \frac{\overrightarrow{\mathrm{PQ}}}{|\overrightarrow{\mathrm{PQ}}|} \).
\( = \frac{1}{3\sqrt{3}}(3\hat {i} + 3\hat {j} + 3\hat {k}) \)
\( = \frac{3}{3\sqrt{3}}(\hat {i} + \hat {j} + \hat {k}) \)
\( = \frac{1}{\sqrt{3}}\hat {i} + \frac{1}{\sqrt{3}}\hat {j} + \frac{1}{\sqrt{3}}\hat {k} \).
In simple words: First, create a vector that goes from point P to point Q by subtracting the coordinates. Then, find the length of this new vector. Divide each part of the vector by its total length to get the unit vector, which is a vector of length 1 pointing in the same direction.

Exam Tip: Pay close attention to the order of points when forming the vector (terminal minus initial). Simplify square roots where possible, such as \( \sqrt{27} = 3\sqrt{3} \), to present the answer in its most reduced form.

Question 9. For given vectors \( \vec{a} = 2\hat {i} – \hat {j} + 2\hat {k} \) and \( \vec{b} = - \hat {i} + \hat {j} – \hat {k} \), find the unit vector in the direction of the vector \( \vec{a} + \vec{b} \).
Answer: First, we need to find the sum of the two vectors, \( \vec{a} + \vec{b} \).
\( \vec{a} + \vec{b} = (2\hat {i} – \hat {j} + 2\hat {k}) + (-\hat {i} + \hat {j} – \hat {k}) \)
\( = (2 - 1)\hat {i} + (-1 + 1)\hat {j} + (2 - 1)\hat {k} \)
\( = 1\hat {i} + 0\hat {j} + 1\hat {k} \)
\( = \hat {i} + \hat {k} \).
Next, we calculate the magnitude of the resultant vector \( \vec{a} + \vec{b} \).
\( |\vec{a} + \vec{b}| = \sqrt{1^{2} + 0^{2} + 1^{2}} = \sqrt{1 + 0 + 1} = \sqrt{2} \).
Finally, the unit vector in the direction of \( \vec{a} + \vec{b} \) is \( \frac{\vec{a} + \vec{b}}{|\vec{a} + \vec{b}|} \).
\( = \frac{1}{\sqrt{2}}(\hat {i} + \hat {k}) \)
\( = \frac{1}{\sqrt{2}}\hat {i} + \frac{1}{\sqrt{2}}\hat {k} \).
In simple words: First, add the two vectors together by adding their corresponding parts. Then, find the length of this new combined vector. Finally, divide each part of the combined vector by its total length to get the unit vector.

Exam Tip: When a component's coefficient is zero (like \( 0\hat {j} \) here), it can be omitted in the final vector expression, but it must be included in the magnitude calculation to ensure accuracy.

Question 10. Find a vector in the direction of \( 5\hat {i} – \hat {j} + 2\hat {k} \) which has magnitude 8 units.
Answer: The given vector is \( \vec{a} = 5\hat {i} - \hat {j} + 2\hat {k} \).
First, we find the magnitude of \( \vec{a} \).
\( |\vec{a}| = \sqrt{5^{2} + (-1)^{2} + 2^{2}} = \sqrt{25 + 1 + 4} = \sqrt{30} \).
Next, we find the unit vector in the direction of \( \vec{a} \).
\( \hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{1}{\sqrt{30}}(5\hat {i} - \hat {j} + 2\hat {k}) \).
To find a vector with magnitude 8 in this direction, we multiply the unit vector by 8.
Vector \( = 8 \times \hat{a} \)
\( = 8 \times \frac{1}{\sqrt{30}}(5\hat {i} - \hat {j} + 2\hat {k}) \)
\( = \frac{40}{\sqrt{30}}\hat {i} - \frac{8}{\sqrt{30}}\hat {j} + \frac{16}{\sqrt{30}}\hat {k} \).
In simple words: To create a vector that points the same way but has a specific length, first make the original vector into a unit vector (length 1). Then, just multiply that unit vector by the desired length.

Exam Tip: The process involves two key steps: first, normalize the given vector to a unit vector, and second, multiply this unit vector by the desired magnitude. This ensures the new vector has the correct length while maintaining the original direction.

Question 11. Show that the vectors \( 2\hat {i} – 3\hat {j} + 4\hat {k} \) and \( - 4\hat {i} + 6\hat {j} – 8\hat {k} \) are collinear.
Answer: Let \( \vec{a} = 2\hat {i} – 3\hat {j} + 4\hat {k} \) and \( \vec{b} = -4\hat {i} + 6\hat {j} – 8\hat {k} \).
We can observe that \( \vec{b} \) can be written as a scalar multiple of \( \vec{a} \).
\( \vec{b} = -4\hat {i} + 6\hat {j} – 8\hat {k} \)
\( = -2(2\hat {i} – 3\hat {j} + 4\hat {k}) \)
\( = -2\vec{a} \).
Since \( \vec{b} = -2\vec{a} \), it means that \( \vec{b} \) is a scalar multiple of \( \vec{a} \) (here, the scalar \( \lambda = -2 \)).
When one vector is a scalar multiple of another, the vectors are collinear. In this particular instance, they have opposite directions because the scalar is negative.
In simple words: Two vectors are on the same line (collinear) if one can be made by just multiplying the other by a number. If that number is positive, they point the same way; if it's negative, they point in opposite directions, but they are still collinear.

Exam Tip: To show collinearity, demonstrate that one vector can be expressed as \( \lambda \) times the other vector (i.e., \( \vec{b} = \lambda\vec{a} \)). The value of \( \lambda \) can be any real number, including negative values, which simply indicates an opposite direction but still collinearity.

Question 12. Find the direction cosines of the vectors \( \hat {i} + 2\hat {j} + 3\hat {k} \).
Answer: For a vector \( x\hat {i} + y\hat {j} + z\hat {k} \), the direction cosines (l, m, n) are given by:
\( l = \frac{x}{\sqrt{x^2 + y^2 + z^2}} \)
\( m = \frac{y}{\sqrt{x^2 + y^2 + z^2}} \)
\( n = \frac{z}{\sqrt{x^2 + y^2 + z^2}} \)
Here, for the vector \( \vec{a} = \hat {i} + 2\hat {j} + 3\hat {k} \), we have \( x = 1, y = 2, z = 3 \).
First, calculate the magnitude: \( \sqrt{x^2 + y^2 + z^2} = \sqrt{1^{2} + 2^{2} + 3^{2}} = \sqrt{1 + 4 + 9} = \sqrt{14} \).
Now, we can find the direction cosines:
\( l = \frac{1}{\sqrt{14}} \)
\( m = \frac{2}{\sqrt{14}} \)
\( n = \frac{3}{\sqrt{14}} \)
Thus, the direction cosines of \( \vec{a} \) are \( \frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}} \).
In simple words: Direction cosines tell you how much a vector "leans" along each of the x, y, and z axes. You find them by dividing each of the vector's components by its total length.

Exam Tip: Direction cosines are essentially the components of the unit vector in the direction of the given vector. Ensure the denominator (magnitude) is calculated accurately, as any error will propagate through all three direction cosines.

Question 13. Find the direction cosines of the vector joining the points A(1, 2, – 3) and B(- 1, – 2, 1), directed from A to B.
Answer: Given points are \( A(1, 2, -3) \) and \( B(-1, -2, 1) \). The vector is directed from A to B.
Let \( (x_1, y_1, z_1) = (1, 2, -3) \) and \( (x_2, y_2, z_2) = (-1, -2, 1) \).
The vector joining points A and B, \( \overrightarrow{\mathrm{AB}} \), is calculated as:
\( \overrightarrow{\mathrm{AB}} = (x_2 - x_1)\hat {i} + (y_2 - y_1)\hat {j} + (z_2 - z_1)\hat {k} \)
\( = (-1 - 1)\hat {i} + (-2 - 2)\hat {j} + (1 - (-3))\hat {k} \)
\( = -2\hat {i} - 4\hat {j} + (1 + 3)\hat {k} \)
\( = -2\hat {i} - 4\hat {j} + 4\hat {k} \).
Next, we find the magnitude of \( \overrightarrow{\mathrm{AB}} \).
\( |\overrightarrow{\mathrm{AB}}| = \sqrt{(-2)^{2} + (-4)^{2} + 4^{2}} \)
\( = \sqrt{4 + 16 + 16} = \sqrt{36} = 6 \).
Finally, the direction cosines are given by \( \frac{x}{|\overrightarrow{\mathrm{AB}}|} \), \( \frac{y}{|\overrightarrow{\mathrm{AB}}|} \), \( \frac{z}{|\overrightarrow{\mathrm{AB}}|} \).
\( l = \frac{-2}{6} = -\frac{1}{3} \)
\( m = \frac{-4}{6} = -\frac{2}{3} \)
\( n = \frac{4}{6} = \frac{2}{3} \)
Hence, the direction cosines of \( \overrightarrow{\mathrm{AB}} \) are \( (-\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}) \).
In simple words: First, create the vector from point A to point B by subtracting coordinates. Then, find the length of this vector. To get the direction cosines, divide each number in the vector by its total length.

Exam Tip: Always make sure to subtract the coordinates of the initial point from the terminal point for vector formation. Simplify the fractions for the direction cosines to their lowest terms.

Question 14. Show that the vector \( \hat {i} + \hat {j} + \hat {k} \) is equally inclined to the axes OX, OY and OZ.
Answer: For a vector \( x\hat {i} + y\hat {j} + z\hat {k} \), the direction cosines are \( l = \frac{x}{\sqrt{x^2+y^2+z^2}} \), \( m = \frac{y}{\sqrt{x^2+y^2+z^2}} \), and \( n = \frac{z}{\sqrt{x^2+y^2+z^2}} \).
Let the given vector be \( \vec{a} = \hat {i} + \hat {j} + \hat {k} \).
Here, \( x = 1, y = 1, z = 1 \).
The magnitude of \( \vec{a} \) is \( |\vec{a}| = \sqrt{1^{2} + 1^{2} + 1^{2}} = \sqrt{1 + 1 + 1} = \sqrt{3} \).
The direction cosines of \( \vec{a} \) are:
\( l = \frac{1}{\sqrt{3}} \)
\( m = \frac{1}{\sqrt{3}} \)
\( n = \frac{1}{\sqrt{3}} \)
Since all direction cosines (l, m, n) are equal, the vector \( \vec{a} \) makes equal angles with the OX, OY, and OZ axes. This implies it is equally inclined to all three axes.
In simple words: A vector is equally tilted toward all the main axes (X, Y, and Z) if its direction cosines are all the same. Here, all three numbers are \( 1/\sqrt{3} \), which shows it's equally inclined.

Exam Tip: A vector is equally inclined to the axes if and only if its direction cosines are equal. This also means that the angles the vector makes with each axis are the same (e.g., \( \cos\alpha = \cos\beta = \cos\gamma \)).

Question 15. Find the position vector of a point R which divides the line segment joining the points, whose positive vectors are P \( (\hat {i} + 2\hat {j} – \hat {k}) \) and Q\( (-\hat {i} + \hat {j} + \hat {k} ) \), in the ratio 2 : 1
(i) internally,
(ii) externally.
Answer: Let the position vectors of points P and Q be \( \vec{a} = \hat {i} + 2\hat {j} – \hat {k} \) and \( \vec{b} = -\hat {i} + \hat {j} + \hat {k} \), respectively. The ratio is \( m:n = 2:1 \), so \( m = 2 \) and \( n = 1 \).
(i) For internal division, the position vector of R is given by the section formula:
\( \vec{R} = \frac{m\vec{b} + n\vec{a}}{m + n} \)
\( \vec{R} = \frac{2(-\hat {i} + \hat {j} + \hat {k}) + 1(\hat {i} + 2\hat {j} – \hat {k})}{2 + 1} \)
\( = \frac{(-2\hat {i} + 2\hat {j} + 2\hat {k}) + (\hat {i} + 2\hat {j} – \hat {k})}{3} \)
\( = \frac{(-2 + 1)\hat {i} + (2 + 2)\hat {j} + (2 – 1)\hat {k}}{3} \)
\( = \frac{-1\hat {i} + 4\hat {j} + 1\hat {k}}{3} \)
\( = -\frac{1}{3}\hat {i} + \frac{4}{3}\hat {j} + \frac{1}{3}\hat {k} \).
(ii) For external division, the position vector of R' is given by the formula:
\( \vec{R}' = \frac{m\vec{b} - n\vec{a}}{m - n} \)
\( \vec{R}' = \frac{2(-\hat {i} + \hat {j} + \hat {k}) - 1(\hat {i} + 2\hat {j} – \hat {k})}{2 - 1} \)
\( = \frac{(-2\hat {i} + 2\hat {j} + 2\hat {k}) - (\hat {i} + 2\hat {j} – \hat {k})}{1} \)
\( = (-2 - 1)\hat {i} + (2 - 2)\hat {j} + (2 - (-1))\hat {k} \)
\( = -3\hat {i} + 0\hat {j} + 3\hat {k} \)
\( = -3\hat {i} + 3\hat {k} \).
In simple words: To find a point that divides a line segment, use special formulas. For internal division, you add the vectors with weighted ratios. For external division, you subtract them. The key is to correctly apply the ratio (m:n) and the signs in the formula.

Exam Tip: Clearly distinguish between internal and external division formulas. A common mistake is using the internal division formula for external division, or vice versa. Remember, for internal division, the denominator is \( m+n \), and for external, it's \( m-n \).

Question 16. Find the mid-point of the vector joining the points P(2, 3, 4) and Q(4, 1, – 2).
Answer: Let the position vectors of points P and Q be \( \vec{a} = 2\hat {i} + 3\hat {j} + 4\hat {k} \) and \( \vec{b} = 4\hat {i} + \hat {j} – 2\hat {k} \), respectively.
The formula for the mid-point of a vector joining two points is \( \frac{\vec{a} + \vec{b}}{2} \).
Required mid-point \( = \frac{(2\hat {i} + 3\hat {j} + 4\hat {k}) + (4\hat {i} + \hat {j} – 2\hat {k})}{2} \)
\( = \frac{(2 + 4)\hat {i} + (3 + 1)\hat {j} + (4 - 2)\hat {k}}{2} \)
\( = \frac{6\hat {i} + 4\hat {j} + 2\hat {k}}{2} \)
\( = 3\hat {i} + 2\hat {j} + \hat {k} \).
In simple words: To find the middle point of a line segment between two points, simply add their position vectors together and then divide the entire result by 2. This gives you the average position.

Exam Tip: Finding the mid-point is a special case of internal division where the ratio is 1:1. It's essentially averaging the corresponding coordinates (or vector components) of the two end points.

Question 17. 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 \( \vec{c} = \hat {i} – 3\hat {j} – 5\hat {k} \), respectively form the vertices of a right angled triangle.
Answer: To show that points A, B, C form a right-angled triangle, we need to calculate the vectors representing the sides and then check if the square of the longest side's magnitude equals the sum of the squares of the other two sides' magnitudes (Pythagorean theorem).
First, find the vectors representing the sides of the triangle:
\( \overrightarrow{\mathrm{AB}} = \vec{b} - \vec{a} = (2\hat {i} – \hat {j} + \hat {k}) - (3\hat {i} – 4\hat {j} – 4\hat {k}) \)
\( = (2-3)\hat {i} + (-1-(-4))\hat {j} + (1-(-4))\hat {k} \)
\( = -\hat {i} + 3\hat {j} + 5\hat {k} \).
Magnitude squared: \( |\overrightarrow{\mathrm{AB}}|^2 = (-1)^2 + 3^2 + 5^2 = 1 + 9 + 25 = 35 \).

\( \overrightarrow{\mathrm{BC}} = \vec{c} - \vec{b} = (\hat {i} – 3\hat {j} – 5\hat {k}) - (2\hat {i} – \hat {j} + \hat {k}) \)
\( = (1-2)\hat {i} + (-3-(-1))\hat {j} + (-5-1)\hat {k} \)
\( = -\hat {i} - 2\hat {j} - 6\hat {k} \).
Magnitude squared: \( |\overrightarrow{\mathrm{BC}}|^2 = (-1)^2 + (-2)^2 + (-6)^2 = 1 + 4 + 36 = 41 \).

\( \overrightarrow{\mathrm{CA}} = \vec{a} - \vec{c} = (3\hat {i} – 4\hat {j} – 4\hat {k}) - (\hat {i} – 3\hat {j} – 5\hat {k}) \)
\( = (3-1)\hat {i} + (-4-(-3))\hat {j} + (-4-(-5))\hat {k} \)
\( = 2\hat {i} - \hat {j} + \hat {k} \).
Magnitude squared: \( |\overrightarrow{\mathrm{CA}}|^2 = 2^2 + (-1)^2 + 1^2 = 4 + 1 + 1 = 6 \).

Now, we check if the Pythagorean theorem holds true for these magnitudes:
We have \( |\overrightarrow{\mathrm{AB}}|^2 = 35 \), \( |\overrightarrow{\mathrm{BC}}|^2 = 41 \), and \( |\overrightarrow{\mathrm{CA}}|^2 = 6 \).
Notice that \( |\overrightarrow{\mathrm{AB}}|^2 + |\overrightarrow{\mathrm{CA}}|^2 = 35 + 6 = 41 \).
This is equal to \( |\overrightarrow{\mathrm{BC}}|^2 \).
Since \( |\overrightarrow{\mathrm{AB}}|^2 + |\overrightarrow{\mathrm{CA}}|^2 = |\overrightarrow{\mathrm{BC}}|^2 \), the triangle ABC is a right-angled triangle, with the right angle at point A (opposite to the side \( \overrightarrow{\mathrm{BC}} \)).
In simple words: To prove a right-angled triangle using vectors, first find the vectors for each side. Then, calculate the square of the length of each side. If the sum of the squares of the two shorter sides equals the square of the longest side, then it's a right-angled triangle. The right angle is at the vertex opposite the longest side.

Exam Tip: When using the Pythagorean theorem, ensure you correctly identify the longest side (hypotenuse) by comparing the magnitudes. The right angle will always be at the vertex opposite this longest side.

Question 18. In triangle ABC, which of the following is not true?
(A) \( \overrightarrow{\mathrm{AB}} + \overrightarrow{\mathrm{BC}} + \overrightarrow{\mathrm{CA}} = \vec{0} \)
(B) \( \overrightarrow{\mathrm{AB}} + \overrightarrow{\mathrm{BC}} – \overrightarrow{\mathrm{AC}} = \vec{0} \)
(C) \( \overrightarrow{\mathrm{AB}} + \overrightarrow{\mathrm{BC}} – \overrightarrow{\mathrm{CA}} = \vec{0} \)
(D) \( \overrightarrow{\mathrm{AB}} – \overrightarrow{\mathrm{CB}} + \overrightarrow{\mathrm{CA}} = \vec{0} \)
Answer: (C) \( \overrightarrow{\mathrm{AB}} + \overrightarrow{\mathrm{BC}} – \overrightarrow{\mathrm{CA}} = \vec{0} \)
In simple words: According to the triangle law of vector addition, if you follow the vectors around a triangle in order (head to tail), their sum should be zero. This means \( \overrightarrow{\mathrm{AB}} + \overrightarrow{\mathrm{BC}} = \overrightarrow{\mathrm{AC}} \). So, \( \overrightarrow{\mathrm{AB}} + \overrightarrow{\mathrm{BC}} + \overrightarrow{\mathrm{CA}} = \vec{0} \) is always true because \( \overrightarrow{\mathrm{CA}} \) is \( -\overrightarrow{\mathrm{AC}} \). Option (C) states \( \overrightarrow{\mathrm{AB}} + \overrightarrow{\mathrm{BC}} – \overrightarrow{\mathrm{CA}} = \vec{0} \), which becomes \( \overrightarrow{\mathrm{AC}} - (-\overrightarrow{\mathrm{AC}}) = \vec{0} \), or \( 2\overrightarrow{\mathrm{AC}} = \vec{0} \), which is only true if \( \overrightarrow{\mathrm{AC}} = \vec{0} \), which is not generally true for a triangle.

Exam Tip: The triangle law of vector addition states that if two vectors represent two sides of a triangle taken in order, their sum represents the third side taken in the opposite order. Thus, for vectors forming a closed loop (like a triangle), their sum taken in cyclic order is always the zero vector.

Question 19. If \( \vec{a} \) and \( \vec{b} \) are two collinear vectors, then which of the following is incorrect:
(A) \( \vec{b} = \lambda\vec{a} \), for some scalar
(B) \( \vec{a} = \pm \vec{b} \)
(C) the respective components of \( \vec{a} \) and \( \vec{b} \) are proportional.
(D) vectors \( \vec{a} \) and \( \vec{b} \) have same direction, but different magnitude.
Answer: (D) vectors \( \vec{a} \) and \( \vec{b} \) have same direction, but different magnitude.
In simple words: Collinear vectors lie on the same line. This means one is a scalar multiple of the other (true for A). Their components are proportional (true for C). They can point in the same direction or opposite directions. Option (D) incorrectly says they *must* have the same direction; they could point oppositely if the scalar \( \lambda \) is negative.

Exam Tip: Collinear vectors are defined by the condition \( \vec{b} = \lambda\vec{a} \). If \( \lambda > 0 \), they have the same direction. If \( \lambda < 0 \), they have opposite directions. If \( \lambda = 0 \), one vector is the zero vector. Therefore, assuming they *must* have the same direction is incorrect.