CBSE Class 12 Mathematics Vector Algebra Assignment Set 06

Very Short Answer Type Questions (VSA)

Question. If a unit vector \( \vec{a} \) makes angles \( \frac{\pi}{3} \) with \( \hat{i} \), \( \frac{\pi}{4} \) with \( \hat{j} \) and an acute angle \( \theta \) with \( \hat{k} \), then find the value of \( \theta \).
Answer: We have, \( l = \cos \frac{\pi}{3} = \frac{1}{2} \), \( m = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \) and \( n = \cos \theta \).
Now, \( l^2 + m^2 + n^2 = 1 \)
\( \Rightarrow \left(\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + n^2 = 1 \)
\( \Rightarrow \frac{1}{4} + \frac{1}{2} + n^2 = 1 \Rightarrow n^2 = \frac{1}{4} \Rightarrow n = \pm \frac{1}{2} \)
\( \Rightarrow \cos\theta = \pm \frac{1}{2} \)
But \( \theta \) is an acute angle (given).
\( \therefore \theta = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \).

Question. Find the sum of the following vectors: \( \vec{a} = \hat{i} - 3\hat{k} \), \( \vec{b} = 2\hat{j} - \hat{k} \), \( \vec{c} = 2\hat{i} - 3\hat{j} + 2\hat{k} \).
Answer: Required sum \( = \vec{a} + \vec{b} + \vec{c} \)
\( = (\hat{i} - 3\hat{k}) + (2\hat{j} - \hat{k}) + (2\hat{i} - 3\hat{j} + 2\hat{k}) \)
\( = 3\hat{i} - \hat{j} - 2\hat{k} \).

Question. Find a unit vector in the direction of the sum of the vectors \( \vec{a} = 2\hat{i} + 2\hat{j} - 5\hat{k} \) and \( \vec{b} = 4\hat{i} - 3\hat{j} + 2\hat{k} \).
Answer: Let \( \vec{c} \) be the sum of vectors \( \vec{a} \) and \( \vec{b} \).
\( \vec{c} = \vec{a} + \vec{b} = (2\hat{i} + 2\hat{j} - 5\hat{k}) + (4\hat{i} - 3\hat{j} + 2\hat{k}) = 6\hat{i} - \hat{j} - 3\hat{k} \)
Magnitude of \( \vec{c} \):
\( |\vec{c}| = \sqrt{6^2 + (-1)^2 + (-3)^2} = \sqrt{36 + 1 + 9} = \sqrt{46} \)
The unit vector in the direction of \( \vec{c} \) is:
\( \hat{c} = \frac{\vec{c}}{|\vec{c}|} = \frac{6\hat{i} - \hat{j} - 3\hat{k}}{\sqrt{46}} \).

Question. \( L \) and \( M \) are two points with position vectors \( 2\vec{a} - \vec{b} \) and \( \vec{a} + 2\vec{b} \) respectively. What is the position vector of a point \( N \) which divides the line segment \( LM \) in the ratio \( 2 : 1 \) externally?
Answer: The position vector of a point \( N \) which divides the line segment \( LM \) joining points with position vectors \( \vec{OL} = 2\vec{a} - \vec{b} \) and \( \vec{OM} = \vec{a} + 2\vec{b} \) externally in the ratio \( 2 : 1 \) is given by:
\( \vec{ON} = \frac{2(\vec{a} + 2\vec{b}) - 1(2\vec{a} - \vec{b})}{2 - 1} \)
\( = \frac{2\vec{a} + 4\vec{b} - 2\vec{a} + \vec{b}}{1} \)
\( = 5\vec{b} \).

Question. Find the value of \( |\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 \) if \( |\vec{a}| = 5 \) and \( |\vec{b}| = 4 \).
Answer: We know by Lagrange's identity:
\( |\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \)
Given \( |\vec{a}| = 5 \) and \( |\vec{b}| = 4 \).
Substituting these values, we get:
\( |\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = (5)^2 (4)^2 = 25 \times 16 = 400 \).

Question. Find the area of a parallelogram whose adjacent sides are represented by the vectors \( \hat{i} - 3\hat{k} \) and \( 2\hat{j} + \hat{k} \).
Answer: Let the adjacent sides be represented by vectors \( \vec{a} = \hat{i} - 3\hat{k} \) and \( \vec{b} = 2\hat{j} + \hat{k} \).
The cross product \( \vec{a} \times \vec{b} \) is:
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & -3 \\ 0 & 2 & 1 \end{vmatrix} = \hat{i}(0 - (-6)) - \hat{j}(1 - 0) + \hat{k}(2 - 0) = 6\hat{i} - \hat{j} + 2\hat{k} \)
Area of the parallelogram \( = |\vec{a} \times \vec{b}| = \sqrt{6^2 + (-1)^2 + 2^2} = \sqrt{36 + 1 + 4} = \sqrt{41} \) sq. units.

Question. Find the projection of the vector \( \vec{a} = 2\hat{i} + 3\hat{j} + 2\hat{k} \) on the vector \( \vec{b} = 2\hat{i} + 2\hat{j} + \hat{k} \).
Answer: Projection of the vector \( \vec{a} \) on vector \( \vec{b} \) is given by:
\( \text{Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{(2\hat{i} + 3\hat{j} + 2\hat{k}) \cdot (2\hat{i} + 2\hat{j} + \hat{k})}{\sqrt{2^2 + 2^2 + 1^2}} \)
\( = \frac{4 + 6 + 2}{\sqrt{4 + 4 + 1}} = \frac{12}{3} = 4 \).

Question. If \( \vec{a} \) and \( \vec{b} \) are unit vectors, then find the angle between \( \vec{a} \) and \( \vec{b} \), given that \( (\sqrt{2}\vec{a} - \vec{b}) \) is a unit vector.
Answer: Let \( \theta \) be the angle between the unit vectors \( \vec{a} \) and \( \vec{b} \).
\( \therefore \cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \vec{a} \cdot \vec{b} \) (since \( |\vec{a}| = 1 = |\vec{b}| \)) ... (1)
Now, we are given that \( |\sqrt{2}\vec{a} - \vec{b}| = 1 \)
\( \Rightarrow |\sqrt{2}\vec{a} - \vec{b}|^2 = 1 \)
\( \Rightarrow (\sqrt{2}\vec{a} - \vec{b}) \cdot (\sqrt{2}\vec{a} - \vec{b}) = 1 \)
\( \Rightarrow 2|\vec{a}|^2 - \sqrt{2}\vec{a} \cdot \vec{b} - \vec{b} \cdot \sqrt{2}\vec{a} + |\vec{b}|^2 = 1 \)
\( \Rightarrow 2(1)^2 - 2\sqrt{2}(\vec{a} \cdot \vec{b}) + (1)^2 = 1 \) (since \( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \))
\( \Rightarrow 3 - 2\sqrt{2}\vec{a} \cdot \vec{b} = 1 \)
\( \Rightarrow 2\sqrt{2}\vec{a} \cdot \vec{b} = 2 \)
\( \Rightarrow \vec{a} \cdot \vec{b} = \frac{1}{\sqrt{2}} \)
\( \Rightarrow \cos\theta = \frac{1}{\sqrt{2}} \) [By using (1)]
\( \therefore \theta = \pi/4 \).

Question. Find the angle between x-axis and the vector \( \hat{i} + \hat{j} + \hat{k} \).
Answer: Here, \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \) and vector along x-axis is \( \hat{i} \).
\( \therefore \) Angle \( \theta \) between \( \vec{a} \) and \( \hat{i} \) is given by:
\( \cos \theta = \frac{\vec{a} \cdot \hat{i}}{|\vec{a}| |\hat{i}|} = \frac{(\hat{i} + \hat{j} + \hat{k}) \cdot \hat{i}}{\sqrt{1^2 + 1^2 + 1^2} \cdot \sqrt{1^2}} = \frac{1}{\sqrt{3} \cdot 1} = \frac{1}{\sqrt{3}} \)
\( \Rightarrow \theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \).

Question. If \( \vec{a} \) and \( \vec{b} \) are two vectors such that \( |\vec{a} + \vec{b}| = |\vec{a}| \), then prove that vector \( 2\vec{a} + \vec{b} \) is perpendicular to vector \( \vec{b} \).
Answer: Here, \( |\vec{a} + \vec{b}| = |\vec{a}| \)
Squaring both sides:
\( \Rightarrow |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 \)
\( \Rightarrow (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = \vec{a} \cdot \vec{a} \)
\( \Rightarrow \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} = \vec{a} \cdot \vec{a} \)
\( \Rightarrow 2\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} = 0 \) [since \( \vec{b} \cdot \vec{a} = \vec{a} \cdot \vec{b} \)]
\( \Rightarrow (2\vec{a} + \vec{b}) \cdot \vec{b} = 0 \)
\( \Rightarrow (2\vec{a} + \vec{b}) \perp \vec{b} \).

Short Answer Type Questions (SA-I)

Question. \( X \) and \( Y \) are two points with position vectors \( 3\vec{a} + \vec{b} \) and \( \vec{a} - 3\vec{b} \) respectively. Write the position vector of a point \( Z \) which divides the line segment \( XY \) in the ratio \( 2 : 1 \) externally.
Answer: Position vector which divides the line segment joining points with position vectors \( 3\vec{a} + \vec{b} \) and \( \vec{a} - 3\vec{b} \) in the ratio \( 2 : 1 \) externally is given by:
\( \vec{OZ} = \frac{2(\vec{a} - 3\vec{b}) - 1(3\vec{a} + \vec{b})}{2 - 1} \)
\( = \frac{2\vec{a} - 6\vec{b} - 3\vec{a} - \vec{b}}{1} \)
\( = -\vec{a} - 7\vec{b} \).

Question. Find a unit vector perpendicular to each of the vectors \( \vec{a} \) and \( \vec{b} \) where \( \vec{a} = 5\hat{i} + 6\hat{j} - 2\hat{k} \) and \( \vec{b} = 7\hat{i} + 6\hat{j} + 2\hat{k} \).
Answer: Here, \( \vec{a} = 5\hat{i} + 6\hat{j} - 2\hat{k} \) and \( \vec{b} = 7\hat{i} + 6\hat{j} + 2\hat{k} \).
Vector perpendicular to both \( \vec{a} \) and \( \vec{b} \) is:
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 6 & -2 \\ 7 & 6 & 2 \end{vmatrix} \)
\( = \hat{i}(12 - (-12)) - \hat{j}(10 - (-14)) + \hat{k}(30 - 42) \)
\( = 24\hat{i} - 24\hat{j} - 12\hat{k} = 12(2\hat{i} - 2\hat{j} - \hat{k}) \)
\( \therefore \) Unit vector perpendicular to both \( \vec{a} \) and \( \vec{b} \):
\( \hat{n} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \frac{12(2\hat{i} - 2\hat{j} - \hat{k})}{\sqrt{24^2 + (-24)^2 + (-12)^2}} = \frac{12(2\hat{i} - 2\hat{j} - \hat{k})}{\sqrt{576 + 576 + 144}} \)
\( = \frac{12(2\hat{i} - 2\hat{j} - \hat{k})}{\sqrt{1296}} = \frac{12(2\hat{i} - 2\hat{j} - \hat{k})}{36} = \frac{1}{3}(2\hat{i} - 2\hat{j} - \hat{k}) \).

Question. Show that for any two non-zero vectors \( \vec{a} \) and \( \vec{b} \), \( |\vec{a} + \vec{b}| = |\vec{a} - \vec{b}| \) iff \( \vec{a} \) and \( \vec{b} \) are perpendicular vectors.
Answer: For any two non-zero vectors \( \vec{a} \) and \( \vec{b} \), we have:
\( |\vec{a} + \vec{b}| = |\vec{a} - \vec{b}| \Leftrightarrow |\vec{a} + \vec{b}|^2 = |\vec{a} - \vec{b}|^2 \)
\( \Leftrightarrow a^2 + b^2 + 2\vec{a} \cdot \vec{b} = a^2 + b^2 - 2\vec{a} \cdot \vec{b} \)
\( \Leftrightarrow 4\vec{a} \cdot \vec{b} = 0 \Leftrightarrow \vec{a} \cdot \vec{b} = 0 \)
So, \( \vec{a} \) and \( \vec{b} \) are perpendicular vectors.

Question. Show that the vectors \( 2\hat{i} - \hat{j} + \hat{k} \), \( 3\hat{i} + 7\hat{j} + \hat{k} \) and \( 5\hat{i} + 6\hat{j} + 2\hat{k} \) form the sides of a right-angled triangle.
Answer: Let \( A(2\hat{i} - \hat{j} + \hat{k}) \), \( B(3\hat{i} + 7\hat{j} + \hat{k}) \) and \( C(5\hat{i} + 6\hat{j} + 2\hat{k}) \) be the position vectors of the vertices of the triangle.
Then, \( \vec{AB} = (3-2)\hat{i} + (7+1)\hat{j} + (1-1)\hat{k} = \hat{i} + 8\hat{j} \)
\( \vec{AC} = (5-2)\hat{i} + (6+1)\hat{j} + (2-1)\hat{k} = 3\hat{i} + 7\hat{j} + \hat{k} \)
\( \vec{BC} = (5-3)\hat{i} + (6-7)\hat{j} + (2-1)\hat{k} = 2\hat{i} - \hat{j} + \hat{k} \)
Now, the angle between \( \vec{AC} \) and \( \vec{BC} \) is given by:
\( \cos \theta = \frac{\vec{AC} \cdot \vec{BC}}{|\vec{AC}| |\vec{BC}|} = \frac{6 - 7 + 1}{\sqrt{9 + 49 + 1} \sqrt{4 + 1 + 1}} = \frac{0}{\sqrt{59} \sqrt{6}} = 0 \)
\( \Rightarrow \cos \theta = 0 \Rightarrow \vec{AC} \perp \vec{BC} \).
So, \( A, B, C \) are the vertices of a right-angled triangle.

Question. If two vectors \( \vec{a} \) and \( \vec{b} \) are such that \( |\vec{a}| = 2 \), \( |\vec{b}| = 1 \) and \( \vec{a} \cdot \vec{b} = 1 \), then find the value of \( (3\vec{a} - 5\vec{b}) \cdot (2\vec{a} + 7\vec{b}) \).
Answer: We have \( |\vec{a}| = 2 \), \( |\vec{b}| = 1 \) and \( \vec{a} \cdot \vec{b} = 1 \).
Now, \( (3\vec{a} - 5\vec{b}) \cdot (2\vec{a} + 7\vec{b}) \)
\( = 6|\vec{a}|^2 + 21\vec{a} \cdot \vec{b} - 10\vec{b} \cdot \vec{a} - 35|\vec{b}|^2 \)
\( = 6|\vec{a}|^2 + 11\vec{a} \cdot \vec{b} - 35|\vec{b}|^2 \)
\( = 6(2)^2 + 11(1) - 35(1)^2 \)
\( = 24 + 11 - 35 = 0 \).

Question. Let \( \vec{a} = \hat{i} + 2\hat{j} - 3\hat{k} \) and \( \vec{b} = 3\hat{i} - \hat{j} + 2\hat{k} \) be two vectors. Show that the vectors \( (\vec{a} + \vec{b}) \) and \( (\vec{a} - \vec{b}) \) are perpendicular to each other.
Answer: Given, \( \vec{a} = \hat{i} + 2\hat{j} - 3\hat{k} \) and \( \vec{b} = 3\hat{i} - \hat{j} + 2\hat{k} \).
Now, \( \vec{a} + \vec{b} = 4\hat{i} + \hat{j} - \hat{k} \)
Also, \( \vec{a} - \vec{b} = -2\hat{i} + 3\hat{j} - 5\hat{k} \)
Now, \( (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = (4\hat{i} + \hat{j} - \hat{k}) \cdot (-2\hat{i} + 3\hat{j} - 5\hat{k}) \)
\( = (4)(-2) + (1)(3) + (-1)(-5) = -8 + 3 + 5 = 0 \).
Hence, \( (\vec{a} + \vec{b}) \) and \( (\vec{a} - \vec{b}) \) are perpendicular to each other.

Question. If \( \theta \) is the angle between two vectors \( \hat{i} - 2\hat{j} + 3\hat{k} \) and \( 3\hat{i} - 2\hat{j} + \hat{k} \), then find \( \sin\theta \).
Answer: Let \( \vec{a} = \hat{i} - 2\hat{j} + 3\hat{k} \), \( \vec{b} = 3\hat{i} - \hat{j} + \hat{k} \).
Now, \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta \)
\( \Rightarrow (\hat{i} - 2\hat{j} + 3\hat{k}) \cdot (3\hat{i} - 2\hat{j} + \hat{k}) = \sqrt{1^2 + (-2)^2 + 3^2} \times \sqrt{3^2 + (-2)^2 + 1^2} \cos\theta \)
\( \Rightarrow 3 + 4 + 3 = \sqrt{14} \times \sqrt{14} \cos\theta \)
\( \Rightarrow 10 = 14 \cos\theta \Rightarrow \cos\theta = \frac{10}{14} = \frac{5}{7} \)
\( \therefore \sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - \frac{25}{49}} = \sqrt{\frac{24}{49}} = \frac{2\sqrt{6}}{7} \).

Question. Find \( |2\vec{a} \cdot (-\vec{b} \times 3\vec{c})| \), where \( \vec{a} = \hat{i} - \hat{j} + 2\hat{k} \), \( \vec{b} = 3\hat{i} + 4\hat{j} - 5\hat{k} \) and \( \vec{c} = 2\hat{i} - \hat{j} + 3\hat{k} \).
Answer: Given, \( \vec{a} = \hat{i} - \hat{j} + 2\hat{k} \), \( \vec{b} = 3\hat{i} + 4\hat{j} - 5\hat{k} \) and \( \vec{c} = 2\hat{i} - \hat{j} + 3\hat{k} \).
\( \therefore 2\vec{a} = 2\hat{i} - 2\hat{j} + 4\hat{k} \)
\( -\vec{b} = -3\hat{i} - 4\hat{j} + 5\hat{k} \)
\( 3\vec{c} = 6\hat{i} - 3\hat{j} + 9\hat{k} \)
Now, \( 2\vec{a} \cdot (-\vec{b} \times 3\vec{c}) = \begin{vmatrix} 2 & -2 & 4 \\ -3 & -4 & 5 \\ 6 & -3 & 9 \end{vmatrix} \)
\( = 2(-36 + 15) + 2(-27 - 30) + 4(9 + 24) \)
\( = 2(-21) - 2(57) + 4(33) \)
\( = -42 - 114 + 132 = -24 \)
\( \therefore |2\vec{a} \cdot (-\vec{b} \times 3\vec{c})| = |-24| = 24 \).

Question. If \( \vec{a} = 2\hat{i} + 3\hat{j} + \hat{k} \), \( \vec{b} = \hat{i} - 2\hat{j} + \hat{k} \) and \( \vec{c} = -3\hat{i} + \hat{j} + 2\hat{k} \), then find \( \vec{a} \cdot (\vec{b} \times \vec{c}) \).
Answer: Given, \( \vec{a} = 2\hat{i} + 3\hat{j} + \hat{k} \), \( \vec{b} = \hat{i} - 2\hat{j} + \hat{k} \) and \( \vec{c} = -3\hat{i} + \hat{j} + 2\hat{k} \).
Now, \( \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ -3 & 1 & 2 \end{vmatrix} \)
\( = \hat{i}(-4 - 1) - \hat{j}(2 + 3) + \hat{k}(1 - 6) = -5\hat{i} - 5\hat{j} - 5\hat{k} \)
\( \dots \vec{a} \cdot (\vec{b} \times \vec{c}) = (2\hat{i} + 3\hat{j} + \hat{k}) \cdot (-5\hat{i} - 5\hat{j} - 5\hat{k}) \)
\( = -10 - 15 - 5 = -30 \).

Question. If \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \), find \( (\vec{r} \times \hat{i}) \cdot (\vec{r} \times \hat{j}) + xy \).
Answer: \( (\vec{r} \times \hat{i}) \cdot (\vec{r} \times \hat{j}) + xy \)
\( = [(x\hat{i} + y\hat{j} + z\hat{k}) \times \hat{i}] \cdot [(x\hat{i} + y\hat{j} + z\hat{k}) \times \hat{j}] + xy \)
\( = (-y\hat{k} + z\hat{j}) \cdot (x\hat{k} - z\hat{i}) + xy \)
\( = -xy + xy = 0 \).

Short Answer Type Questions (SA-II)

Question. The two vectors \( \hat{j} + \hat{k} \) and \( 3\hat{i} - \hat{j} + 4\hat{k} \) represent the two sides \( AB \) and \( AC \), respectively of a \( \Delta ABC \). Find the length of the median through \( A \).
Answer: Take \( A \) to be as origin \( (0, 0, 0) \).
\( \therefore \) Coordinates of \( B \) are \( (0, 1, 1) \) and coordinates of \( C \) are \( (3, -1, 4) \).
Let \( D \) be the mid point of \( BC \) and \( AD \) is a median of \( \Delta ABC \).
\( \therefore \) Coordinates of \( D \) are \( \left( \frac{3}{2}, 0, \frac{5}{2} \right) \).
So, length of \( AD = \sqrt{\left(\frac{3}{2} - 0\right)^2 + (0 - 0)^2 + \left(\frac{5}{2} - 0\right)^2} = \sqrt{\frac{9}{4} + \frac{25}{4}} = \frac{\sqrt{34}}{2} \) units.

Question. Find a vector of magnitude \( 5 \) units and parallel to the resultant of the vectors \( \vec{a} = 2\hat{i} + 3\hat{j} - \hat{k} \) and \( \vec{b} = \hat{i} - 2\hat{j} + \hat{k} \).
Answer: \( \vec{a} = 2\hat{i} + 3\hat{j} - \hat{k} \), \( \vec{b} = \hat{i} - 2\hat{j} + \hat{k} \).
\( \therefore \vec{a} + \vec{b} = (2\hat{i} + 3\hat{j} - \hat{k}) + (\hat{i} - 2\hat{j} + \hat{k}) = 3\hat{i} + \hat{j} \)
\( |\vec{a} + \vec{b}| = \sqrt{3^2 + 1^2} = \sqrt{10} \)
\( \therefore \) A vector of magnitude \( 5 \) units in the direction of \( \vec{a} + \vec{b} \) is:
\( 5 \frac{\vec{a} + \vec{b}}{|\vec{a} + \vec{b}|} = \frac{5(3\hat{i} + \hat{j})}{\sqrt{10}} \).

Question. Let \( \vec{a} = 4\hat{i} + 5\hat{j} - \hat{k} \), \( \vec{b} = \hat{i} - 4\hat{j} + 5\hat{k} \) and \( \vec{c} = 3\hat{i} + \hat{j} - \hat{k} \). Find a vector \( \vec{d} \) which is perpendicular to both \( \vec{c} \) and \( \vec{b} \) and \( \vec{d} \cdot \vec{a} = 21 \).
Answer: Let \( \vec{d} = x\hat{i} + y\hat{j} + z\hat{k} \).
Now, it is given that, \( \vec{d} \) is perpendicular to \( \vec{b} = \hat{i} - 4\hat{j} + 5\hat{k} \) and \( \vec{c} = 3\hat{i} + \hat{j} - \hat{k} \).
\( \therefore \vec{d} \cdot \vec{b} = 0 \text{ and } \vec{d} \cdot \vec{c} = 0 \)
\( \Rightarrow x - 4y + 5z = 0 \) ... (i)
and \( 3x + y - z = 0 \) ... (ii)
Also, \( \vec{d} \cdot \vec{a} = 21 \), where \( \vec{a} = 4\hat{i} + 5\hat{j} - \hat{k} \)
\( \Rightarrow 4x + 5y - z = 21 \) ... (iii)
Eliminating \( z \) from (i) and (ii), we get:
\( 16x + y = 0 \) ... (iv)
Eliminating \( z \) from (ii) and (iii), we get:
\( x + 4y = 21 \) ... (v)
Solving (iv) and (v), we get:
\( x = -\frac{1}{3}, y = \frac{16}{3} \)
Putting the values of \( x \) and \( y \) in (i), we get \( z = \frac{13}{3} \).
\( \therefore \vec{d} = -\frac{1}{3}\hat{i} + \frac{16}{3}\hat{j} + \frac{13}{3}\hat{k} \) is the required vector.

Question. The scalar product of the vector \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \) with a unit vector along the sum of vectors \( \vec{b} = 2\hat{i} + 4\hat{j} - 5\hat{k} \) and \( \vec{c} = \lambda\hat{i} + 2\hat{j} + 3\hat{k} \) is equal to one. Find the value of \( \lambda \) and hence find the unit vector along \( \vec{b} + \vec{c} \).
Answer: Here, \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \); \( \vec{b} = 2\hat{i} + 4\hat{j} - 5\hat{k} \) and \( \vec{c} = \lambda\hat{i} + 2\hat{j} + 3\hat{k} \).
\( \Rightarrow \vec{b} + \vec{c} = (2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k} \)
The unit vector along \( \vec{b} + \vec{c} \) is \( \hat{p} = \frac{\vec{b} + \vec{c}}{|\vec{b} + \vec{c}|} = \frac{(2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{(2 + \lambda)^2 + 6^2 + (-2)^2}} = \frac{(2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{\lambda^2 + 4\lambda + 44}} \)
Also, \( \vec{a} \cdot \hat{p} = 1 \) (Given)
\( \Rightarrow \frac{(2 + \lambda) + 6 - 2}{\sqrt{\lambda^2 + 4\lambda + 44}} = 1 \)
\( \Rightarrow \sqrt{\lambda^2 + 4\lambda + 44} = \lambda + 6 \)
\( \Rightarrow \lambda^2 + 4\lambda + 44 = \lambda^2 + 12\lambda + 36 \)
\( \Rightarrow 8\lambda = 8 \Rightarrow \lambda = 1 \).
\( \therefore \) The required unit vector is:
\( \hat{p} = \frac{(2 + 1)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{1 + 4 + 44}} = \frac{1}{7}(3\hat{i} + 6\hat{j} - 2\hat{k}) \).

Question. 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: Two non-zero vectors are parallel if and only if their cross product is the zero vector.
So, we have to prove that the cross product of \( \vec{a} - \vec{d} \) and \( \vec{b} - \vec{c} \) is the zero vector.
Now, \( (\vec{a} - \vec{d}) \times (\vec{b} - \vec{c}) = \vec{a} \times \vec{b} - \vec{a} \times \vec{c} - \vec{d} \times \vec{b} + \vec{d} \times \vec{c} \)
Since, it is given that \( \vec{a} \times \vec{b} = \vec{c} \times \vec{d} \) and \( \vec{a} \times \vec{c} = \vec{b} \times \vec{d} \),
And, \( -\vec{d} \times \vec{b} = \vec{b} \times \vec{d} \), and \( \vec{d} \times \vec{c} = -\vec{c} \times \vec{d} \).
\( \therefore (\vec{a} - \vec{d}) \times (\vec{b} - \vec{c}) = \vec{c} \times \vec{d} - \vec{b} \times \vec{d} + \vec{b} \times \vec{d} - \vec{c} \times \vec{d} = \vec{0} \).
Hence, \( \vec{a} - \vec{d} \) is parallel to \( \vec{b} - \vec{c} \), where \( \vec{a} \neq \vec{d} \) and \( \vec{b} \neq \vec{c} \).

Question. Dot product of a vector with vectors \( \hat{i} - \hat{j} + \hat{k} \), \( 2\hat{i} + \hat{j} - 3\hat{k} \) and \( \hat{i} + \hat{j} + \hat{k} \) are respectively \( 4 \), \( 0 \) and \( 2 \). Find the vector.
Answer: Let the required vector be \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).
Also let, \( \vec{a} = \hat{i} - \hat{j} + \hat{k} \), \( \vec{b} = 2\hat{i} + \hat{j} - 3\hat{k} \) and \( \vec{c} = \hat{i} + \hat{j} + \hat{k} \).
Given, \( \vec{r} \cdot \vec{a} = 4 \), \( \vec{r} \cdot \vec{b} = 0 \), and \( \vec{r} \cdot \vec{c} = 2 \).
\( \Rightarrow x - y + z = 4 \) ... (i)
\( 2x + y - 3z = 0 \) ... (ii)
\( x + y + z = 2 \) ... (iii)
Now, (iii) - (i) \( \Rightarrow 2y = -2 \Rightarrow y = -1 \).
From (ii) and (iii):
\( 2x - 3z - 1 = 0 \), and \( x + z - 3 = 0 \Rightarrow x = 2, z = 1 \).
\( \therefore \) The required vector is \( \vec{r} = 2\hat{i} - \hat{j} + \hat{k} \).

Question. Find the values of \( \lambda \) for which the angle between the vectors \( \vec{a} = 2\lambda^2\hat{i} + 4\lambda\hat{j} + \hat{k} \) and \( \vec{b} = 7\hat{i} - 2\hat{j} + \lambda\hat{k} \) is obtuse.
Answer: Here, \( \vec{a} = 2\lambda^2\hat{i} + 4\lambda\hat{j} + \hat{k} \) and \( \vec{b} = 7\hat{i} - 2\hat{j} + \lambda\hat{k} \).
If \( \theta \) is the angle between the vectors \( \vec{a} \) and \( \vec{b} \), then:
\( \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \)
For \( \theta \) to be obtuse, \( \cos \theta < 0 \Rightarrow \vec{a} \cdot \vec{b} < 0 \)
\( \Rightarrow (2\lambda^2\hat{i} + 4\lambda\hat{j} + \hat{k}) \cdot (7\hat{i} - 2\hat{j} + \lambda\hat{k}) < 0 \)
\( \Rightarrow 2\lambda^2(7) + 4\lambda(-2) + 1 \cdot \lambda < 0 \)
\( \Rightarrow 14\lambda^2 - 7\lambda < 0 \Rightarrow 7\lambda(2\lambda - 1) < 0 \)
\( \Rightarrow \text{Either } \lambda < 0, 2\lambda - 1 > 0 \text{ or } \lambda > 0, 2\lambda - 1 < 0 \)
\( \Rightarrow \text{Either } \lambda < 0, \lambda > \frac{1}{2} \text{ or } \lambda > 0, \lambda < \frac{1}{2} \)
The first alternative is impossible.
\( \therefore \lambda > 0, \lambda < \frac{1}{2} \text{ i.e., } 0 < \lambda < \frac{1}{2} \text{ i.e., } \lambda \in \left(0, \frac{1}{2}\right) \).

Question. Using vectors, find the area of the triangle with vertices \( A(1, 1, 2) \), \( B(2, 3, 5) \) and \( C(1, 5, 5) \).
Answer: Given, \( \Delta ABC \) with vertices \( A(1, 1, 2) \), \( B(2, 3, 5) \) and \( C(1, 5, 5) \).
Now, \( \vec{AB} = (2-1)\hat{i} + (3-1)\hat{j} + (5-2)\hat{k} = \hat{i} + 2\hat{j} + 3\hat{k} \),
and \( \vec{AC} = (1-1)\hat{i} + (5-1)\hat{j} + (5-2)\hat{k} = 4\hat{j} + 3\hat{k} \).
\( \therefore \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 4 & 3 \end{vmatrix} = -6\hat{i} - 3\hat{j} + 4\hat{k} \).
Hence, area of \( \Delta ABC = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} \sqrt{(-6)^2 + (-3)^2 + 4^2} = \frac{1}{2} \sqrt{61} \) sq. units.

Question. If \( \vec{a} \), \( \vec{b} \) and \( \vec{c} \) are three vectors such that \( |\vec{a}| = 3 \), \( |\vec{b}| = 4 \), and \( |\vec{c}| = 5 \) and each one of them is perpendicular to the sum of the other two, then find \( |\vec{a} + \vec{b} + \vec{c}| \).
Answer: Given, \( |\vec{a}| = 3 \), \( |\vec{b}| = 4 \), \( |\vec{c}| = 5 \) ... (i)
and \( \vec{a} \cdot (\vec{b} + \vec{c}) = 0 \), \( \vec{b} \cdot (\vec{c} + \vec{a}) = 0 \), \( \vec{c} \cdot (\vec{a} + \vec{b}) = 0 \)
\( \Rightarrow \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c} + \vec{b} \cdot \vec{a} + \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} = 0 \)
\( \Rightarrow 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \) ... (ii)
Now, \( |\vec{a} + \vec{b} + \vec{c}|^2 = (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) \)
\( = \vec{a} \cdot (\vec{a} + \vec{b} + \vec{c}) + \vec{b} \cdot (\vec{a} + \vec{b} + \vec{c}) + \vec{c} \cdot (\vec{a} + \vec{b} + \vec{c}) \)
\( = \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} + \vec{c} \cdot \vec{c} \)
\( = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \)
\( = 3^2 + 4^2 + 5^2 + 0 \) [Using (i) and (ii)]
\( = 50 \).
\( \dots |\vec{a} + \vec{b} + \vec{c}| = \sqrt{50} = 5\sqrt{2} \).

Question. If \( \vec{a} = \hat{i} + 2\hat{j} + 3\hat{k} \) and \( \vec{b} = 2\hat{i} + 4\hat{j} - 5\hat{k} \) represent two adjacent sides of a parallelogram, find unit vectors parallel to the diagonals of the parallelogram.
Answer: Let \( \vec{a} = \hat{i} + 2\hat{j} + 3\hat{k} \) and \( \vec{b} = 2\hat{i} + 4\hat{j} - 5\hat{k} \).
Then diagonal \( \vec{AC} \) of the parallelogram is \( \vec{p} = \vec{a} + \vec{b} = (\hat{i} + 2\hat{j} + 3\hat{k}) + (2\hat{i} + 4\hat{j} - 5\hat{k}) = 3\hat{i} + 6\hat{j} - 2\hat{k} \).
Therefore, unit vector parallel to it is:
\( \hat{p} = \frac{3\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{9 + 36 + 4}} = \frac{1}{7}(3\hat{i} + 6\hat{j} - 2\hat{k}) \).
Now, diagonal \( \vec{BD} \) of the parallelogram is \( \vec{p'} = \vec{b} - \vec{a} = (2\hat{i} + 4\hat{j} - 5\hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) = \hat{i} + 2\hat{j} - 8\hat{k} \).
Therefore, unit vector parallel to it is:
\( \hat{p'} = \frac{\hat{i} + 2\hat{j} - 8\hat{k}}{\sqrt{1 + 4 + 64}} = \frac{1}{\sqrt{69}}(\hat{i} + 2\hat{j} - 8\hat{k}) \).

Question. Using vectors, find the area of the triangle \( ABC \) with vertices \( A(1, 2, 3) \), \( B(2, -1, 4) \) and \( C(4, 5, -1) \).
Answer: Given, \( \Delta ABC \) with vertices \( A(1, 2, 3) \equiv \hat{i} + 2\hat{j} + 3\hat{k} \), \( B(2, -1, 4) \equiv 2\hat{i} - \hat{j} + 4\hat{k} \), and \( C(4, 5, -1) \equiv 4\hat{i} + 5\hat{j} - \hat{k} \).
Now, \( \vec{AB} = \vec{OB} - \vec{OA} = (2\hat{i} - \hat{j} + 4\hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) = \hat{i} - 3\hat{j} + \hat{k} \).
\( \vec{AC} = \vec{OC} - \vec{OA} = (4\hat{i} + 5\hat{j} - \hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) = 3\hat{i} + 3\hat{j} - 4\hat{k} \).
\( \therefore \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 1 \\ 3 & 3 & -4 \end{vmatrix} = 9\hat{i} + 7\hat{j} + 12\hat{k} \).
Hence, area of \( \Delta ABC = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} |9\hat{i} + 7\hat{j} + 12\hat{k}| = \frac{1}{2} \sqrt{9^2 + 7^2 + 12^2} = \frac{1}{2} \sqrt{81 + 49 + 144} = \frac{1}{2} \sqrt{274} \) sq. units.

Question. If \( \vec{a} = \hat{i} + 2\hat{j} + \hat{k} \), \( \vec{b} = 2\hat{i} + \hat{j} \) and \( \vec{c} = 3\hat{i} - 4\hat{j} - 5\hat{k} \), then find a unit vector perpendicular to both of the vectors \( (\vec{a} - \vec{b}) \) and \( (\vec{c} - \vec{b}) \).
Answer: Here, \( \vec{a} = \hat{i} + 2\hat{j} + \hat{k} \), \( \vec{b} = 2\hat{i} + \hat{j} \), and \( \vec{c} = 3\hat{i} - 4\hat{j} - 5\hat{k} \).
\( \therefore \vec{a} - \vec{b} = (\hat{i} + 2\hat{j} + \hat{k}) - (2\hat{i} + \hat{j}) = -\hat{i} + \hat{j} + \hat{k} \)
\( \vec{c} - \vec{b} = (3\hat{i} - 4\hat{j} - 5\hat{k}) - (2\hat{i} + \hat{j}) = \hat{i} - 5\hat{j} - 5\hat{k} \).
Vector perpendicular to both \( \vec{a} - \vec{b} \) and \( \vec{c} - \vec{b} \) is:
\( (\vec{a} - \vec{b}) \times (\vec{c} - \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 1 \\ 1 & -5 & -5 \end{vmatrix} = (-5 + 5)\hat{i} - (5 - 1)\hat{j} + (5 - 1)\hat{k} = -4\hat{j} + 4\hat{k} \).
\( \therefore \) Unit vector perpendicular to both \( \vec{a} - \vec{b} \) and \( \vec{c} - \vec{b} \):
\( = \frac{-4\hat{j} + 4\hat{k}}{|-4\hat{j} + 4\hat{k}|} = \frac{-4\hat{j} + 4\hat{k}}{\sqrt{(-4)^2 + 4^2}} = \frac{-4\hat{j} + 4\hat{k}}{4\sqrt{2}} = \frac{1}{\sqrt{2}}(-\hat{j} + \hat{k}) \).

Question. Find a unit vector perpendicular to both of the vectors \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \) where \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \), \( \vec{b} = \hat{i} + 2\hat{j} + 3\hat{k} \).
Answer: We have \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} + 3\hat{k} \).
Let \( \vec{r} = \vec{a} + \vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k} \)
and \( \vec{p} = \vec{a} - \vec{b} = -\hat{j} - 2\hat{k} \).
A unit vector perpendicular to both \( \vec{r} \) and \( \vec{p} \) is given as \( \pm \frac{\vec{r} \times \vec{p}}{|\vec{r} \times \vec{p}|} \).
Now, \( \vec{r} \times \vec{p} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 0 & -1 & -2 \end{vmatrix} = -2\hat{i} + 4\hat{j} - 2\hat{k} \).
So, the required unit vector is:
\( = \pm \frac{-2\hat{i} + 4\hat{j} - 2\hat{k}}{\sqrt{(-2)^2 + 4^2 + (-2)^2}} = \pm \frac{-2\hat{i} + 4\hat{j} - 2\hat{k}}{\sqrt{24}} = \pm \frac{-2\hat{i} + 4\hat{j} - 2\hat{k}}{2\sqrt{6}} = \pm \frac{1}{\sqrt{6}}(-\hat{i} + 2\hat{j} - \hat{k}) \).

Question. If the sum of two unit vectors is a unit vector, prove that the magnitude of their difference is \( \sqrt{3} \).
Answer: Given, \( \hat{a} + \hat{b} = \hat{c} \)
\( \Rightarrow (\hat{a} + \hat{b}) \cdot (\hat{a} + \hat{b}) = \hat{c} \cdot \hat{c} \)
\( \Rightarrow \hat{a} \cdot \hat{a} + \hat{a} \cdot \hat{b} + \hat{b} \cdot \hat{a} + \hat{b} \cdot \hat{b} = \hat{c} \cdot \hat{c} \)
\( \Rightarrow 1 + \hat{a} \cdot \hat{b} + \hat{a} \cdot \hat{b} + 1 = 1 \)
\( \Rightarrow 2\hat{a} \cdot \hat{b} = -1 \) ... (i)
Now, \( (\hat{a} - \hat{b})^2 = (\hat{a} - \hat{b}) \cdot (\hat{a} - \hat{b}) \)
\( = \hat{a} \cdot \hat{a} - \hat{a} \cdot \hat{b} - \hat{b} \cdot \hat{a} + \hat{b} \cdot \hat{b} = 1 - 2\hat{a} \cdot \hat{b} + 1 \)
\( = 2 - 2\hat{a} \cdot \hat{b} = 2 - (-1) \) [Using (i)]
\( = 3 \)
\( \therefore |\hat{a} - \hat{b}| = \sqrt{3} \).

Question. 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: Let \( \vec{c} = x\hat{i} + y\hat{j} + z\hat{k} \).
Now we have, \( \vec{a} \times \vec{c} = \vec{b} \)
\( \Rightarrow (\hat{i} + \hat{j} + \hat{k}) \times (x\hat{i} + y\hat{j} + z\hat{k}) = \hat{j} - \hat{k} \)
\( \Rightarrow \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} = \hat{j} - \hat{k} \)
\( \Rightarrow \hat{i}(z - y) - \hat{j}(z - x) + \hat{k}(y - x) = \hat{j} - \hat{k} \)
\( \Rightarrow z - y = 0 \), \( x - z = 1 \) and \( y - x = -1 \)
\( \Rightarrow y = z \), \( x - z = 1 \), \( x - y = 1 \) ... (i)
Also, we have \( \vec{a} \cdot \vec{c} = 3 \)
\( \Rightarrow (\hat{i} + \hat{j} + \hat{k}) \cdot (x\hat{i} + y\hat{j} + z\hat{k}) = 3 \)
\( \Rightarrow x + y + z = 3 \)
\( \Rightarrow x + x - 1 + x - 1 = 3 \) [Using (i)]
\( \Rightarrow 3x - 2 = 3 \Rightarrow x = \frac{5}{3}, y = \frac{2}{3}, z = \frac{2}{3} \).
Hence, \( \vec{c} = \frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k} \).

Long Answer Type Questions (LA)

Question. If \( \vec{a} \), \( \vec{b} \), \( \vec{c} \) are mutually perpendicular vectors of equal magnitudes, show that the vector \( \vec{a} + \vec{b} + \vec{c} \) is equally inclined to \( \vec{a} \), \( \vec{b} \) and \( \vec{c} \). Also, find the angle which \( \vec{a} + \vec{b} + \vec{c} \) makes with \( \vec{a} \) or \( \vec{b} \) or \( \vec{c} \).
Answer: \( |\vec{a}| = |\vec{b}| = |\vec{c}| \) (Given) ... (i)
and \( \vec{a} \cdot \vec{b} = 0 \), \( \vec{b} \cdot \vec{c} = 0 \), \( \vec{c} \cdot \vec{a} = 0 \) ... (ii)
Let \( (\vec{a} + \vec{b} + \vec{c}) \) be inclined to vectors \( \vec{a} \), \( \vec{b} \), \( \vec{c} \) by angles \( \alpha \), \( \beta \) and \( \gamma \) respectively. Then:
\( \cos\alpha = \frac{(\vec{a} + \vec{b} + \vec{c}) \cdot \vec{a}}{|\vec{a} + \vec{b} + \vec{c}| |\vec{a}|} = \frac{\vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{a} + \vec{c} \cdot \vec{a}}{|\vec{a} + \vec{b} + \vec{c}| |\vec{a}|} \)
\( = \frac{|\vec{a}|^2 + 0 + 0}{|\vec{a} + \vec{b} + \vec{c}| |\vec{a}|} \) [Using (ii)]
\( = \frac{|\vec{a}|}{|\vec{a} + \vec{b} + \vec{c}|} \) ... (iii)
Similarly, \( \cos\beta = \frac{|\vec{b}|}{|\vec{a} + \vec{b} + \vec{c}|} \) ... (iv)
and \( \cos\gamma = \frac{|\vec{c}|}{|\vec{a} + \vec{b} + \vec{c}|} \) ... (v)
From (i), (iii), (iv) and (v), we get:
\( \cos\alpha = \cos\beta = \cos\gamma \Rightarrow \alpha = \beta = \gamma \)
Hence, the vector \( \vec{a} + \vec{b} + \vec{c} \) is equally inclined to the vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \).
Also the angle between them is given as:
\( \alpha = \cos^{-1} \left( \frac{|\vec{a}|}{|\vec{a} + \vec{b} + \vec{c}|} \right) \), \( \beta = \cos^{-1} \left( \frac{|\vec{b}|}{|\vec{a} + \vec{b} + \vec{c}|} \right) \), \( \gamma = \cos^{-1} \left( \frac{|\vec{c}|}{|\vec{a} + \vec{b} + \vec{c}|} \right) \).

Question. Show that the points \( A, B, C \) with position vectors \( 2\hat{i} - \hat{j} + \hat{k} \), \( \hat{i} - 3\hat{j} - 5\hat{k} \) and \( 3\hat{i} - 4\hat{j} - 4\hat{k} \) respectively, are the vertices of a right-angled triangle. Hence find the area of the triangle.
Answer: We have, \( A(2\hat{i} - \hat{j} + \hat{k}) \), \( B(\hat{i} - 3\hat{j} - 5\hat{k}) \) and \( C(3\hat{i} - 4\hat{j} - 4\hat{k}) \).
Then, \( \vec{AB} = (1-2)\hat{i} + (-3 - (-1))\hat{j} + (-5-1)\hat{k} = -\hat{i} - 2\hat{j} - 6\hat{k} \)
\( \vec{AC} = (3-2)\hat{i} + (-4 - (-1))\hat{j} + (-4-1)\hat{k} = \hat{i} - 3\hat{j} - 5\hat{k} \)
and \( \vec{BC} = (3-1)\hat{i} + (-4 - (-3))\hat{j} + (-4 - (-5))\hat{k} = 2\hat{i} - \hat{j} + \hat{k} \).
Now angle between \( \vec{AC} \) and \( \vec{BC} \) is given by:
\( \cos \theta = \frac{\vec{AC} \cdot \vec{BC}}{|\vec{AC}| |\vec{BC}|} = \frac{2 + 3 - 5}{\sqrt{1 + 9 + 25} \cdot \sqrt{4 + 1 + 1}} \)
\( \Rightarrow \cos \theta = 0 \Rightarrow \vec{BC} \perp \vec{AC} \).
So, \( A, B, C \) are vertices of a right-angled triangle.
Now area of \( \Delta ABC = \frac{1}{2} |\vec{AC} \times \vec{BC}| \)
\( \vec{AC} \times \vec{BC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & -5 \\ 2 & -1 & 1 \end{vmatrix} = \hat{i}(-3 - 5) - \hat{j}(1 + 10) + \hat{k}(-1 + 6) = -8\hat{i} - 11\hat{j} + 5\hat{k} \).
\( \therefore \text{Area} = \frac{1}{2} |-8\hat{i} - 11\hat{j} + 5\hat{k}| = \frac{1}{2} \sqrt{64 + 121 + 25} = \frac{\sqrt{210}}{2} \) sq. units.

Question. The two adjacent sides of a parallelogram are \( 2\hat{i} - 4\hat{j} - 5\hat{k} \) and \( 2\hat{i} + 2\hat{j} + 3\hat{k} \). Find the two unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram.
Answer: Let \( \vec{a} = 2\hat{i} - 4\hat{j} - 5\hat{k} \) and \( \vec{b} = 2\hat{i} + 2\hat{j} + 3\hat{k} \).
Then diagonal \( \vec{AC} \) of the parallelogram is:
\( \vec{p} = \vec{a} + \vec{b} = 2\hat{i} - 4\hat{j} - 5\hat{k} + 2\hat{i} + 2\hat{j} + 3\hat{k} = 4\hat{i} - 2\hat{j} - 2\hat{k} \).
Therefore, unit vector parallel to it is:
\( \frac{\vec{p}}{|\vec{p}|} = \frac{4\hat{i} - 2\hat{j} - 2\hat{k}}{\sqrt{16 + 4 + 4}} = \frac{2\hat{i} - \hat{j} - \hat{k}}{\sqrt{6}} \).
Now, diagonal \( \vec{BD} \) of the parallelogram is:
\( \vec{p'} = \vec{b} - \vec{a} = 2\hat{i} + 2\hat{j} + 3\hat{k} - (2\hat{i} - 4\hat{j} - 5\hat{k}) = 6\hat{j} + 8\hat{k} \).
Therefore, unit vector parallel to it is:
\( \frac{\vec{p'}}{|\vec{p'}|} = \frac{6\hat{j} + 8\hat{k}}{\sqrt{36 + 64}} = \frac{6\hat{j} + 8\hat{k}}{10} = \frac{3\hat{j} + 4\hat{k}}{5} \).
Now, \( \vec{p} \times \vec{p'} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -2 & -2 \\ 0 & 6 & 8 \end{vmatrix} = \hat{i}(-16 + 12) - \hat{j}(32 - 0) + \hat{k}(24 - 0) = -4\hat{i} - 32\hat{j} + 24\hat{k} \).
\( \therefore \text{Area of parallelogram} = \frac{|\vec{p} \times \vec{p'}|}{2} = \frac{\sqrt{16 + 1024 + 576}}{2} = 2\sqrt{101} \) sq. units.

Question. If \( \vec{a} = 3\hat{i} - \hat{j} \) and \( \vec{b} = 2\hat{i} + \hat{j} - 3\hat{k} \) then express \( \vec{b} \) in the form \( \vec{b} = \vec{b_1} + \vec{b_2} \) where \( \vec{b_1} \parallel \vec{a} \) and \( \vec{b_2} \perp \vec{a} \).
Answer: Here \( \vec{a} = 3\hat{i} - \hat{j} \), \( \vec{b} = 2\hat{i} + \hat{j} - 3\hat{k} \).
We have to express: \( \vec{b} = \vec{b_1} + \vec{b_2} \), where \( \vec{b_1} \parallel \vec{a} \) and \( \vec{b_2} \perp \vec{a} \).
Let \( \vec{b_1} = \lambda\vec{a} = \lambda(3\hat{i} - \hat{j}) \) and \( \vec{b_2} = x\hat{i} + y\hat{j} + z\hat{k} \).
Now, \( \vec{b_2} \perp \vec{a} \Rightarrow \vec{b_2} \cdot \vec{a} = 0 \)
\( \Rightarrow (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (3\hat{i} - \hat{j}) = 0 \)
\( \Rightarrow 3x - y = 0 \) ... (i)
Now, \( \vec{b} = \vec{b_1} + \vec{b_2} \)
\( \Rightarrow 2\hat{i} + \hat{j} - 3\hat{k} = \lambda(3\hat{i} - \hat{j}) + (x\hat{i} + y\hat{j} + z\hat{k}) \)
On comparing, we get:
\( 2 = 3\lambda + x \Rightarrow x = 2 - 3\lambda \)
\( 1 = -\lambda + y \Rightarrow y = 1 + \lambda \) ... (ii)
and \( -3 = z \Rightarrow z = -3 \).
Solving (i) and (ii), we get \( x = \frac{1}{2}, y = \frac{3}{2} \).
\( \therefore 1 = -\lambda + y \Rightarrow 1 = -\lambda + \frac{3}{2} \Rightarrow \lambda = \frac{1}{2} \).
Hence, \( \vec{b_1} = \lambda(3\hat{i} - \hat{j}) = \frac{3}{2}\hat{i} - \frac{1}{2}\hat{j} \)
and \( \vec{b_2} = \frac{1}{2}\hat{i} + \frac{3}{2}\hat{j} - 3\hat{k} \).

Question. If \( \hat{i} + \hat{j} + \hat{k} \), \( 2\hat{i} + 5\hat{j} \), \( 3\hat{i} + 2\hat{j} - 3\hat{k} \) and \( \hat{i} - 6\hat{j} - \hat{k} \) respectively are the position vectors of points \( A, B, C \) and \( D \), then find the angle between the straight lines \( AB \) and \( CD \). Find whether \( \vec{AB} \) and \( \vec{CD} \) are collinear or not.
Answer: Given, position vector of \( A = \hat{i} + \hat{j} + \hat{k} \)
Position vector of \( B = 2\hat{i} + 5\hat{j} \)
Position vector of \( C = 3\hat{i} + 2\hat{j} - 3\hat{k} \)
Position vector of \( D = \hat{i} - 6\hat{j} - \hat{k} \).
\( \therefore \vec{AB} = (2\hat{i} + 5\hat{j}) - (\hat{i} + \hat{j} + \hat{k}) = \hat{i} + 4\hat{j} - \hat{k} \) and
\( \vec{CD} = (\hat{i} - 6\hat{j} - \hat{k}) - (3\hat{i} + 2\hat{j} - 3\hat{k}) = -2\hat{i} - 8\hat{j} + 2\hat{k} \).
Now, \( |\vec{AB}| = \sqrt{(1)^2 + (4)^2 + (-1)^2} = \sqrt{18} \)
\( |\vec{CD}| = \sqrt{(-2)^2 + (-8)^2 + (2)^2} = \sqrt{4 + 64 + 4} = \sqrt{72} = 2\sqrt{18} \).
Let \( \theta \) be the angle between \( \vec{AB} \) and \( \vec{CD} \).
\( \therefore \cos\theta = \frac{\vec{AB} \cdot \vec{CD}}{|\vec{AB}| |\vec{CD}|} = \frac{(\hat{i} + 4\hat{j} - \hat{k}) \cdot (-2\hat{i} - 8\hat{j} + 2\hat{k})}{(\sqrt{18})(2\sqrt{18})} \)
\( = \frac{-2 - 32 - 2}{36} = \frac{-36}{36} = -1 \)
\( \Rightarrow \cos\theta = -1 \Rightarrow \theta = \pi \).
Since, the angle between \( \vec{AB} \) and \( \vec{CD} \) is \( 180^\circ \).
\( \therefore \vec{AB} \) and \( \vec{CD} \) are collinear.