CBSE Class 12 Mathematics HOTs Vector Algebra Set 01

Very Short Answer Type Questions

Question. If \(\vec{a} = 4\hat{i} - \hat{j} + \hat{k}\) and \(\vec{b} = 2\hat{i} - 2\hat{j} + \hat{k}\), then find a unit vector parallel to the vector \(\vec{a} + \vec{b}\). CBSE 2016
Answer: Given vectors, \[\vec{a} = 4\hat{i} - \hat{j} + \hat{k} \text{ and } \vec{b} = 2\hat{i} - 2\hat{j} + \hat{k}\] Now, \(\vec{a} + \vec{b} = (4\hat{i} - \hat{j} + \hat{k}) + (2\hat{i} - 2\hat{j} + \hat{k}) = 6\hat{i} - 3\hat{j} + 2\hat{k}\)
and \(|\vec{a} + \vec{b}| = \sqrt{(6)^2 + (-3)^2 + (2)^2}\) \[= \sqrt{36 + 9 + 4}\] \[= \sqrt{49} = 7 \text{ units}\] \(\therefore\) The unit vector parallel to the vector \(\vec{a} + \vec{b}\) is \[\frac{\vec{a} + \vec{b}}{|\vec{a} + \vec{b}|} = \frac{6\hat{i} - 3\hat{j} + 2\hat{k}}{7}\]

Question. Two vectors \(\hat{j} + \hat{k}\) and \(3\hat{i} - \hat{j} + 4\hat{k}\) represent the two sides \(AB\) and \(AC\), respectively of \(\Delta ABC\). Find the length of the median through \(A\). Competency Based Que
Answer: Given, \(\vec{AB} = \hat{j} + \hat{k}\) and \(\vec{AC} = 3\hat{i} - \hat{j} + 4\hat{k}\)
Clearly, median vector, \(\vec{AD} = \frac{\vec{AB} + \vec{AC}}{2}\) \[= \frac{(\hat{j} + \hat{k}) + (3\hat{i} - \hat{j} + 4\hat{k})}{2}\] \[= \frac{3\hat{i} + 5\hat{k}}{2}\] Now, length of median \(= |\vec{AD}|\) \[= \sqrt{\left(\frac{3}{2}\right)^2 + \left(\frac{5}{2}\right)^2} = \frac{1}{2}\sqrt{9 + 25} = \frac{\sqrt{34}}{2}\] \[= \frac{\sqrt{17} \times \sqrt{2}}{2} = \sqrt{\frac{17}{2}} \text{ units}\]

Question. Find the value of \(p\) for which the vectors \(3\hat{i} + 2\hat{j} + 9\hat{k}\) and \(\hat{i} - 2p\hat{j} + 3\hat{k}\) are parallel. CBSE 2014
Answer: Given, \(3\hat{i} + 2\hat{j} + 9\hat{k}\) and \(\hat{i} - 2p\hat{j} + 3\hat{k}\) are two parallel vectors, so their direction ratios will be proportional. \[\frac{3}{1} = \frac{2}{-2p} = \frac{9}{3}\] \[\Rightarrow -6p = 2 \Rightarrow p = -\frac{2}{6} \Rightarrow p = -\frac{1}{3}\]

Question. Find the angle between X-axis and the vector \(\hat{i} + \hat{j} + \hat{k}\). CBSE 2014C
Answer: Let \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\)
Now, unit vector in the direction of \(\vec{a}\) is \[\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{1^2 + 1^2 + 1^2}}\] \[\Rightarrow \hat{a} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}\] \[\Rightarrow \hat{a} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}\] So, angle between X-axis and the vector \(\hat{i} + \hat{j} + \hat{k}\) is \(\cos\alpha = \frac{1}{\sqrt{3}} \Rightarrow \alpha = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)\)
[\(\because \vec{a} = l\hat{i} + m\hat{j} + n\hat{k}\) and \(\cos\alpha = l \Rightarrow \alpha = \cos^{-1} l\)]

Question. Write the direction ratios of the vector \(3\vec{a} + 2\vec{b}\), where \(\vec{a} = \hat{i} + \hat{j} - 2\hat{k}\) and \(\vec{b} = 2\hat{i} - 4\hat{j} + 5\hat{k}\). CBSE 2015C
Answer: Given, \(\vec{a} = \hat{i} + \hat{j} - 2\hat{k}\) and \(\vec{b} = 2\hat{i} - 4\hat{j} + 5\hat{k}\)
Now, \(3\vec{a} + 2\vec{b} = 3(\hat{i} + \hat{j} - 2\hat{k}) + 2(2\hat{i} - 4\hat{j} + 5\hat{k})\) \[= 3\hat{i} + 3\hat{j} - 6\hat{k} + 4\hat{i} - 8\hat{j} + 10\hat{k}\] \[= 7\hat{i} - 5\hat{j} + 4\hat{k}\] Therefore, direction ratios of \(3\vec{a} + 2\vec{b}\) are \(7, -5, 4\).

Question. Write the value of cosine of the angle which the vector \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\) makes with Y-axis. CBSE 2014C
Answer: Given, \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\)
Now, unit vector in the direction of \(\vec{a}\) is \[\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{(1)^2 + (1)^2 + (1)^2}}\] \[= \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}\] \(\therefore\) Cosine of the angle which given vector makes with Y-axis is \(\frac{1}{\sqrt{3}}\).

Question. Let \(l_i, m_i, n_i; i = 1, 2, 3\) be the direction cosines of three mutually perpendicular vectors in space. Show that \(AA' = I_3\), where \(A = \begin{bmatrix} l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3 \end{bmatrix}\). CBSE Sample Paper 2017
Answer: Given, \(A = \begin{bmatrix} l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3 \end{bmatrix}\), then \(A' = \begin{bmatrix} l_1 & l_2 & l_3 \\ m_1 & m_2 & m_3 \\ n_1 & n_2 & n_3 \end{bmatrix}\)
Also given, \(l_i, m_i, n_i; (i = 1, 2, 3)\) are direction cosines of mutually perpendicular vectors, so \[l_i^2 + m_i^2 + n_i^2 = \sum l_i^2 = 1, \text{ for each } i = 1, 2, 3 \text{ ...(i)}\] and \(l_i l_j + m_i m_j + n_i n_j = \sum l_i l_j = 0 (i \neq j)\) for each \(i, j = 1, 2, 3 \text{ ...(ii)}\)
Now, \(AA' = \begin{bmatrix} l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3 \end{bmatrix} \begin{bmatrix} l_1 & l_2 & l_3 \\ m_1 & m_2 & m_3 \\ n_1 & n_2 & n_3 \end{bmatrix}\) \[= \begin{bmatrix} \sum l_1^2 & \sum l_1 l_2 & \sum l_1 l_3 \\ \sum l_2 l_1 & \sum l_2^2 & \sum l_2 l_3 \\ \sum l_3 l_1 & \sum l_3 l_2 & \sum l_3^2 \end{bmatrix}\] \[= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \text{ [using Eqs. (i) and (ii)]}\] \[= I_3\]

Question. Write a unit vector in the direction of vector \(\vec{PQ}\), where \(P\) and \(Q\) are the points \((1, 3, 0)\) and \((4, 5, 6)\), respectively. Competency Based Que
Answer: Given points are \(P(1, 3, 0)\) and \(Q(4, 5, 6)\).
Here, \(x_1 = 1, y_1 = 3, z_1 = 0\) and \(x_2 = 4, y_2 = 5, z_2 = 6\)
So, vector \(\vec{PQ} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}\) \[= (4-1)\hat{i} + (5-3)\hat{j} + (6-0)\hat{k}\] \[= 3\hat{i} + 2\hat{j} + 6\hat{k}\] \(\therefore\) Magnitude of given vector \[|\vec{PQ}| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7 \text{ units}\] Hence, the unit vector in the direction of \(\vec{PQ}\) is \[\frac{\vec{PQ}}{|\vec{PQ}|} = \frac{3\hat{i} + 2\hat{j} + 6\hat{k}}{7} = \frac{3}{7}\hat{i} + \frac{2}{7}\hat{j} + \frac{6}{7}\hat{k}\]

Question. \(L\) and \(M\) are two points with position vectors \(2\vec{a} - \vec{b}\) and \(\vec{a} + 2\vec{b}\), respectively. Write the position vector of a point \(N\), which divides the line segment \(LM\) in the ratio \(2 : 1\) externally. Competency Based Que
Answer: Given position vectors are \(\vec{OL} = 2\vec{a} - \vec{b}\) and \(\vec{OM} = \vec{a} + 2\vec{b}\).
Now, \(\vec{ON}\) is the position vector of point \(N\), which divides the join of points, with position vectors \(\vec{OL}\) and \(\vec{OM}\), externally in the ratio \(2 : 1\).
\(\therefore \vec{ON} = \frac{2\vec{OM} - 1\vec{OL}}{2-1}\) [by external section formula] \[= \frac{2(\vec{a} + 2\vec{b}) - (2\vec{a} - \vec{b})}{1}\] \[= 2\vec{a} + 4\vec{b} - 2\vec{a} + \vec{b} = 5\vec{b}\]

Question. Find \(|\vec{x}|\), if for a unit vector \(\hat{a}\), \((\vec{x} - \hat{a}) \cdot (\vec{x} + \hat{a}) = 15\). CBSE 2013
Answer: Given, \(\hat{a}\) is a unit vector. \[\therefore |\hat{a}| = 1\] Now, \((\vec{x} - \hat{a}) \cdot (\vec{x} + \hat{a}) = 15\) \[\Rightarrow |\vec{x}|^2 - \vec{x} \cdot \hat{a} + \hat{a} \cdot \vec{x} - |\hat{a}|^2 = 15\] \[\Rightarrow |\vec{x}|^2 - 1 = 15\] \[\Rightarrow |\vec{x}|^2 = 16\] \[\Rightarrow |\vec{x}| = 4\]

Question. For what value of \(\lambda\), the vectors \(\hat{i} + 2\lambda\hat{j} + \hat{k}\) and \(2\hat{i} + \hat{j} - 3\hat{k}\) are perpendicular? CBSE 2011C
Answer: Let \(\vec{a} = \hat{i} + 2\lambda\hat{j} + \hat{k}\) and \(\vec{b} = 2\hat{i} + \hat{j} - 3\hat{k}\)
Given, \(\vec{a}\) and \(\vec{b}\) are perpendicular. \[\vec{a} \cdot \vec{b} = 0\] \[\Rightarrow (\hat{i} + 2\lambda\hat{j} + \hat{k}) \cdot (2\hat{i} + \hat{j} - 3\hat{k}) = 0\] \[\Rightarrow (1)(2) + (2\lambda)(1) + (1)(-3) = 0\] \[\Rightarrow 2 + 2\lambda - 3 = 0\] \[\Rightarrow 2\lambda = 1\] \[\Rightarrow \lambda = \frac{1}{2}\]

Question. Find \(\lambda\), when projection of \(\vec{a} = \lambda\hat{i} + \hat{j} + 4\hat{k}\) on \(\vec{b} = 2\hat{i} + 6\hat{j} + 3\hat{k}\) is 4 units. CBSE 2012
Answer: Given, \(\vec{a} = \lambda\hat{i} + \hat{j} + 4\hat{k}\) and \(\vec{b} = 2\hat{i} + 6\hat{j} + 3\hat{k}\)
Also given, projection of \(\vec{a}\) on \(\vec{b} = 4\) \[\Rightarrow \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = 4\] \[\Rightarrow \frac{(\lambda\hat{i} + \hat{j} + 4\hat{k}) \cdot (2\hat{i} + 6\hat{j} + 3\hat{k})}{\sqrt{4 + 36 + 9}} = 4\] \[\Rightarrow \frac{2\lambda + 6 + 12}{7} = 4\] \[\Rightarrow 2\lambda + 18 = 28\] \[\Rightarrow 2\lambda = 10\] \[\Rightarrow \lambda = 5\]

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 \(\vec{b}\). CBSE 2013
Answer: Given, \[|\vec{a} + \vec{b}| = |\vec{a}|\] \[\Rightarrow |\vec{a} + \vec{b}|^2 = |\vec{a}|^2\] \[\Rightarrow |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = |\vec{a}|^2\] \[\Rightarrow |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = 0 \text{ ...(i)}\] Now, \((2\vec{a} + \vec{b}) \cdot \vec{b} = 2\vec{a} \cdot \vec{b} + |\vec{b}|^2 = 0\) [using Eq. (i)]
Hence, \(2\vec{a} + \vec{b}\) is perpendicular to \(\vec{b}\).

Question. If \(\vec{a}\) and \(\vec{b}\) are unit vectors, then what is the angle between \(\vec{a}\) and \(\vec{b}\) so that \(\sqrt{2}\vec{a} - \vec{b}\) is a unit vector? CBSE 2015C
Answer: Given, \(\vec{a}\) and \(\vec{b}\) are unit vectors. \[\therefore |\vec{a}| = 1 \text{ and } |\vec{b}| = 1\] Also given, \(\sqrt{2}\vec{a} - \vec{b}\) is a unit vector. \[\therefore |\sqrt{2}\vec{a} - \vec{b}| = 1\] \[\Rightarrow |\sqrt{2}\vec{a} - \vec{b}|^2 = 1\] \[\Rightarrow 2|\vec{a}|^2 + |\vec{b}|^2 - 2\sqrt{2}\vec{a} \cdot \vec{b} = 1\] \[\Rightarrow 2 + 1 - 2\sqrt{2}\vec{a} \cdot \vec{b} = 1\] \[\Rightarrow 2\sqrt{2}\vec{a} \cdot \vec{b} = 2\] \[\Rightarrow \vec{a} \cdot \vec{b} = \frac{1}{\sqrt{2}}\] \[\Rightarrow |\vec{a}||\vec{b}|\cos\theta = \frac{1}{\sqrt{2}}\] \[\Rightarrow (1)(1)\cos\theta = \frac{1}{\sqrt{2}} \Rightarrow \cos\theta = \frac{1}{\sqrt{2}}\] \[\therefore \theta = \frac{\pi}{4}\]

Question. Find the angle between the unit vectors \(\hat{a}\) and \(\hat{b}\), given that \(|\hat{a} + \hat{b}| = 1\). CBSE Sample Paper 2021
Answer: We have, \(|\hat{a} + \hat{b}| = 1\)
Let \(\theta\) be the angle between \(\hat{a}\) and \(\hat{b}\).
Now, \(|\hat{a} + \hat{b}| = 1\) \[\Rightarrow |\hat{a} + \hat{b}|^2 = 1\] \[\Rightarrow (\hat{a} + \hat{b}) \cdot (\hat{a} + \hat{b}) = 1\] \[\Rightarrow |\hat{a}|^2 + |\hat{b}|^2 + 2\hat{a} \cdot \hat{b} = 1\] \[\Rightarrow 1 + 1 + 2|\hat{a}||\hat{b}|\cos\theta = 1 \text{ [}\because |\hat{a}| = |\hat{b}| = 1\text{]}\] \[\Rightarrow 2\cos\theta = -1\] \[\Rightarrow \cos\theta = -\frac{1}{2}\] \[\Rightarrow \theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\]

Question. Find the projection of the vector \(\hat{i} - \hat{j}\) on the vector \(\hat{i} + \hat{j}\). CBSE 2020
Answer: Let \(\vec{a} = \hat{i} - \hat{j}\) and \(\vec{b} = \hat{i} + \hat{j}\)
We know that projection of \(\vec{a}\) on \(\vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\) \[= \frac{(\hat{i} - \hat{j}) \cdot (\hat{i} + \hat{j})}{|\hat{i} + \hat{j}|} = \frac{1 - 1}{\sqrt{2}} = 0\]

Short Answer Type Questions

Question. If the points with position vectors \(10\hat{i} + 3\hat{j}\), \(12\hat{i} - 5\hat{j}\) and \(\lambda\hat{i} + 11\hat{j}\) are collinear, then find the value of \(\lambda\).
Answer: Let A, B and C be the points with position vectors \(10\hat{i} + 3\hat{j}\), \(12\hat{i} - 5\hat{j}\) and \(\lambda\hat{i} + 11\hat{j}\), respectively.
Then, \(\vec{AB} = \text{Position vector of B} - \text{Position vector of A}\)
\(= 12\hat{i} - 5\hat{j} - 10\hat{i} - 3\hat{j} = 2\hat{i} - 8\hat{j}\)
and \(\vec{BC} = \text{Position vector of C} - \text{Position vector of B}\)
\(= \lambda\hat{i} + 11\hat{j} - 12\hat{i} + 5\hat{j} = (\lambda - 12)\hat{i} + 16\hat{j}\)
Since, A, B and C are collinear.
\(\therefore \vec{AB} = \mu \vec{BC}\)
\(\Rightarrow 2\hat{i} - 8\hat{j} = \mu[(\lambda - 12)\hat{i} + 16\hat{j}]\)
On comparing the coefficients of \(\hat{i}\) and \(\hat{j}\), we get
\(2 = \mu(\lambda - 12)\) and \(-8 = \mu 16 \Rightarrow \mu = -\frac{1}{2}\)
\(\Rightarrow 2 = -\frac{1}{2}(\lambda - 12)\)
\(\Rightarrow -\lambda + 12 = 4 \Rightarrow \lambda = 8\)

Question. The position vectors of points A, B and C are \(\lambda\hat{i} + 3\hat{j}\), \(12\hat{i} + \mu\hat{j}\) and \(11\hat{i} - 3\hat{j}\), respectively. If C divides the line segment joining A and B in the ratio \(3 : 1\), then find the value of \(\lambda\) and \(\mu\).
Answer: Given that C divides the line segment joining A and B in the ratio \(3 : 1\).
\(\therefore 11\hat{i} - 3\hat{j} = \frac{3 \times (12\hat{i} + \mu\hat{j}) + 1 \times (\lambda\hat{i} + 3\hat{j})}{3 + 1}\)
\(\Rightarrow 11\hat{i} - 3\hat{j} = \frac{(36 + \lambda)\hat{i} + (3\mu + 3)\hat{j}}{4}\)
\(\Rightarrow 44\hat{i} - 12\hat{j} = (36 + \lambda)\hat{i} + (3\mu + 3)\hat{j}\)
On equating the corresponding coefficients of \(\hat{i}\) and \(\hat{j}\), we get
\(36 + \lambda = 44\) and \(3\mu + 3 = -12\)
\(\Rightarrow \lambda = 44 - 36 = 8\)
and \(3\mu = -15 \Rightarrow \mu = -5\)

Question. If a unit vector \(\vec{a}\) makes angles \(\frac{\pi}{4}\) with \(\hat{i}\), \(\frac{\pi}{3}\) with \(\hat{j}\) and an acute angle \(\theta\) with \(\hat{k}\), then find the components of \(\vec{a}\) and the angle \(\theta\).
Answer: We know that if a vector \(\vec{a}\) makes angles \(\alpha, \beta\) and \(\gamma\) with \(\hat{i}, \hat{j}\) and \(\hat{k}\) respectively, then
\(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\)
It is given that \(\alpha = \frac{\pi}{4}\), \(\beta = \frac{\pi}{3}\) and \(\gamma = \theta\) an acute angle.
\(\therefore \cos^2 \frac{\pi}{4} + \cos^2 \frac{\pi}{3} + \cos^2 \theta = 1\)
\(\Rightarrow \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{2}\right)^2 + \cos^2 \theta = 1\)
\(\Rightarrow \frac{1}{2} + \frac{1}{4} + \cos^2 \theta = 1 \Rightarrow \cos^2 \theta = \frac{1}{4}\)
\(\Rightarrow \cos \theta = \frac{1}{2}\) \([\because \theta \text{ is an acute angle, therefore } \cos \theta > 0]\)
\(\Rightarrow \theta = \frac{\pi}{3}\)
\(\therefore \gamma = \frac{\pi}{3}\)
Now, \(\vec{a} = |\vec{a}|\{(\cos \alpha)\hat{i} + (\cos \beta)\hat{j} + (\cos \gamma)\hat{k}\}\)
\(\Rightarrow \vec{a} = \cos \frac{\pi}{4} \hat{i} + \cos \frac{\pi}{3} \hat{j} + \cos \frac{\pi}{3} \hat{k}\) \([\because |\vec{a}| = 1, \alpha = \frac{\pi}{4}, \beta = \frac{\pi}{3} \text{ and } \gamma = \frac{\pi}{3}]\)
\(\Rightarrow \vec{a} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{2}\hat{j} + \frac{1}{2}\hat{k}\)
Thus, the components of \(\vec{a}\) are \(\left(\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}\right)\).

Question. Find the projection of \(\vec{b} + \vec{c}\) on \(\vec{a}\), where \(\vec{a} = 2\hat{i} - 2\hat{j} + \hat{k}\), \(\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}\) and \(\vec{c} = 2\hat{i} - \hat{j} + 4\hat{k}\).
Answer: Given, \(\vec{a} = 2\hat{i} - 2\hat{j} + \hat{k}\), \(\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}\) and \(\vec{c} = 2\hat{i} - \hat{j} + 4\hat{k}\)
Now, \((\vec{b} + \vec{c}) = (\hat{i} + 2\hat{j} - 2\hat{k}) + (2\hat{i} - \hat{j} + 4\hat{k}) = 3\hat{i} + \hat{j} + 2\hat{k}\)
\(\therefore\) Projection of \((\vec{b} + \vec{c})\) on \(\vec{a} = \frac{(\vec{b} + \vec{c}) \cdot \vec{a}}{|\vec{a}|}\)
\(= \frac{(3\hat{i} + \hat{j} + 2\hat{k}) \cdot (2\hat{i} - 2\hat{j} + \hat{k})}{\sqrt{4 + 4 + 1}}\)
\(= \frac{3(2) + 1(-2) + 2(1)}{\sqrt{9}} = \frac{6 - 2 + 2}{3} = 2\)

Long Answer Type Questions

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: We have, \(\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}\) and \(\vec{b} = 2\hat{i} + 4\hat{j} - 5\hat{k}\)
So, the diagonals of the parallelogram whose adjacent sides are \(\vec{a}\) and \(\vec{b}\) are given by
\(\vec{p} = \vec{a} + \vec{b}\) and \(\vec{q} = \vec{b} - \vec{a}\)
Now, \(\vec{p} = (\hat{i} + 2\hat{j} + 3\hat{k}) + (2\hat{i} + 4\hat{j} - 5\hat{k}) = 3\hat{i} + 6\hat{j} - 2\hat{k}\)
and \(\vec{q} = (2\hat{i} + 4\hat{j} - 5\hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) = \hat{i} + 2\hat{j} - 8\hat{k}\)
\(\therefore \hat{p} = \frac{\vec{p}}{|\vec{p}|} = \frac{3\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{9 + 36 + 4}} = \frac{3\hat{i} + 6\hat{j} - 2\hat{k}}{7} = \frac{3}{7}\hat{i} + \frac{6}{7}\hat{j} - \frac{2}{7}\hat{k}\)
and \(\hat{q} = \frac{\vec{q}}{|\vec{q}|} = \frac{\hat{i} + 2\hat{j} - 8\hat{k}}{\sqrt{1 + 4 + 64}} = \frac{\hat{i} + 2\hat{j} - 8\hat{k}}{\sqrt{69}} = \frac{1}{\sqrt{69}}\hat{i} + \frac{2}{\sqrt{69}}\hat{j} - \frac{8}{\sqrt{69}}\hat{k}\)

Question. Show that the points A, B and 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, \(\vec{AB} = \text{Position vector of B} - \text{Position vector of A}\)
\(= (\hat{i} - 3\hat{j} - 5\hat{k}) - (2\hat{i} - \hat{j} + \hat{k}) = -\hat{i} - 2\hat{j} - 6\hat{k}\)
\(\vec{BC} = (3\hat{i} - 4\hat{j} - 4\hat{k}) - (\hat{i} - 3\hat{j} - 5\hat{k}) = 2\hat{i} - \hat{j} + \hat{k}\)
and \(\vec{CA} = (2\hat{i} - \hat{j} + \hat{k}) - (3\hat{i} - 4\hat{j} - 4\hat{k}) = -\hat{i} + 3\hat{j} + 5\hat{k}\)
Here, \(\vec{AB} + \vec{BC} + \vec{CA} = \vec{0}\)
\(\Rightarrow\) A, B and C are the vertices of a triangle.
Now, \(\vec{BC} \cdot \vec{CA} = (2\hat{i} - \hat{j} + \hat{k}) \cdot (-\hat{i} + 3\hat{j} + 5\hat{k})\)
\(= -2 - 3 + 5 = 0\)
\(\Rightarrow \vec{BC} \perp \vec{CA} \Rightarrow \angle C = 90^\circ\)
Now, area of \(\Delta ABC = \frac{1}{2} |\vec{CA} \times \vec{BC}|\)
\(= \frac{1}{2} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 3 & 5 \\ 2 & -1 & 1 \end{vmatrix}\)
\(= \frac{1}{2} |8\hat{i} + 11\hat{j} - 5\hat{k}|\)
\(= \frac{1}{2} \sqrt{64 + 121 + 25} = \frac{1}{2} \sqrt{210}\) 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 ABCD be the given parallelogram with \(\vec{AB} = 2\hat{i} - 4\hat{j} - 5\hat{k}\) and \(\vec{AD} = 2\hat{i} + 2\hat{j} + 3\hat{k}\).
Clearly, the diagonal \(\vec{AC}\) is given by \(\vec{AB} + \vec{AD} = 4\hat{i} - 2\hat{j} - 2\hat{k}\)
and the diagonal \(\vec{BD}\) is given by \(\vec{BC} + \vec{BA} = \vec{AD} - \vec{AB} = 6\hat{j} + 8\hat{k}\) [using parallelogram law of addition]
Since, the unit vector along \(\vec{AC}\) is given by
\(\frac{\vec{AC}}{|\vec{AC}|} = \frac{4\hat{i} - 2\hat{j} - 2\hat{k}}{\sqrt{16 + 4 + 4}} = \frac{4\hat{i} - 2\hat{j} - 2\hat{k}}{\sqrt{24}} = \frac{4\hat{i} - 2\hat{j} - 2\hat{k}}{2\sqrt{6}} = \frac{1}{\sqrt{6}} (2\hat{i} - \hat{j} - \hat{k})\)
and the unit vector along \(\vec{BD}\) is given by
\(\frac{\vec{BD}}{|\vec{BD}|} = \frac{6\hat{j} + 8\hat{k}}{\sqrt{36 + 64}} = \frac{6\hat{j} + 8\hat{k}}{10} = \frac{1}{5}(3\hat{j} + 4\hat{k})\)
Since, area of parallelogram \(ABCD = \frac{1}{2} |\vec{AC} \times \vec{BD}|\)
Here, \(\vec{AC} \times \vec{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -2 & -2 \\ 0 & 6 & 8 \end{vmatrix} = \hat{i}(-16 + 12) - \hat{j}(32 - 0) + \hat{k}(24 - 0) = -4\hat{i} - 32\hat{j} + 24\hat{k}\)
and \(|\vec{AC} \times \vec{BD}| = \sqrt{(-4)^2 + (-32)^2 + (24)^2} = \sqrt{4^2 (1 + 8^2 + 6^2)} = 4\sqrt{1 + 64 + 36} = 4\sqrt{101}\)
\(\therefore\) Area of parallelogram \(ABCD = \frac{1}{2} \times 4\sqrt{101} = 2\sqrt{101}\) sq units

Question. If \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\), then find \((\vec{r} \times \hat{i}) \cdot (\vec{r} \times \hat{j}) + xy\).
Answer: Given, \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\)
Now, \(\vec{r} \times \hat{i} = (x\hat{i} + y\hat{j} + z\hat{k}) \times \hat{i} = x(\hat{i} \times \hat{i}) + y(\hat{j} \times \hat{i}) + z(\hat{k} \times \hat{i}) = -y\hat{k} + z\hat{j}\)
and \((\vec{r} \times \hat{j}) = (x\hat{i} + y\hat{j} + z\hat{k}) \times \hat{j} = x(\hat{i} \times \hat{j}) + y(\hat{j} \times \hat{j}) + z(\hat{k} \times \hat{j}) = x\hat{k} - z\hat{i}\)
\(\therefore (\vec{r} \times \hat{i}) \cdot (\vec{r} \times \hat{j}) = (-y\hat{k} + z\hat{j}) \cdot (x\hat{k} - z\hat{i}) = -yx = -xy\)
\(\therefore (\vec{r} \times \hat{i}) \cdot (\vec{r} \times \hat{j}) + xy = -xy + xy = 0\)