CBSE Previous Year Question: Vectors

CBSE Previous Year Questions 2025

Multiple Choice Questions (MCQ)

Q1. If vector $\vec{a} = 3\hat{i} + 2\hat{j} - \hat{k}$ and vector $\vec{b} = \hat{i} - \hat{j} + \hat{k}$, then which of the following is correct? (1 Mark)
(A) $\vec{a} \parallel \vec{b}$
(B) $\vec{a} \perp \vec{b}$
(C) $|\vec{b}| > |\vec{a}|$
(D) $|\vec{a}| = |\vec{b}|$ 

Ans: (B) \( \vec{a} \perp \vec{b} \)

Q2. If $\vec{a} + \vec{b} + \vec{c} = 0$, $|\vec{a}| = \sqrt{37}$, $|\vec{b}| = 3$ and $|\vec{c}| = 4$, then angle between $\vec{b}$ and $\vec{c}$ is  (1 Mark)
(A) $\frac{\pi}{6}$
(B) $\frac{\pi}{4}$
(C) $\frac{\pi}{3}$
(D) $\frac{\pi}{2}$

Ans: (C) \( \frac{\pi}{3} \)

Q3. If $\vec{\alpha} = \hat{i} - 4\hat{j} + 9\hat{k}$ and $\vec{\beta} = 2\hat{i} - 8\hat{j} + \lambda\hat{k}$ are two mutually parallel vectors, then $\lambda$ is equal to: (1 Mark)
(A) -18
(B) 18
(C) \(-\frac{34}{9}\)
(D) \(\frac{34}{9}\) 

Ans: (B) 18

Q4. The unit vector perpendicular to the vectors $\hat{i} - \hat{j}$ and $\hat{i} + \hat{j}$ is  (1 Mark)
(A) $\hat{k}$
(B) $-\hat{k} + \hat{j}$
(C) $\frac{\hat{i} - \hat{j}} { \sqrt{2}} $
(D) $\frac{\hat{i} + \hat{j}} {\sqrt{2}}$

Ans: (A) $\hat{k}$

Q5. The projection vector of vector $\vec{a}$ on vector $\vec{b}$ is     (1 Mark)
(A) $\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right)\vec{b}$
(B) $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$
(C) $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}$
(D) $\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}\right)\vec{b}$

Ans: (A) $\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right)\vec{b}$

Q6. Let $\vec{p}$ and $\vec{q}$ be two unit vectors and $\alpha$ be the angle between them. Then $(\vec{p} + \vec{q})$ will be a unit vector for what value of $\alpha$ ?  (1 Mark)
(A) $\frac{\pi}{4}$
(B) $\frac{\pi}{3}$
(C) $\frac{\pi}{2}$
(D) $\frac{2\pi}{3}$

Ans: (D) $\frac{2\pi}{3}$

Q7. The values of $x$ for which the angle between the vectors $\vec{a} = 2x^2\hat{i} + 4x\hat{j} + \hat{k}$ and $\vec{b} = 7\hat{i} - 2\hat{j} + x\hat{k}$ is obtuse, is :  (1 Mark)
(A) 0 or $\frac{1}{2}$
(B) $x > \frac{1}{2}$
(C) $(0, \frac{1}{2})$
(D) $[0, \frac{1}{2}]$

Ans: (C) $(0, \frac{1}{2})$

Q8. If $\vec{PQ} \times \vec{PR} = 4\hat{i} + 8\hat{j} - 8\hat{k}$, then the area ($\Delta PQR$) is  (1 Mark)
(A) 2 sq units
(B) 4 sq units
(C) 6 sq units
(D) 12 sq units

Ans: (C) 6 sq units

Q9. If \(\vec{p}\) and \(\vec{q}\) are unit vectors, then which of the following values of \(\vec{p} \cdot \vec{q}\) is not possible? (1 Mark)
(A) \(\frac{-1}{2}\)
(B) \(\frac{1}{\sqrt{2}}\)
(C) \(\frac{\sqrt{3}}{2}\)
(D) \(\sqrt{3}\)

Ans: (D) \(\sqrt{3}\)

Q10. If projection of \(\vec{a} = \alpha\hat{i} + \hat{j} + 4\hat{k}\) on \(\vec{b} = 2\hat{i} + 6\hat{j} + 3\hat{k}\) is 4 units, then \(\alpha\) is (1 Mark)
(A) -13
(B) -5
(C) 13
(D) 5

Ans: (D) 5

Q11. If the sides AB and AC of \(\triangle ABC\) are represented by vectors \(\hat{j} + \hat{k}\) and \(3\hat{i} - \hat{j} + 4\hat{k}\) respectively, then the length of the median through A on BC is : (1 Mark)
(A) \(2\sqrt{2}\) units
(B) \(\sqrt{18}\) units
(C) \(\frac{\sqrt{34}}{2}\) units
(D) \(\frac{\sqrt{48}}{2}\) units

Ans: (C) \(\frac{\sqrt{34}}{2}\) units

Q12. Let \(|\vec{a}| = 5\) and \(-2 \le \lambda \le 1\). Then, the range of \(|\lambda \vec{a}|\) is:  (1 Mark)
(A) \([0, 10]\)
(B) \([-2, 5]\)
(C) \([-2, 1]\)
(D) \([-10, 5]\)

Ans: (A) \([0, 10]\)

Q13. A student tries to tie ropes, parallel to each other from one end of the wall to the other. If one rope is along the vector $3\hat{i} + 15\hat{j} + 6\hat{k}$ and the other is along the vector $2\hat{i} + 10\hat{j} + \lambda\hat{k}$, then the value of $\lambda$ is : (1 Mark)
(A) 6
(B) 1
(C) $\frac{1}{4}$
(D) 4

Ans: (D) 4

Q14. If $|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|$ for any two vectors, then vectors $\vec{a}$ and $\vec{b}$ are :  (1 Mark)
(A) orthogonal vectors
(B) parallel to each other
(C) unit vectors
(D) collinear vectors

Ans: (A) orthogonal vectors

Q15. The number of vector(s) of unit length perpendicular to the vectors $\vec{a} = 2\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = \hat{j} + \hat{k}$ is (are) : (1 Mark)
(A) one
(B) two
(C) three
(D) infinite

Ans: (B) two

Q16. Two statements are given, one labelled Assertion (A) and the other labelled Reason (R).   (1 Mark)
Select the correct answer from the codes (A), (B), (C) and (D) as given below. 

Assertion (A): If \(|\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = 256\) and \(|\vec{b}| = 8\), then \(|\vec{a}| = 2\).
Reason (R): \(\sin^2 \theta + \cos^2 \theta = 1\) and \(|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta\) and \(\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta\).

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). 
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). 
(C) Assertion (A) is true, but Reason (R) is false. 
(D) Assertion (A) is false, but Reason (R) is true.

Ans: (A)

Q17. Two statements are given, one labelled Assertion (A) and the other labelled Reason (R).   (1 Mark)
Select the correct answer from the codes (A), (B), (C) and (D) as given below.
Assertion (A) : The vectors $\vec{a} = 4\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = -2\hat{i} + 3\hat{j} - 5\hat{k}$ are mutually perpendicular vectors.
Reason (R) : Two vectors $\vec{a}$ and $\vec{b}$ are perpendicular to each other, if $\vec{a} \cdot \vec{b} = 0$.

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). 
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). 
(C) Assertion (A) is true, but Reason (R) is false. 
(D) Assertion (A) is false, but Reason (R) is true.

Ans: (A)

Very Short Answer Questions

Q18. The diagonals of a parallelogram are given by $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} - \hat{k}$. Find the area of the parallelogram.  (2 Marks)

Ans: 

\( \vec{a} \times \vec{b} = \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 3 & -1 \end{bmatrix} = -2\hat{i} + 3\hat{j} + 7\hat{k} \)

Area of parallelogram = \( \frac{1}{2} |\vec{a} \times \vec{b}| \)

\( = \frac{1}{2} \sqrt{(-2)^2 + 3^2 + 7^2} = \frac{\sqrt{62}}{2} \)

Q19. Two friends while flying kites from different locations, find the strings of their kites crossing each other. The strings can be represented by vectors $\vec{a} = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = 2\hat{i} - 2\hat{j} + 4\hat{k}$. Determine the angle formed between the kite strings. Assume there is no slack in the strings.  (2 Marks)

Ans: 

Let the required angle between the kite strings be \( \theta \).

Then, \( \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \)

\( \Rightarrow \cos \theta = \frac{\left(3\hat{i} + \hat{j} + 2\hat{k}\right) \left(2\hat{i} - 2\hat{j} + 4\hat{k}\right)}{\sqrt{9+1+4\sqrt{4+4+16}}} = \frac{12}{\sqrt{336}} = \frac{3}{\sqrt{21}} \)

\( \Rightarrow \theta = \cos^{-1}\left(\frac{12}{\sqrt{336}}\right) \) or \( \cos^{-1}\left(\frac{3}{\sqrt{21}}\right) \)

Q20. Find a vector of magnitude 21 units in the direction opposite to that of $\overrightarrow{AB}$ where A and B are the points A(2, 1, 3) and B(8, -1, 0) respectively.   (2 Marks)

Ans: 

\( \vec{BA} = -6\hat{i} + 2\hat{j} + 3\hat{k} \)

Required unit vector of magnitude 21

\( = 21 \times \left( \frac{-6\hat{i} + 2\hat{j} + 3\hat{k}}{\sqrt{36 + 4 + 9}} \right) \)

\( = 3(-6\hat{i} + 2\hat{j} + 3\hat{k}) \) or \( -18\hat{i} + 6\hat{j} + 9\hat{k} \)

Q21. A vector $\vec{a}$ makes equal angles with all the three axes. If the magnitude of the vector is $5\sqrt{3}$ units, then find $\vec{a}$.  (2 Marks)

Ans: 

Let \( \alpha \) be the angle which the vector \( \vec{a} \) makes with all the three axes.

Then \( 3\cos^2\alpha = 1 \)

\( \Rightarrow \cos\alpha = \frac{1}{\sqrt{3}} \)

The unit vector along the vector \( \vec{a} = \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k}) \)

\( \vec{a} = 5(\hat{i} + \hat{j} + \hat{k}) \)

Q22.  If $\vec{\alpha}$ and $\vec{\beta}$ are position vectors of two points P and Q respectively, then find the position vector of a point R in QP produced such that $QR = \frac{3}{2}QP$.   (2 Marks)

Ans: 

\( \frac{QR}{QP} = \frac{3}{2} \)

Hence, R divides PQ, externally, in the ratio 1:3.

The Position vector of R = \( \vec{x} = \frac{\vec{\beta} - 3\vec{a}}{1-3} = \frac{3\vec{\alpha} - \vec{\beta}}{2} \)

Q23. If $\vec{a}$ and $\vec{b}$ are two non-collinear vectors, then find $x$, such that $\vec{\alpha} = (x-2)\vec{a} + \vec{b}$ and $\vec{\beta} = (3+2x)\vec{a} - 2\vec{b}$ are collinear.   (2 Marks)

Ans: 

\( \vec{a} \) and \( \vec{p} \) are collinear

\( \Rightarrow \frac{x-2}{3+2x} = \frac{1}{-2} \Rightarrow x = \frac{1}{4} \)

Q24. Find a vector of magnitude 5 which is perpendicular to both the vectors \(3\hat{i} - 2\hat{j} + \hat{k}\) and \(4\hat{i} + 3\hat{j} - 2\hat{k}\).  (2 Marks)

Ans: 

Let \( \vec{a} = 3\hat{i} - 2\hat{j} + \hat{k} \), \( \vec{b} = 4\hat{i} + 3\hat{j} - 2\hat{k} \)

\( \vec{a} \times \vec{b} = \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -2 & 1 \\ 4 & 3 & -2 \end{bmatrix} = \hat{i} + 10\hat{j} + 17\hat{k} \)

\( |\vec{a} \times \vec{b}| = \sqrt{1^2 + 10^2 + 17^2} = \sqrt{390} \)

Unit vector \( \hat{n} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \frac{1}{\sqrt{390}} (1\hat{i} + 10\hat{j} + 17\hat{k}) \)

∴ Required vector = \( \frac{5}{\sqrt{390}} (1\hat{i} + 10\hat{j} + 17\hat{k}) \)

Q25. Let \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) be three vectors such that \(\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c}\) and \(\vec{a} \times \vec{b} = \vec{a} \times \vec{c}\), \(\vec{a} \neq 0\). Show that \(\vec{b} = \vec{c}\).   (2 Marks)

Ans: 

\( \vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} \Rightarrow \vec{a} \cdot (\vec{b} - \vec{c}) = 0 \)

⇒ either \( \vec{b} = \vec{c} \) or \( \vec{a} \perp (\vec{b} - \vec{c}) \), since \( \vec{a} \neq 0 \)

Also, \( \vec{a} \times \vec{b} = \vec{a} \times \vec{c} \Rightarrow \vec{a} \times (\vec{b} - \vec{c}) = 0 \)

⇒ either \( \vec{b} = \vec{c} \) or \( \vec{a} \parallel (\vec{b} - \vec{c}) \), since \( \vec{a} \neq 0 \)

Since vectors \( \vec{a} \) and \( (\vec{b} - \vec{c}) \) cannot be \( \parallel \) and \( \perp \) simultaneously

Hence \( \vec{b} = \vec{c} \)

Q26. A man needs to hang two lanterns on a straight wire whose end points have coordinates A (4, 1, -2) and B (6, 2, -3). Find the coordinates of the points where he hangs the lanterns such that these points trisect the wire AB.  (2 Marks)

Ans: 

Let P and Q trisect the wire AB.

P divides AB in the ratio 1:2 then, coordinate of point P = \( \left(\frac{14}{3}, \frac{4}{3}, -\frac{7}{3}\right) \)

Q divides AB in the ratio 2:1 then, coordinate of point Q = (16/3, 5/3, -8/3)

Q27. Let \(\vec{p} = 2\hat{i} - 3\hat{j} - \hat{k}\), \(\vec{q} = -3\hat{i} + 4\hat{j} + \hat{k}\) and \(\vec{r} = \hat{i} + \hat{j} + 2\hat{k}\). Express \(\vec{r}\) in the form of \(\vec{r} = \lambda\vec{p} + \mu\vec{q}\) and hence find the values of \(\lambda\) and \(\mu\).  (2 Marks)

Ans: 

\( \vec{r} = \lambda \vec{p} + \mu \vec{q} \)

\( \Rightarrow 1 = 2\lambda - 3\mu, 1 = -3\lambda + 4\mu, 2 = -\lambda + \mu \)

\( \Rightarrow \lambda = -7, \mu = -5 \)

Q28. If \(|\vec{a}| = 2\), \(|\vec{b}| = 3\) and \(\vec{a} \cdot \vec{b} = 4\), then evaluate \(|\vec{a} + 2\vec{b}|\).  (2Marks)

Ans: 

\( (\vec{a} + 2\vec{b})^2 = |\vec{a}|^2 + |2\vec{b}|^2 + 4\vec{a} \cdot \vec{b} \)

= 56

\( \Rightarrow |\vec{a} + 2\vec{b}| = \sqrt{56} \)

Q29. If $\vec{a}$ and $\vec{b}$ are position vectors of point A and point B respectively, find the position vector of point C on BA produced such that $BC = 3BA$.  (2 Marks)

Ans: 

C divides BA in the ratio 3 : 2 externally
Required vector = \( \vec{c} = \frac{3\vec{a} - 2\vec{b}}{3-2} = 3\vec{a} - 2\vec{b} \)

Q30. Vector $\vec{r}$ is inclined at equal angles to the three axes x, y and z. If magnitude of $\vec{r}$ is $5\sqrt{3}$ units, then find $\vec{r}$. (2 Marks)

Ans: 

Unit vector equally inclined along coordinate axes is \( \frac{\hat{i}}{\sqrt{3}} + \frac{\hat{j}}{\sqrt{3}} + \frac{\hat{k}}{\sqrt{3}} \)

\( \vec{r} = 5\sqrt{3} \left( \frac{\hat{i}}{\sqrt{3}} + \frac{\hat{j}}{\sqrt{3}} + \frac{\hat{k}}{\sqrt{3}} \right) = 5\hat{i} + 5\hat{j} + 5\hat{k} \quad (\text{or } -5\hat{i} - 5\hat{j} - 5\hat{k}) \)

Q31. If $|\vec{a} + \vec{b}| = 60$, $|\vec{a} - \vec{b}| = 40$ and $|\vec{b}| = 46$, then find $|\vec{a}|$.  (2 Marks)

Ans: 

\( |\vec{a} + \vec{b}|^2 + |\vec{a} - \vec{b}|^2 = 2(|\vec{a}|^2 + |\vec{b}|^2) \)

\( (60)^2 + (40)^2 = 2(|\vec{a}|^2 + (46)^2) \)

\( \frac{(3600 + 1600)}{2} = (|\vec{a}|^2 + (46)^2) \)

2600 - 2116 = \( |\vec{a}|^2 \)

484 = \( |\vec{a}|^2 \)

 \( |\vec{a}| \) = 22

Q32. Using vectors, find the value of K such that the points (K, -11, 2), (0, -2, 2) and (2, 4, 2) are collinear.  (2 Marks)

Ans: 

Let A, B, C be the points (k, -11, 2), (0, -2, 2) and (2, 4, 2) respectively

\( \vec{AB} = -k\hat{i} + 9\hat{j} \)

\( \vec{BC} = 2\hat{i} + 6\hat{j} \)

Since \( \vec{AB} \) is parallel to \( \vec{BC} \)

\( \frac{-k}{2} = \frac{9}{6} \)

\( \Rightarrow k = -3 \)

Short Answer Questions

Q33. If \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\) such that \(|\vec{a}| = 3\), \(|\vec{b}| = 5\), \(|\vec{c}| = 7\), then find the angle between \(\vec{a}\) and \(\vec{b}\). (3 Marks)

Ans: 

Given \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \) \( \Rightarrow |\vec{a} + \vec{b}| = |\vec{c}| \)

\( \Rightarrow |\vec{a} + \vec{b}|^2 = |\vec{c}|^2 \Rightarrow |\vec{a}|^2 + |\vec{b}|^2 + 2 \vec{a} \cdot \vec{b} = |\vec{c}|^2 \)

\( \Rightarrow 9 + 25 + 2 \vec{a} \cdot \vec{b} = 49 \)

\( \Rightarrow 2 |\vec{a}| |\vec{b}| \cos \theta = 15 \)

\( \Rightarrow \cos \theta = \frac{1}{2} \therefore \theta = \frac{\pi}{3} \)

Q34. If \(\vec{a}\) and \(\vec{b}\) are unit vectors inclined with each other at an angle \(\theta\), then prove that \(\frac{1}{2} |\vec{a} - \vec{b}| = \sin \frac{\theta}{2}\).  (3 Marks)

Ans: 

\( |\vec{a}| = |\vec{b}| = 1 \)

\( |\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2 \vec{a} \cdot \vec{b} \)

= 1 + 1 - 2 \( |\vec{a}| |\vec{b}| \cos \theta \)

= \(2 - 2 \cos \theta\)

\(= 2\left(2\sin^2\frac{\theta}{2}\right)=4\sin^2\frac{\theta}{2}\)

\( \Rightarrow \sin \frac{\theta}{2} = \frac{1}{2} |\vec{a} - \vec{b}| \)

Q35. The scalar product of the vector $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$ with a unit vector along 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 1. Find the value of $\lambda$.  (3 Marks)

Ans: 

Step A: Find the sum of $\mathbf{b}$ and $\mathbf{c}$  

$$\vec{d} = \vec{b} + \vec{c} = (2 - \lambda)\hat{i} + (-4 - 2)\hat{j} + (5 - 3)\hat{k}$$

$$\vec{d} = (2 - \lambda)\hat{i} - 6\hat{j} + 2\hat{k}$$

Step B: Find the unit vector $\hat{d}$  

$$\hat{d} = \frac{\vec{d}}{|\vec{d}|} = \frac{(2 - \lambda)\hat{i} - 6\hat{j} + 2\hat{k}}{\sqrt{(2 - \lambda)^2 + (-6)^2 + 2^2}} = \frac{(2 - \lambda)\hat{i} - 6\hat{j} + 2\hat{k}}{\sqrt{(2 - \lambda)^2 + 40}}$$

Step C: Solve the scalar product $\mathbf{a} \cdot \hat{d} = 1$

Given $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$:

$$\frac{(1)(2 - \lambda) + (-1)(-6) + (2)(2)}{\sqrt{(2 - \lambda)^2 + 40}} = 1$$

$$\frac{2 - \lambda + 6 + 4}{\sqrt{(2 - \lambda)^2 + 40}} = 1 \implies 12 - \lambda = \sqrt{(2 - \lambda)^2 + 40}$$

Step D: Square both sides and solve for $\lambda$

$$(12 - \lambda)^2 = (2 - \lambda)^2 + 40$$

$$144 - 24\lambda + \lambda^2 = 4 - 4\lambda + \lambda^2 + 40$$

$$144 - 24\lambda = 44 - 4\lambda$$

$$100 = 20\lambda$$

$$\mathbf{\lambda = 5}$$

Q36. 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 the vectors $(\vec{a} - \vec{b})$ and $(\vec{c} - \vec{b})$.  (3 Marks)

Ans: 

let \( \vec{p} = \vec{a} - \vec{b} = -\hat{t} + \hat{j} + \hat{k} \)

\( \vec{q} = \vec{c} - \vec{b} = \hat{t} - 5\hat{j} - 5\hat{k} \)

Vector perpendicular to \( \vec{p} \) and \( \vec{q} = \vec{p} \times \vec{q} = \begin{bmatrix} \hat{t} & \hat{j} & \hat{k} \\ -1 & 1 & 1 \\ 1 & -5 & -5 \end{bmatrix} = -4\hat{j} + 4\hat{k} \)

Unit Vector perpendicular to \( \vec{p} \) and \( \vec{q} = \frac{-4\hat{j} + 4\hat{k}}{\sqrt{4^2 + 4^2}} \)

= \( \frac{-\hat{j} + \hat{k}}{\sqrt{2}} \quad \text{or} \quad \frac{-1}{\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k} \)

Long Answer Questions

Q37. Show that the area of a parallelogram whose diagonals are represented by \(\vec{a}\) and \(\vec{b}\) is given by \(\frac{1}{2} |\vec{a} \times \vec{b}|\). Also find the area of a parallelogram whose diagonals are \(2\hat{i} - \hat{j} + \hat{k}\) and \(\hat{i} + 3\hat{j} - \hat{k}\).  (5 Marks)

Ans: 

Let ABCD be the parallelogram with diagonals \( \vec{AB} = \vec{a} \) and \( \vec{BD} = \vec{b} \).

\( \therefore \vec{AB} = \frac{1}{2}(\vec{a} - \vec{b}) \) and \( \vec{AD} = \frac{1}{2}(\vec{a} + \vec{b}) \)

Area of ABCD

= \( |\vec{AB} \times \vec{AD}| \)

= \( \left| \frac{1}{2} (\vec{a} - \vec{b}) \times \frac{1}{2} (\vec{a} + \vec{b}) \right| \)

= \( \frac{1}{4} |\vec{a} \times \vec{a} + \vec{a} \times \vec{b} - \vec{b} \times \vec{a} - \vec{b} \times \vec{b}| \)

= \( \frac{1}{4} |\vec{a} \times \vec{b} + \vec{a} \times \vec{b}| \)

(\( \because \vec{a} \times \vec{a} = \vec{0} \))

= \( \frac{1}{4} |2(\vec{a} \times \vec{b})| \)

= \( \frac{1}{2} |\vec{a} \times \vec{b}| \)

Given \( \vec{a} = 2\hat{i} - \hat{j} + \hat{k}, \vec{b} = \hat{i} + 3\hat{j} - \hat{k} \)

$\text{Area} = \frac{1}{2} |\mathbf{a} \times \mathbf{b}| = \frac{1}{2} \sqrt{(-2)^2 + 3^2 + 7^2} = \frac{1}{2} \sqrt{4 + 9 + 49} = \dfrac{\sqrt{62}}{2}$