Class 12 Vector Algebra PYQs

Q1 2008 Board

00:00

Find the scalar product $(\hat{i} + \hat{j} + \hat{k}) \cdot (2\hat{i} - \hat{j} + \hat{k})$. 1 Mark

Step 1: Apply Dot Product Formula

$\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z$.

Step 2: Calculate

$= (1)(2) + (1)(-1) + (1)(1) = 2 - 1 + 1 = 2$.

2

Q2 2008 Board

00:00

Find the angle between the vectors $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = \hat{i} - \hat{j}$. 2 Marks

Step 1: Find Dot Product

$\vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) = 0$.

Step 2: Conclude Angle

Since $\vec{a} \cdot \vec{b} = 0$, the vectors are perpendicular.

Angle $\theta = 90^\circ$ or $\pi/2$.

π/2

Q3 2009 Board

00:00

Find the projection of vector $2\hat{i} + 3\hat{j} + 6\hat{k}$ on the vector $\hat{i} + 2\hat{j} + 2\hat{k}$. 2 Marks

Step 1: Label Vectors

Let $\vec{a} = 2\hat{i} + 3\hat{j} + 6\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 2\hat{k}$.

Step 2: Apply Projection Formula

Projection of $\vec{a}$ on $\vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.

Step 3: Calculate Dot Product

$\vec{a} \cdot \vec{b} = (2)(1) + (3)(2) + (6)(2) = 2 + 6 + 12 = 20$.

Step 4: Calculate Magnitude of b

$|\vec{b}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3$.

Step 5: Final Result

Projection $= 20/3$.

20/3

Q4 2010 Board

00:00

Find the scalar product $(2\hat{i} - \hat{j} + \hat{k}) \cdot (\hat{i} + 2\hat{j} + 3\hat{k})$. 1 Mark

$= (2)(1) + (-1)(2) + (1)(3) = 2 - 2 + 3 = 3$.

3

Q5 2012 Board

00:00

Find the angle between the vectors $\hat{i} + 2\hat{j} + 2\hat{k}$ and $2\hat{i} - \hat{j} + 2\hat{k}$. 2 Marks

Step 1: Calculate Dot Product

$\vec{a} \cdot \vec{b} = (1)(2) + (2)(-1) + (2)(2) = 2 - 2 + 4 = 4$.

Step 2: Calculate Magnitudes

$|\vec{a}| = 3, |\vec{b}| = 3$.

Step 3: Apply Cosine Formula

$\cos \theta = \frac{4}{3 \times 3} = 4/9$.

$\theta = \cos^{-1}(4/9)$.

cos⁻¹(4/9)

Q6 2015 Board

00:00

Show that the vectors $\hat{i} + \hat{j}$ and $\hat{i} - \hat{j}$ are orthogonal. 2 Marks

Step 1: Calculate Dot Product

$\vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) = 0$.

Step 2: Conclude

Since the scalar product is zero, the vectors are orthogonal.

Vectors are orthogonal

Q7 2019 Board

00:00

Prove that $\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}$. 3 Marks

Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$.

Let $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$ and $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$.

$\vec{b} + \vec{c} = (b_1+c_1)\hat{i} + (b_2+c_2)\hat{j} + (b_3+c_3)\hat{k}$.

$\vec{a} \cdot (\vec{b} + \vec{c}) = a_1(b_1+c_1) + a_2(b_2+c_2) + a_3(b_3+c_3)$.

$= (a_1b_1 + a_2b_2 + a_3b_3) + (a_1c_1 + a_2c_2 + a_3c_3) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}$.

Hence Proved

Q8 2018 Board

00:00

Find the projection of the vector $2\hat{i} + 3\hat{j} + 4\hat{k}$ on the vector $\hat{i} - \hat{j} + 2\hat{k}$. 2 Marks

$\vec{a} \cdot \vec{b} = (2)(1) + (3)(-1) + (4)(2) = 2 - 3 + 8 = 7$.

$|\vec{b}| = \sqrt{1^2 + (-1)^2 + 2^2} = \sqrt{6}$.

Projection $= 7/\sqrt{6}$.

7/√6

Q9 2017 Board

00:00

Find the angle between the vectors $\hat{i} + \hat{j} + \hat{k}$ and $\hat{i} - \hat{j} + \hat{k}$. 2 Marks

$\vec{a} \cdot \vec{b} = 1 - 1 + 1 = 1$.

$|\vec{a}| = \sqrt{3}, |\vec{b}| = \sqrt{3}$.

$\cos \theta = 1/3 \Rightarrow \theta = \cos^{-1}(1/3)$.

cos⁻¹(1/3)

Q10 2014 Board

00:00

Find the projection of the vector $3\hat{i} + 4\hat{j} + 5\hat{k}$ on the vector $\hat{i} + \hat{j} + \hat{k}$. 2 Marks

$\vec{a} \cdot \vec{b} = 3 + 4 + 5 = 12$.

$|\vec{b}| = \sqrt{3}$.

Projection $= 12/\sqrt{3} = 4\sqrt{3}$.

4√3

Q11 2016 Board

00:00

If vectors $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - 2\hat{j} + 3\hat{k}$ are perpendicular, find the value of $\vec{a} \cdot \vec{b}$. 1 Mark

By definition, if two vectors are perpendicular, their scalar product is zero.

$\vec{a} \cdot \vec{b} = 0$.

0

Q12 2019 Board

00:00

Find the scalar projection of the vector $2\hat{i} + 3\hat{j} + 6\hat{k}$ on the vector $\hat{i} + 2\hat{j} + 2\hat{k}$. 2 Marks

Scalar projection $= \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.

$\vec{a} \cdot \vec{b} = 2 + 6 + 12 = 20$.

$|\vec{b}| = 3$.

Scalar projection $= 20/3$.

20/3

Q13 2026 Board

00:00

The value of p for which vectors $\hat{i}$ + $2\hat{j}$ + $3\hat{k}$ and $2\hat{i}$ − $p\hat{j}$ + $\hat{k}$ are perpendicular to each other is ______. 1 Mark