CUET PG MCA Vector Previous Year Questions (PYQs) – Page 1 of 2

Solution:

Given two unit vectors \( \vec{a} \) and \( \vec{b} \), and the vectors \( \vec{a} + 2\vec{b} \) and \( 5\vec{a} - 4\vec{b} \) are perpendicular, we use the condition for perpendicular vectors: $$ (\vec{a} + 2\vec{b}) \cdot (5\vec{a} - 4\vec{b}) = 0 $$ Expanding the dot product: $$ (\vec{a} + 2\vec{b}) \cdot (5\vec{a} - 4\vec{b}) = \vec{a} \cdot 5\vec{a} + \vec{a} \cdot (-4\vec{b}) + 2\vec{b} \cdot 5\vec{a} + 2\vec{b} \cdot (-4\vec{b}) $$ Using properties of dot products and knowing \( \vec{a} \) and \( \vec{b} \) are unit vectors (\( \vec{a} \cdot \vec{a} = 1 \) and \( \vec{b} \cdot \vec{b} = 1 \)): $$ 5(\vec{a} \cdot \vec{a}) - 4(\vec{a} \cdot \vec{b}) + 10(\vec{b} \cdot \vec{a}) - 8(\vec{b} \cdot \vec{b}) = 0 $$ Simplifying: $$ 5(1) - 4(\vec{a} \cdot \vec{b}) + 10(\vec{a} \cdot \vec{b}) - 8(1) = 0 $$ $$ 5 - 8 + 6(\vec{a} \cdot \vec{b}) = 0 $$ $$ -3 + 6(\vec{a} \cdot \vec{b}) = 0 $$ $$ 6(\vec{a} \cdot \vec{b}) = 3 $$ $$ \vec{a} \cdot \vec{b} = \frac{1}{2} $$ The dot product \( \vec{a} \cdot \vec{b} = \cos \theta \), where \( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \): $$ \cos \theta = \frac{1}{2} $$ Therefore, the angle \( \theta \) is: $$ \theta = \cos^{-1} \left( \frac{1}{2} \right) = 60^\circ $$
Final Answer:
$$ \boxed{60^\circ} $$

Solution:

Given vectors:

• \(\vec{a} = \hat{i} - \hat{j}\)
• \(\vec{b} = \hat{i} + \hat{j} + \hat{k}\)
• And \((\vec{a} \times \vec{c}) + \vec{b} = 0\)
• \(\vec{a} \cdot \vec{c} = 4\)

From \((\vec{a} \times \vec{c}) + \vec{b} = 0\), we get: \[ \vec{a} \times \vec{c} = -\vec{b} \] Let \(\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}\). The cross product \(\vec{a} \times \vec{c}\) is: \[ \vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ x & y & z \end{vmatrix} \] Expanding this determinant: \[ \vec{a} \times \vec{c} = (z \hat{i} + z \hat{j} + (x + y) \hat{k}) \] Setting \(\vec{a} \times \vec{c} = -\vec{b}\), we get: \[ z = -1, \quad z = -1, \quad x + y = -1 \] Therefore: \[ x + y = -1 \] Now, from \(\vec{a} \cdot \vec{c} = 4\): \[ \vec{a} \cdot \vec{c} = 1 \cdot x + (-1) \cdot y = 4 \] Simplifying: \[ x - y = 4 \] Solving the system of equations: \[ x + y = -1 \] \[ x - y = 4 \] Adding the two equations: \[ 2x = 3 \quad \Rightarrow \quad x = \frac{3}{2} \] Substituting into \(x + y = -1\): \[ \frac{3}{2} + y = -1 \quad \Rightarrow \quad y = -\frac{5}{2} \] Now, \(\vec{c} = \frac{3}{2} \hat{i} - \frac{5}{2} \hat{j} - \hat{k}\). To find \(|\vec{c}|^2\), we compute: \[ |\vec{c}|^2 = \left( \frac{3}{2} \right)^2 + \left( -\frac{5}{2} \right)^2 + (-1)^2 = \frac{9}{4} + \frac{25}{4} + 1 \] \[ |\vec{c}|^2 = \frac{9 + 25 + 4}{4} = \frac{38}{4} = 9.5 \]
Final Answer:
$$ \boxed{9.5} $$

$|\vec a - \vec b| = \sqrt{2-2\cos 2\theta}$

Given:

$\sqrt{2-2\cos 2\theta} < 1$

Squaring:

$2-2\cos 2\theta < 1$

$\cos 2\theta > \dfrac{1}{2}$

So,

$0 \le 2\theta < \dfrac{\pi}{3}$

$\Rightarrow 0 \le \theta < \dfrac{\pi}{6}$

From given options, valid interval is included in

$\left[0,\dfrac{\pi}{2}\right]$