GATE BT 2024 | Question: 22

Question

Let $\overrightarrow{O R}$ be the vector that is perpendicular to the vectors $\overrightarrow{O P}=2 \hat{\imath}-3 \hat{\jmath}+\hat{k}$ and $\overrightarrow{O Q}=-2 \hat{\imath}+\hat{\jmath}+\hat{k}$. If the length of the vector $\overrightarrow{O R}$ is $\alpha \sqrt{3}$, then $\alpha$ is $\_\_\_\_\_\_\_$.

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

The vector $\vec{OR}$ perpendicular to two vectors $\vec{OP}$ and $\vec{OQ}$ is found using the cross product
\[ \vec{OR} = \vec{OP} \times \vec{OQ} \]
The magnitude of this vector is given by $|\vec{OR}| = \sqrt{x^2 + y^2 + z^2}$.

Given $\vec{OP} = 2\hat{i} - 3\hat{j} + \hat{k}$ and $\vec{OQ} = -2\hat{i} + \hat{j} + \hat{k}$:
\[ \vec{OR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 1 \\ -2 & 1 & 1 \end{vmatrix} \]
\[ \vec{OR} = \hat{i}[(-3)(1) - (1)(1)] - \hat{j}[(2)(1) - (1)(-2)] + \hat{k}[(2)(1) - (-3)(-2)] \]
\[ \vec{OR} = \hat{i}[-4] - \hat{j}[4] + \hat{k}[-4] = -4\hat{i} - 4\hat{j} - 4\hat{k} \]

Finding the value of } $\alpha$:
Calculate the length (magnitude) of $\vec{OR}$:
\[ |\vec{OR}| = \sqrt{(-4)^2 + (-4)^2 + (-4)^2} = \sqrt{16 + 16 + 16} = \sqrt{48} \]
\[ |\vec{OR}| = \sqrt{16 \times 3} = 4\sqrt{3} \]
Given the length is $\alpha\sqrt{3}$, comparing both values:
\[ \alpha\sqrt{3} = 4\sqrt{3} \implies \alpha = 4 \]