Q66 / 180 · COMEDK 2025 (10 May Shift 1) · Mathematics
A line $L_1$ passing through the point A with position vector $\vec{a}=4 \hat{\imath}+2 \hat{\jmath}+2 \hat{k}$ is parallel to the vector $\vec{b}=2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k}$. The length of the perpendicular drawn from a point P with position vector $\vec{p}=\hat{\imath}+2 \hat{\jmath}+3 \hat{k}$ to $L_1$ is
- Correct answerA$\sqrt{10}$
- B$\sqrt{15}$
- C$0$
- D$2 \sqrt{3}$
Solution
The line $L_1$ passes through point $A$ with position vector $\vec{a} = 4\hat{i} + 2\hat{j} + 2\hat{k}$ and is parallel to the vector $\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}$.
The position vector of point $P$ is $\vec{p} = \hat{i} + 2\hat{j} + 3\hat{k}$.
The vector $\vec{AP} = \vec{p} - \vec{a} = (1-4)\hat{i} + (2-2)\hat{j} + (3-2)\hat{k} = -3\hat{i} + 0\hat{j} + \hat{k}$.
The length of the perpendicular from $P$ to the line $L_1$ is given by the formula $d = \dfrac{|\vec{AP} \times \vec{b}|}{|\vec{b}|}$.
First, calculate the cross product $\vec{AP} \times \vec{b}$:
$\vec{AP} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 0 & 1 \\ 2 & 3 & 6 \end{vmatrix} = \hat{i}(0 - 3) - \hat{j}(-18 - 2) + \hat{k}(-9 - 0) = -3\hat{i} + 20\hat{j} - 9\hat{k}$.
Calculate the magnitude $|\vec{AP} \times \vec{b}| = \sqrt{(-3)^2 + 20^2 + (-9)^2} = \sqrt{9 + 400 + 81} = \sqrt{490} = 7\sqrt{10}$.
Calculate the magnitude $|\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.
Therefore, $d = \dfrac{7\sqrt{10}}{7} = \sqrt{10}$.
Answer: $\sqrt{10}$
The position vector of point $P$ is $\vec{p} = \hat{i} + 2\hat{j} + 3\hat{k}$.
The vector $\vec{AP} = \vec{p} - \vec{a} = (1-4)\hat{i} + (2-2)\hat{j} + (3-2)\hat{k} = -3\hat{i} + 0\hat{j} + \hat{k}$.
The length of the perpendicular from $P$ to the line $L_1$ is given by the formula $d = \dfrac{|\vec{AP} \times \vec{b}|}{|\vec{b}|}$.
First, calculate the cross product $\vec{AP} \times \vec{b}$:
$\vec{AP} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 0 & 1 \\ 2 & 3 & 6 \end{vmatrix} = \hat{i}(0 - 3) - \hat{j}(-18 - 2) + \hat{k}(-9 - 0) = -3\hat{i} + 20\hat{j} - 9\hat{k}$.
Calculate the magnitude $|\vec{AP} \times \vec{b}| = \sqrt{(-3)^2 + 20^2 + (-9)^2} = \sqrt{9 + 400 + 81} = \sqrt{490} = 7\sqrt{10}$.
Calculate the magnitude $|\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.
Therefore, $d = \dfrac{7\sqrt{10}}{7} = \sqrt{10}$.
Answer: $\sqrt{10}$
Hence, the correct answer is option: 'A'.