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Question #41642

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The Question

Let $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ , $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$ , and $\vec{c} = \hat{i} + \hat{j} + 2\hat{k}$ . Let $\hat{n} = l\hat{i} + m\hat{j} + n\hat{k}$ be a unit vector parallel to $\vec{v} = \vec{a} \times (\vec{b} \times \vec{c}) - 4\hat{i} - 3\hat{j} + \hat{k}$ such that $m > 0$ . A plane $\Pi$ passes through the point $P(3, 5, -2)$ and has $\hat{n}$ as its normal vector. The perpendicular distance from the point $Q(l, m, n)$ to the plane $\Pi$ is:

Options

A.
$1$
B.
$4$
C.
$\frac{39 - \sqrt{42}}{\sqrt{42}}$
D.
$\frac{15 - \sqrt{6}}{\sqrt{6}}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Distance from point to plane|Vector Triple Product Expansion|Direction Cosines constraint

Why This Question Is Tricky

Common Mistakes

This question tests the expansion of the Vector Triple Product (BAC-CAB rule) and the strict application of the Direction Cosines constraint ( $l^2+m^2+n^2=1$ ).

A common trap is expanding the VTP incorrectly as $(\vec{a} \cdot \vec{b})\vec{c} - (\vec{a} \cdot \vec{c})\vec{b}$ , which leads to a completely different normal vector and distance (Option C).

Another major trap is failing to normalize $\vec{v}$ before identifying the coordinates of $Q(l, m, n)$ .

If a student uses the raw components of $\vec{v}$ $(0, 4, 0)$ for $Q$ , they will incorrectly calculate the distance as 1 (Option A), bypassing the fundamental definition of direction cosines.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Three-Dimensional Geometry - Class 12
Concept	Distance from point to plane
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free