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3D Geometry question

2025 · 22 Jan · Shift 2 · Q31

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3D Geometry
2025 · 22 Jan · Shift 2 · Q31

3D Geometry question

2025 · 22 Jan · Shift 2 · Q31

JEE MainMathematics3D GeometryMCQ+4 / −1

The perpendicular distance, of the line

\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}

from the point

\mathrm{P}(2,-10,1)

, is :

A
$6$
B
$4 \sqrt{3}$
C
$3 \sqrt{5}$
D
$5 \sqrt{2}$

View written solutionFree

Correct answer: C

Write the line in vector form

The line is

\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}=\lambda

So a point on the line is $A(1,-2,-3)$ and its direction vector is $\vec d=(2,-1,2).$

The given point is $P(2,-10,1).$

Use the formula for distance of a point from a line in 3D

If $A$ is a point on the line and $\vec d$ is the direction vector, then

\text{Distance} = \frac{\left|\overrightarrow{AP}\times \vec d\right|}{|\vec d|}.

First,

\overrightarrow{AP}=P-A=(2-1,-10-(-2),1-(-3))=(1,-8,4).

Compute the cross product

\overrightarrow{AP}\times \vec d= \begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & -8 & 4\\ 2 & -1 & 2 \end{vmatrix}

Expanding,

=\hat i\big((-8)(2)-4(-1)\big)-\hat j\big((1)(2)-4(2)\big)+\hat k\big((1)(-1)-(-8)(2)\big)

=\hat i(-16+4)-\hat j(2-8)+\hat k(-1+16)

=(-12,6,15).

So,

\left|\overrightarrow{AP}\times \vec d\right|= \sqrt{(-12)^2+6^2+15^2} =\sqrt{144+36+225} =\sqrt{405}=9\sqrt{5}.

Compute the magnitude of direction vector

|\vec d|=\sqrt{2^2+(-1)^2+2^2}= \sqrt{4+1+4}=3.

Find the distance

\text{Distance}= rac{9\sqrt{5}}{3}=3\sqrt{5}.

Match with the options

$3\sqrt{5}$ corresponds to Option C.

Comparison with stored correct answer

Stored correct answer: C

Our derived answer: C

So they agree.

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