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3D Geometry question

2024 · 9 Apr · Shift 2 · Q56

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3D Geometry
2024 · 9 Apr · Shift 2 · Q56

3D Geometry question

2024 · 9 Apr · Shift 2 · Q56

JEE MainMathematics3D GeometryNumerical+4 / −1

The square of the distance of the image of the point

(6,1,5)

in the line

\frac{x-1}{3}=\frac{y}{2}=\frac{z-2}{4}

, from the origin is

\underline{\hspace{2cm}}

.

Numerical answer

View written solutionFree

Correct answer: 62

Interpret the problem

We need the image of the point $P(6,1,5)$ in the line

\frac{x-1}{3}=\frac{y}{2}=\frac{z-2}{4}.

"Image in the line" means the reflection of the point across the given line in 3D.

Write the line in parametric form

Let the common value be $t$ . Then

x=1+3t,\qquad y=2t,\qquad z=2+4t.

So the line passes through

and has direction vector

Find the foot of the perpendicular from $P$ to the line

Let the foot be $M=A+t\vec d$ . Then

M=(1+3t,\,2t,\,2+4t).

Since $PM\perp \vec d$ ,

(M-P)\cdot \vec d=0.

Now,

P=(6,1,5),

so

M-P=(1+3t-6,\,2t-1,\,2+4t-5)=(-5+3t,\,2t-1,\,-3+4t).

Dot with $(3,2,4)$ :

(-5+3t)3+(2t-1)2+(-3+4t)4=0.

Simplify:

-15+9t+4t-2-12+16t=0

29t-29=0

t=1.

Hence,

M=(1+3,\,2,\,2+4)=(4,2,6).

Use midpoint property of reflection

If $P'$ is the image of $P$ in the line, then the line is the perpendicular bisector of $PP'$ , so $M$ is the midpoint of $PP'$ .

Thus,

P'=2M-P.

So,

P'=(2\cdot 4-6,\,2\cdot 2-1,\,2\cdot 6-5)=(2,3,7).

Find the square of the distance of $P'$ from the origin

Required value is

OP'^2=2^2+3^2+7^2=4+9+49=62.

Compare with stored answer

Our derived answer is $62$ , which matches the stored correct answer.

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