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3D Geometry question

2013 · Shift 1 · Q28

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PYQ Library
JEE Advanced
Mathematics
3D Geometry
2013 · Shift 1 · Q28

3D Geometry question

2013 · Shift 1 · Q28

JEE AdvancedMathematics3D GeometryMCQ+3 / −1

Perpendiculars are drawn from points on the line

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

to the plane

x+y+z=3

. The foot of perpendiculars lie on the line

A
$\frac{x}{5}=\frac{y-1}{8}=\frac{z-2}{-13}$
B
$\frac{x}{2}=\frac{y-1}{3}=\frac{z-2}{-5}$
C
$\frac{x}{4}=\frac{y-1}{3}=\frac{z-2}{-7}$
D
$\frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5}$

View written solutionFree

Correct answer: D

Parametrize the given line

The line

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

has parametric form

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

So a general point on the line is

P(2t-2,-t-1,3t).

Equation of the perpendicular to the plane

The plane is

x+y+z=3.

Its normal vector is

\vec n=(1,1,1).

Hence the perpendicular from any point to the plane will be along direction $(1,1,1)$ .

So if $Q$ is the foot of perpendicular from $P$ , then

Q=P+\lambda(1,1,1).

Thus

Q=(2t-2+\lambda, -t-1+\lambda, 3t+\lambda).

Use the fact that $Q$ lies on the plane

Since $Q$ lies on $x+y+z=3$ ,

(2t-2+\lambda)+(-t-1+\lambda)+(3t+\lambda)=3.

Simplifying,

4t-3+3\lambda=3,

so

3\lambda=6-4t

\lambda=2-\frac{4t}{3}.

Coordinates of the foot $Q$

Substitute $\lambda$ into coordinates of $Q$ :

x=2t-2+2-\frac{4t}{3}=\frac{2t}{3},

y=-t-1+2-\frac{4t}{3}=1-\frac{7t}{3},

z=3t+2-\frac{4t}{3}=2+\frac{5t}{3}.

So

Q\left(\frac{2t}{3},\;1-\frac{7t}{3},\;2+\frac{5t}{3}\right).

Write the locus as a line

Let

s=\frac{t}{3}.

Then

x=2s,\qquad y=1-7s,\qquad z=2+5s.

Hence the line is

\frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5}.

Match with the options

This is exactly Option D.

Verification with stored answer

Stored correct answer: D

Our derived answer: D

They match.

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