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3D Geometry question

2025 · 29 Jan · Shift 2 · Q43
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  5. /2025 · 29 Jan · Shift 2 · Q43

3D Geometry question

2025 · 29 Jan · Shift 2 · Q43

JEE MainMathematics3D GeometryMCQ+4 / −1
Let P be the foot of the perpendicular from the point (1,2,2)(1,2,2)(1,2,2) on the line L:x−11=y+1−1=z−22\mathrm{L}: \frac{x-1}{1}=\frac{y+1}{-1}=\frac{z-2}{2}L:1x−1​=−1y+1​=2z−2​. Let the line r⃗=(−i^+j^−2k^)+λ(i^−j^+k^),λ∈R\vec{r}=(-\hat{i}+\hat{j}-2 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}), \lambda \in \mathbf{R}r=(−i^+j^​−2k^)+λ(i^−j^​+k^),λ∈R, intersect the line L at Q . Then 2(PQ)22(\mathrm{PQ})^22(PQ)2 is equal to :
  1. A
    25
  2. B
    27
  3. C
    19
  4. D
    29
View written solutionFree

Correct answer: B

  1. Write line LLL in parametric form

Given

x−11=y+1−1=z−22=t\frac{x-1}{1}=\frac{y+1}{-1}=\frac{z-2}{2}=t1x−1​=−1y+1​=2z−2​=t

so

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

Thus a point on LLL is A=(1,−1,2),A=(1,-1,2),A=(1,−1,2), and its direction vector is d⃗=(1,−1,2).\vec d=(1,-1,2).d=(1,−1,2).

  1. Find the foot of perpendicular PPP from S=(1,2,2)S=(1,2,2)S=(1,2,2) to line LLL

A general point on LLL is P=(1+t,−1−t,2+2t).P=(1+t,-1-t,2+2t).P=(1+t,−1−t,2+2t). Then SP→=P−S=(t,−3−t,2t).\overrightarrow{SP}=P-S=(t,-3-t,2t).SP=P−S=(t,−3−t,2t). Since PPP is the foot of the perpendicular, SP→\overrightarrow{SP}SP must be perpendicular to the direction vector d⃗=(1,−1,2)\vec d=(1,-1,2)d=(1,−1,2).

So,

(t,−3−t,2t)⋅(1,−1,2)=0.(t,-3-t,2t)\cdot(1,-1,2)=0.(t,−3−t,2t)⋅(1,−1,2)=0.

Compute:

t+(3+t)+4t=0t+(3+t)+4t=0t+(3+t)+4t=0 6t+3=06t+3=06t+3=0 t=−12.t=-\frac12.t=−21​.

Hence,

P=(1−12,−1+12,2−1)=(12,−12,1).P=\left(1-\frac12,-1+\frac12,2-1\right)=\left(\frac12,-\frac12,1\right).P=(1−21​,−1+21​,2−1)=(21​,−21​,1).
  1. Find the intersection point QQQ of the given line with LLL

The second line is

r⃗=(−1,1,−2)+λ(1,−1,1).\vec r=(-1,1,-2)+\lambda(1,-1,1).r=(−1,1,−2)+λ(1,−1,1).

So its parametric form is

x=−1+λ,y=1−λ,z=−2+λ.x=-1+\lambda,\qquad y=1-\lambda,\qquad z=-2+\lambda.x=−1+λ,y=1−λ,z=−2+λ.

At intersection with LLL,

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

From (1),

λ=t+2.\lambda=t+2.λ=t+2.

Substitute into (3):

2+2t=−2+(t+2)=t.2+2t=-2+(t+2)=t.2+2t=−2+(t+2)=t.

Thus,

2+2t=t  ⟹  t=−2.2+2t=t \implies t=-2.2+2t=t⟹t=−2.

Then

λ=0.\lambda=0.λ=0.

So the intersection point is

Q=(−1,1,−2).Q=(-1,1,-2).Q=(−1,1,−2).
  1. Find PQPQPQ
PQ→=Q−P=(−1−12, 1+12, −2−1)=(−32,32,−3).\overrightarrow{PQ}=Q-P=\left(-1-\frac12,\,1+\frac12,\,-2-1\right)=\left(-\frac32,\frac32,-3\right).PQ​=Q−P=(−1−21​,1+21​,−2−1)=(−23​,23​,−3).

Therefore,

(PQ)2=(−32)2+(32)2+(−3)2(PQ)^2=\left(-\frac32\right)^2+\left(\frac32\right)^2+(-3)^2(PQ)2=(−23​)2+(23​)2+(−3)2 =94+94+9=184+9=92+9=272.=\frac94+\frac94+9=\frac{18}{4}+9=\frac92+9=\frac{27}{2}.=49​+49​+9=418​+9=29​+9=227​.

Hence,

2(PQ)2=2⋅272=27.2(PQ)^2=2\cdot\frac{27}{2}=27.2(PQ)2=2⋅227​=27.
  1. Match with options
2(PQ)2=272(PQ)^2=272(PQ)2=27

So the correct option is:

B: 27

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