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3D Geometry question

2024 · 4 Apr · Shift 2 · Q35
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  5. /2024 · 4 Apr · Shift 2 · Q35

3D Geometry question

2024 · 4 Apr · Shift 2 · Q35

JEE MainMathematics3D GeometryMCQ+4 / −1
Let P\mathrm{P}P be the point of intersection of the lines x−21=y−45=z−21\frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}1x−2​=5y−4​=1z−2​ and x−32=y−23=z−32\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}2x−3​=3y−2​=2z−3​. Then, the shortest distance of P\mathrm{P}P from the line 4x=2y=z4 x=2 y=z4x=2y=z is
  1. A
    3147\frac{3 \sqrt{14}}{7}7314​​
  2. B
    5147\frac{5 \sqrt{14}}{7}7514​​
  3. C
    147\frac{\sqrt{14}}{7}714​​
  4. D
    6147\frac{6 \sqrt{14}}{7}7614​​
View written solutionFree

Correct answer: A

  1. Write the two lines in parametric form

The first line is

x−21=y−45=z−21=λ\frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}=\lambda1x−2​=5y−4​=1z−2​=λ

So,

x=2+λ,y=4+5λ,z=2+λx=2+\lambda,\quad y=4+5\lambda,\quad z=2+\lambdax=2+λ,y=4+5λ,z=2+λ

The second line is

x−32=y−23=z−32=μ\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}=\mu2x−3​=3y−2​=2z−3​=μ

So,

x=3+2μ,y=2+3μ,z=3+2μx=3+2\mu,\quad y=2+3\mu,\quad z=3+2\mux=3+2μ,y=2+3μ,z=3+2μ
  1. Find their point of intersection PPP

At the intersection point,

2+λ=3+2μ...(1)2+\lambda=3+2\mu \quad ...(1)2+λ=3+2μ...(1) 4+5λ=2+3μ...(2)4+5\lambda=2+3\mu \quad ...(2)4+5λ=2+3μ...(2) 2+λ=3+2μ...(3)2+\lambda=3+2\mu \quad ...(3)2+λ=3+2μ...(3)

Equation (3) is same as (1), so solve (1) and (2).

From (1):

λ−2μ=1\lambda-2\mu=1λ−2μ=1

So,

λ=1+2μ\lambda=1+2\muλ=1+2μ

Substitute into (2):

4+5(1+2μ)=2+3μ4+5(1+2\mu)=2+3\mu4+5(1+2μ)=2+3μ 4+5+10μ=2+3μ4+5+10\mu=2+3\mu4+5+10μ=2+3μ 9+10μ=2+3μ9+10\mu=2+3\mu9+10μ=2+3μ 7μ=−77\mu=-77μ=−7 μ=−1\mu=-1μ=−1

Then,

λ=1+2(−1)=−1\lambda=1+2(-1)=-1λ=1+2(−1)=−1

Hence point of intersection is

P=(2−1, 4+5(−1), 2−1)=(1,−1,1)P=(2-1,\,4+5(-1),\,2-1)=(1,-1,1)P=(2−1,4+5(−1),2−1)=(1,−1,1)
  1. Write the given line in standard form

Given line:

4x=2y=z4x=2y=z4x=2y=z

Let the common value be ttt. Then

4x=t,2y=t,z=t4x=t,\quad 2y=t,\quad z=t4x=t,2y=t,z=t

So,

x=t4,y=t2,z=tx=\frac t4,\quad y=\frac t2,\quad z=tx=4t​,y=2t​,z=t

This is a line through the origin with direction ratios

(14,12,1)\left(\frac14,\frac12,1\right)(41​,21​,1)

which are proportional to

(1,2,4)(1,2,4)(1,2,4)

So the line passes through O=(0,0,0)O=(0,0,0)O=(0,0,0) and has direction vector

d⃗=(1,2,4)\vec d=(1,2,4)d=(1,2,4)
  1. Use distance formula from a point to a line in 3D

Distance from point PPP to line through origin with direction vector d⃗\vec dd is

Distance=∣OP⃗×d⃗∣∣d⃗∣\text{Distance}=\frac{|\vec{OP}\times \vec d|}{|\vec d|}Distance=∣d∣∣OP×d∣​

Here,

OP⃗=(1,−1,1)\vec{OP}=(1,-1,1)OP=(1,−1,1)

and

d⃗=(1,2,4)\vec d=(1,2,4)d=(1,2,4)

Compute cross product:

OP⃗×d⃗=∣i^j^k^1−11124∣\vec{OP}\times \vec d= \begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & -1 & 1\\ 1 & 2 & 4 \end{vmatrix}OP×d=​i^11​j^​−12​k^14​​ =i^((−1)(4)−1⋅2)−j^(1⋅4−1⋅1)+k^(1⋅2−(−1)⋅1)=\hat i((-1)(4)-1\cdot 2)-\hat j(1\cdot 4-1\cdot 1)+\hat k(1\cdot 2-(-1)\cdot 1)=i^((−1)(4)−1⋅2)−j^​(1⋅4−1⋅1)+k^(1⋅2−(−1)⋅1) =i^(−4−2)−j^(4−1)+k^(2+1)=\hat i(-4-2)-\hat j(4-1)+\hat k(2+1)=i^(−4−2)−j^​(4−1)+k^(2+1) =(−6,−3,3)=(-6,-3,3)=(−6,−3,3)

So,

∣OP⃗×d⃗∣=(−6)2+(−3)2+32=36+9+9=54=36|\vec{OP}\times \vec d|=\sqrt{(-6)^2+(-3)^2+3^2}= \sqrt{36+9+9}=\sqrt{54}=3\sqrt{6}∣OP×d∣=(−6)2+(−3)2+32​=36+9+9​=54​=36​

Also,

∣d⃗∣=12+22+42=21|\vec d|=\sqrt{1^2+2^2+4^2}=\sqrt{21}∣d∣=12+22+42​=21​

Therefore,

Distance=3621=327\text{Distance}=\frac{3\sqrt6}{\sqrt{21}}=3\sqrt{\frac{2}{7}}Distance=21​36​​=372​​

Rationalizing,

Distance=3147\text{Distance}=\frac{3\sqrt{14}}{7}Distance=7314​​
  1. Match with options
3147\frac{3\sqrt{14}}{7}7314​​

This is Option A.

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