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3D Geometry question

2024 · 9 Apr · Shift 2 · Q46
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3D Geometry question

2024 · 9 Apr · Shift 2 · Q46

JEE MainMathematics3D GeometryMCQ+4 / −1
Consider the line L\mathrm{L}L passing through the points (1,2,3)(1,2,3)(1,2,3) and (2,3,5)(2,3,5)(2,3,5). The distance of the point (113,113,193)\left(\frac{11}{3}, \frac{11}{3}, \frac{19}{3}\right)(311​,311​,319​) from the line L\mathrm{L}L along the line 3x−112=3y−111=3z−192\frac{3 x-11}{2}=\frac{3 y-11}{1}=\frac{3 z-19}{2}23x−11​=13y−11​=23z−19​ is equal to
  1. A
    6
  2. B
    3
  3. C
    5
  4. D
    4
View written solutionFree

Correct answer: B

  1. Equation of the given line LLL

The line LLL passes through (1,2,3)(1,2,3)(1,2,3) and (2,3,5)(2,3,5)(2,3,5).

So its direction vector is

(2−1,3−2,5−3)=(1,1,2).(2-1, 3-2, 5-3)=(1,1,2).(2−1,3−2,5−3)=(1,1,2).

Hence a parametric form of LLL is

(x,y,z)=(1,2,3)+λ(1,1,2).(x,y,z)=(1,2,3)+\lambda(1,1,2).(x,y,z)=(1,2,3)+λ(1,1,2).

So,

x=1+λ,y=2+λ,z=3+2λ.x=1+\lambda, \quad y=2+\lambda, \quad z=3+2\lambda.x=1+λ,y=2+λ,z=3+2λ.
  1. Equation of the line along which distance is measured

Given:

3x−112=3y−111=3z−192.\frac{3x-11}{2}=\frac{3y-11}{1}=\frac{3z-19}{2}.23x−11​=13y−11​=23z−19​.

Let the common value be ttt. Then

3x−11=2t,3y−11=t,3z−19=2t.3x-11=2t, \quad 3y-11=t, \quad 3z-19=2t.3x−11=2t,3y−11=t,3z−19=2t.

Thus,

x=11+2t3,y=11+t3,z=19+2t3.x=\frac{11+2t}{3}, \quad y=\frac{11+t}{3}, \quad z=\frac{19+2t}{3}.x=311+2t​,y=311+t​,z=319+2t​.

This line passes through the point

P=(113,113,193)P=\left(\frac{11}{3},\frac{11}{3},\frac{19}{3}\right)P=(311​,311​,319​)

when t=0t=0t=0, and has direction vector

(23,13,23)\left(\frac{2}{3},\frac{1}{3},\frac{2}{3}\right)(32​,31​,32​)

which is proportional to (2,1,2)(2,1,2)(2,1,2).

So the required distance is the length from PPP to the point where this line meets LLL.

  1. Find intersection of the two lines

A general point on the second line is

(11+2t3,11+t3,19+2t3).\left(\frac{11+2t}{3},\frac{11+t}{3},\frac{19+2t}{3}\right).(311+2t​,311+t​,319+2t​).

This must lie on LLL, so for some λ\lambdaλ,

1+λ=11+2t3...(1)1+\lambda=\frac{11+2t}{3} \quad ...(1)1+λ=311+2t​...(1) 2+λ=11+t3...(2)2+\lambda=\frac{11+t}{3} \quad ...(2)2+λ=311+t​...(2) 3+2λ=19+2t3...(3)3+2\lambda=\frac{19+2t}{3} \quad ...(3)3+2λ=319+2t​...(3)

From (1):

3+3λ=11+2t3+3\lambda=11+2t3+3λ=11+2t 3λ=8+2t3\lambda=8+2t3λ=8+2t λ=8+2t3.\lambda=\frac{8+2t}{3}. λ=38+2t​.

From (2):

6+3λ=11+t6+3\lambda=11+t6+3λ=11+t 3λ=5+t3\lambda=5+t3λ=5+t λ=5+t3.\lambda=\frac{5+t}{3}.λ=35+t​.

Equating:

8+2t3=5+t3\frac{8+2t}{3}=\frac{5+t}{3}38+2t​=35+t​ 8+2t=5+t8+2t=5+t8+2t=5+t t=−3.t=-3.t=−3.

Then

λ=5+(−3)3=23.\lambda=\frac{5+(-3)}{3}=\frac{2}{3}.λ=35+(−3)​=32​.

Check in (3):

3+2⋅23=133,3+2\cdot\frac{2}{3}=\frac{13}{3},3+2⋅32​=313​,

while

19+2(−3)3=133,\frac{19+2(-3)}{3}=\frac{13}{3},319+2(−3)​=313​,

so it is consistent.

Thus the intersection point is obtained on the second line at t=−3t=-3t=−3.

  1. Compute the distance from PPP to this intersection along the given line

On the second line,

P=(113,113,193)P=\left(\frac{11}{3},\frac{11}{3},\frac{19}{3}\right)P=(311​,311​,319​)

corresponds to t=0t=0t=0.

The intersection point corresponds to t=−3t=-3t=−3.

The direction vector of the line is

(23,13,23).\left(\frac{2}{3},\frac{1}{3},\frac{2}{3}\right).(32​,31​,32​).

So the displacement from t=0t=0t=0 to t=−3t=-3t=−3 is

−3(23,13,23)=(−2,−1,−2).-3\left(\frac{2}{3},\frac{1}{3},\frac{2}{3}\right)=(-2,-1,-2).−3(32​,31​,32​)=(−2,−1,−2).

Hence the required distance is

(−2)2+(−1)2+(−2)2=4+1+4=9=3.\sqrt{(-2)^2+(-1)^2+(-2)^2} =\sqrt{4+1+4} =\sqrt{9}=3.(−2)2+(−1)2+(−2)2​=4+1+4​=9​=3.
  1. Match with options

Thus the distance is

3.\boxed{3}.3​.

So the correct option is B.

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