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3D Geometry question

2024 · 27 Jan · Shift 1 · Q37
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  5. /2024 · 27 Jan · Shift 1 · Q37

3D Geometry question

2024 · 27 Jan · Shift 1 · Q37

JEE MainMathematics3D GeometryMCQ+4 / −1
The distance, of the point (7,−2,11)(7,-2,11)(7,−2,11) from the line x−61=y−40=z−83\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}1x−6​=0y−4​=3z−8​ along the line x−52=y−1−3=z−56\frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}2x−5​=−3y−1​=6z−5​, is :
  1. A
    12
  2. B
    18
  3. C
    21
  4. D
    14
View written solutionFree

Correct answer: D

  1. Interpret the given lines

The line from which distance is to be measured is

x−61=y−40=z−83.\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}.1x−6​=0y−4​=3z−8​.

Since the denominator of yyy is 000, this means y=4y=4y=4 always. So this line can be written as

x=6+λ,y=4,z=8+3λ.x=6+\lambda,\quad y=4,\quad z=8+3\lambda.x=6+λ,y=4,z=8+3λ.

A point on this line is A=(6,4,8),A=(6,4,8),A=(6,4,8), and its direction vector is d⃗1=(1,0,3).\vec d_1=(1,0,3).d1​=(1,0,3).

The distance is to be measured along the line

x−52=y−1−3=z−56.\frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}.2x−5​=−3y−1​=6z−5​.

Let this line pass through the given point P=(7,−2,11)P=(7,-2,11)P=(7,−2,11) and meet the first line at some point QQQ. Then the required distance is PQPQPQ.

Notice that P=(7,−2,11)P=(7,-2,11)P=(7,−2,11) satisfies the second line:

7−52=1,−2−1−3=1,11−56=1.\frac{7-5}{2}=1,\quad \frac{-2-1}{-3}=1,\quad \frac{11-5}{6}=1.27−5​=1,−3−2−1​=1,611−5​=1.

So indeed PPP lies on the second line.

  1. Equation of the second line through P=(7,−2,11)P=(7,-2,11)P=(7,−2,11)

Its direction vector is

d⃗2=(2,−3,6).\vec d_2=(2,-3,6).d2​=(2,−3,6).

Hence parametric form through PPP is

x=7+2t,y=−2−3t,z=11+6t.x=7+2t,\quad y=-2-3t,\quad z=11+6t.x=7+2t,y=−2−3t,z=11+6t.
  1. Find intersection of the two lines

At the intersection point QQQ, coordinates must satisfy both lines.

From the first line, we must have y=4.y=4.y=4. From the second line, −2−3t=4.-2-3t=4.−2−3t=4. So, −3t=6  ⟹  t=−2.-3t=6 \implies t=-2.−3t=6⟹t=−2.

Now substitute into the second line:

x=7+2(−2)=3,x=7+2(-2)=3,x=7+2(−2)=3, z=11+6(−2)=−1.z=11+6(-2)=-1.z=11+6(−2)=−1.

So the candidate intersection point is Q=(3,4,−1).Q=(3,4,-1).Q=(3,4,−1).

Check whether this lies on the first line: For the first line, x=6+λ=3  ⟹  λ=−3.x=6+\lambda=3 \implies \lambda=-3.x=6+λ=3⟹λ=−3. Then z=8+3(−3)=8−9=−1,z=8+3(-3)=8-9=-1,z=8+3(−3)=8−9=−1, which matches. So the lines intersect at Q=(3,4,−1).Q=(3,4,-1).Q=(3,4,−1).

  1. Compute the required distance

The distance of point P=(7,−2,11)P=(7,-2,11)P=(7,−2,11) from the first line along the second line is simply PQ.PQ.PQ. Now,

PQ⃗=Q−P=(3−7, 4−(−2), −1−11)=(−4,6,−12).\vec{PQ}=Q-P=(3-7,\,4-(-2),\,-1-11)=(-4,6,-12).PQ​=Q−P=(3−7,4−(−2),−1−11)=(−4,6,−12).

Therefore,

PQ=(−4)2+62+(−12)2=16+36+144=196=14.PQ=\sqrt{(-4)^2+6^2+(-12)^2} =\sqrt{16+36+144} =\sqrt{196}=14.PQ=(−4)2+62+(−12)2​=16+36+144​=196​=14.
  1. Option check

Thus the required distance is 14.14.14. So the correct option is:

D: 14

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They agree.

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