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3D Geometry question

2025 · 29 Jan · Shift 2 · Q28
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  5. /2025 · 29 Jan · Shift 2 · Q28

3D Geometry question

2025 · 29 Jan · Shift 2 · Q28

JEE MainMathematics3D GeometryMCQ+4 / −1
Let a straight line LLL pass through the point P(2,−1,3)P(2, -1, 3)P(2,−1,3) and be perpendicular to the lines x−12=y+11=z−3−2\frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 3}{-2}2x−1​=1y+1​=−2z−3​ and x−31=y−23=z+24\frac{x - 3}{1} = \frac{y - 2}{3} = \frac{z + 2}{4}1x−3​=3y−2​=4z+2​. If the line LLL intersects the yzyzyz-plane at the point QQQ, then the distance between the points PPP and QQQ is:
  1. A
    10\sqrt{10}10​
  2. B
    222
  3. C
    232\sqrt{3}23​
  4. D
    333
View written solutionFree

Correct answer: D

  1. Identify direction vectors of the given lines

The lines are

x−12=y+11=z−3−2\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-3}{-2}2x−1​=1y+1​=−2z−3​

and

x−31=y−23=z+24.\frac{x-3}{1}=\frac{y-2}{3}=\frac{z+2}{4}.1x−3​=3y−2​=4z+2​.

So their direction vectors are:

d⃗1=(2,1,−2),d⃗2=(1,3,4).\vec d_1=(2,1,-2), \qquad \vec d_2=(1,3,4).d1​=(2,1,−2),d2​=(1,3,4).

Since line LLL is perpendicular to both lines, its direction vector must be perpendicular to both d⃗1\vec d_1d1​ and d⃗2\vec d_2d2​. Hence it is parallel to

d⃗1×d⃗2.\vec d_1 \times \vec d_2.d1​×d2​.
  1. Find the direction vector of line LLL using cross product
d⃗1×d⃗2=∣i^j^k^21−2134∣\vec d_1 \times \vec d_2= \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 1 & -2 \\ 1 & 3 & 4 \end{vmatrix}d1​×d2​=​i^21​j^​13​k^−24​​ =i^(1⋅4−(−2)⋅3)−j^(2⋅4−(−2)⋅1)+k^(2⋅3−1⋅1)=\hat i(1\cdot 4-(-2)\cdot 3)-\hat j(2\cdot 4-(-2)\cdot 1)+\hat k(2\cdot 3-1\cdot 1)=i^(1⋅4−(−2)⋅3)−j^​(2⋅4−(−2)⋅1)+k^(2⋅3−1⋅1) =i^(4+6)−j^(8+2)+k^(6−1)=(10,−10,5).=\hat i(4+6)-\hat j(8+2)+\hat k(6-1) =(10,-10,5).=i^(4+6)−j^​(8+2)+k^(6−1)=(10,−10,5).

This simplifies to

(2,−2,1).(2,-2,1).(2,−2,1).

So line LLL passes through P(2,−1,3)P(2,-1,3)P(2,−1,3) with direction vector (2,−2,1)(2,-2,1)(2,−2,1).

  1. Write the parametric equation of line LLL
(x,y,z)=(2,−1,3)+t(2,−2,1).(x,y,z)=(2,-1,3)+t(2,-2,1).(x,y,z)=(2,−1,3)+t(2,−2,1).

Thus,

x=2+2t,y=−1−2t,z=3+t.x=2+2t, \qquad y=-1-2t, \qquad z=3+t.x=2+2t,y=−1−2t,z=3+t.
  1. Find the point where LLL intersects the yzyzyz-plane

The yzyzyz-plane is given by

x=0.x=0.x=0.

So,

2+2t=0  ⟹  t=−1.2+2t=0 \implies t=-1.2+2t=0⟹t=−1.

Hence the point of intersection is

Q=(0,−1−2(−1),3+(−1))=(0,1,2).Q=(0, -1-2(-1), 3+(-1))=(0,1,2).Q=(0,−1−2(−1),3+(−1))=(0,1,2).
  1. Find the distance PQPQPQ
PQ=(0−2)2+(1−(−1))2+(2−3)2PQ=\sqrt{(0-2)^2+(1-(-1))^2+(2-3)^2}PQ=(0−2)2+(1−(−1))2+(2−3)2​ =(−2)2+22+(−1)2=4+4+1=9=3.=\sqrt{(-2)^2+2^2+(-1)^2} = \sqrt{4+4+1}=\sqrt{9}=3.=(−2)2+22+(−1)2​=4+4+1​=9​=3.
  1. Compare with the options

Thus,

PQ=3.PQ=3.PQ=3.

So the correct option is:

D: 333

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer is also D. So they agree.

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