The shortest distance between the lines (x-3)/(2)=(y+15)/(-7)=(z-9)/(5) and (x+1)/(2)=(y-1)/(1)=(z-9)/(-3) is

Correct answer: B. 4 3 aligned & (x-3)/(2)=(y+15)/(-7)=(z-9)/(5) & (x+1)/(2)=(y-1)/(1)=(z-9)/(-3) & S.D = |( a_2 · a_1) ·( b_1 · b_2)| | b_1 × b_2| & a_1=3,-15,9 & b_1=2,-7,5 & a_2=-1,1,9 & b_2=2,1,-3 & a_2-a_1=-4,16,0 & b_1 × b_2=| array ccc i & j & k 2 & -7 & 5 2 & 1 & -3 array |= i (16)- j (-16)+ k (16) & 16( i + j + k ) & | b_1 × b_2|=16 3 & ( a_2- a_1) ·( b_1- b_2)=16[-4+16]=(16)(12) & S.D. = (16)(12) 16 3 =4 3 & aligned

JEE Main 20246 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 6 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Three Dimensional Geometry

JEE Main 2024 · 6 April · Morning Shift · Mathematics · Three Dimensional Geometry

The shortest distance between the lines...

The shortest distance between the lines

\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}

and

\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}

Options

A $8 \sqrt{3}$
B $4 \sqrt{3}$
C $5 \sqrt{3}$
D $6 \sqrt{3}$

Correct answer

B. $4 \sqrt{3}$

Step-by-step solution

\begin{aligned} & \frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5} & \frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3} \\ & \text{ S.D }=\frac{\left|\left(\overline{\mathrm{a}_2 \cdot \overline{\mathrm{a}_1\right) \cdot\left(\overline{\mathrm{b}_1 \cdot \overline{\mathrm{b}_2\right)\right|}{\left|\overline{\mathrm{b}_1 \times \overline{\mathrm{b}_2\right|} \\ & a_1=3,-15,9 \\ & \mathrm{ b}_1=2,-7,5 \\ & \mathrm{a}_2=-1,1,9 \\ & \mathrm{ b}_2=2,1,-3 \\ & a_2-a_1=-4,16,0 \\ & \overline{\mathrm{b}_1 \times \overline{\mathrm{b}_2=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & -7 & 5 \\ 2 & 1 & -3\end{array}\right|=\hat{\mathrm{i}}(16)-\hat{\mathrm{j}}(-16)+\hat{\mathrm{k}}(16) \\ & 16(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) \\ & \left|\overline{\mathrm{b}_1 \times \overline{\mathrm{b}_2\right|=16 \sqrt{3} \\ & \therefore\left(\overline{\mathrm{a}_2-\overline{\mathrm{a}_1\right) \cdot\left(\overline{\mathrm{b}_1-\overline{\mathrm{b}_2\right)=16[-4+16]=(16)(12) \\ & \text{ S.D. }=\frac{(16)(12)}{16 \sqrt{3}}=4 \sqrt{3} \\ & \end{aligned}

Practice Three Dimensional Geometry on Quantrex Academy →

More from Three Dimensional Geometry

Let P be the plane such that it contains the straight line (x-1)/(2) = (y-3)/(3) =… JEE Advanced 2026 Let L be the straight line joining the points P(1, 2, -1) and Q(2, 3, 1) . Let S be the… JEE Advanced 2026 Let the foot of perpendicular from the point (λ, 2, 3) on the line (x-4)/(1) = (y-9)/(2)… JEE Main 2026 Let a line L_1 pass through the origin and be perpendicular to the lines L_2: r = (3+t) i… JEE Main 2026 The shortest distance between the lines (x-4)/(1) = (y-3)/(2) = (z-2)/(-3) and (x+2)/(2)… JEE Main 2026 Let the image of the point P(0, -5, 0) in the line (x-1)/(2) = (y)/(1) = (z+1)/(-2) be… JEE Main 2026 Let the image of the point P(1, 6, a) in the line L: (x)/(1) = (y-1)/(2) = (z-a+1)/(b)… JEE Main 2026 Let a line L be perpendicular to both the lines L_1: (x+1)/(3) = (y+3)/(5) = (z+5)/(7)… JEE Main 2026 Let a triangle PQR be such that P and Q lie on the line (x+3)/(8) = (y-4)/(2) = (z+1)/(2)… JEE Main 2026 If the distance of the point (a, 2, 5) from the image of the point (1, 2, 7) in the line… JEE Main 2026 The square of the distance of the point P(5, 6, 7) from the line (x-2)/(2) = (y-5)/(3) =… JEE Main 2026 The square of the distance of the point of intersection of the lines r = ( i + j - k ) +… JEE Main 2026 The shortest distance between the lines r =((1)/(3) i +2 j +(8)/(3) k )+λ(2 i -5 j +6 k )… JEE Main 2026 If (2α+1, α^2-3α, (α-1)/(2)) is the image of (α, 2α, 1) in the line… JEE Main 2026 A line with direction ratios 1, -1, 2 intersects the lines (x)/(2) = (y)/(3) = (z+1)/(3)… JEE Main 2026 The square of the distance of the point (-2, -8, 6) from the line (x-1)/(1) = (y-1)/(2) =… JEE Main 2026 Full Three Dimensional Geometry list Three Dimensional Geometry JEE Maths PYQs JEE Mathematics PYQs All JEE Main PYQs