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3D Geometry question

2023 · 24 Jan · Shift 1 · Q38
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  5. /2023 · 24 Jan · Shift 1 · Q38

3D Geometry question

2023 · 24 Jan · Shift 1 · Q38

JEE MainMathematics3D GeometryNumerical+4 / −1
The shortest distance between the lines x−23=y+12=z−62{{x - 2} \over 3} = {{y + 1} \over 2} = {{z - 6} \over 2}3x−2​=2y+1​=2z−6​ and x−63=1−y2=z+80{{x - 6} \over 3} = {{1 - y} \over 2} = {{z + 8} \over 0}3x−6​=21−y​=0z+8​ is equal to ‾\underline{\hspace{2cm}}​
Numerical answer
View written solutionFree

Correct answer: 14

  1. Write the lines in vector form

For the first line, x−23=y+12=z−62=t\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-6}{2}=t3x−2​=2y+1​=2z−6​=t So, x=2+3t,y=−1+2t,z=6+2tx=2+3t,\quad y=-1+2t,\quad z=6+2tx=2+3t,y=−1+2t,z=6+2t Hence line L1L_1L1​ passes through A(2,−1,6)A(2,-1,6)A(2,−1,6) and has direction vector d⃗1=(3,2,2).\vec d_1=(3,2,2).d1​=(3,2,2).

For the second line, x−63=1−y2=z+80=s\frac{x-6}{3}=\frac{1-y}{2}=\frac{z+8}{0}=s3x−6​=21−y​=0z+8​=s From z+80\dfrac{z+8}{0}0z+8​, we get z=−8z=-8z=−8 (always), and x=6+3s,1−y=2s⇒y=1−2s.x=6+3s,\quad 1-y=2s \Rightarrow y=1-2s.x=6+3s,1−y=2s⇒y=1−2s. Thus line L2L_2L2​ passes through B(6,1,−8)B(6,1,-8)B(6,1,−8) and has direction vector d⃗2=(3,−2,0).\vec d_2=(3,-2,0).d2​=(3,−2,0).

  1. Use formula for shortest distance between two skew lines

The shortest distance is D=∣(AB→)⋅(d⃗1×d⃗2)∣∣d⃗1×d⃗2∣.D=\frac{|(\overrightarrow{AB})\cdot(\vec d_1\times \vec d_2)|}{|\vec d_1\times \vec d_2|}.D=∣d1​×d2​∣∣(AB)⋅(d1​×d2​)∣​.

Here, AB→=B−A=(6−2,  1−(−1),  −8−6)=(4,2,−14).\overrightarrow{AB}=B-A=(6-2,\;1-(-1),\;-8-6)=(4,2,-14).AB=B−A=(6−2,1−(−1),−8−6)=(4,2,−14).

  1. Compute the cross product
\begin{vmatrix} \hat i & \hat j & \hat k\\ 3 & 2 & 2\\ 3 & -2 & 0 \end{vmatrix}$$ $$=\hat i(2\cdot 0-2\cdot(-2)) - \hat j(3\cdot 0-2\cdot 3) + \hat k(3\cdot(-2)-2\cdot 3)$$ $$=(4,6,-12).$$ Its magnitude is $$|\vec d_1\times \vec d_2|=\sqrt{4^2+6^2+(-12)^2}= \sqrt{16+36+144}=\sqrt{196}=14.$$ 4. **Compute the scalar triple product** $$(\overrightarrow{AB})\cdot(\vec d_1\times \vec d_2) =(4,2,-14)\cdot(4,6,-12)$$ $$=4\cdot 4+2\cdot 6+(-14)(-12)$$ $$=16+12+168=196.$$ 5. **Find the distance** $$D=\frac{|196|}{14}=14.$$ 6. **Final answer** The shortest distance is $$\boxed{14}.$$
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