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3D Geometry question

2023 · 10 Apr · Shift 1 · Q25
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  5. /2023 · 10 Apr · Shift 1 · Q25

3D Geometry question

2023 · 10 Apr · Shift 1 · Q25

JEE MainMathematics3D GeometryMCQ+4 / −1
The shortest distance between the lines x+21=y−2=z−52{{x + 2} \over 1} = {y \over { - 2}} = {{z - 5} \over 2}1x+2​=−2y​=2z−5​ and x−41=y−12=z+30{{x - 4} \over 1} = {{y - 1} \over 2} = {{z + 3} \over 0}1x−4​=2y−1​=0z+3​ is :
  1. A
    8
  2. B
    7
  3. C
    6
  4. D
    9
View written solutionFree

Correct answer: D

  1. Write the lines in vector/parametric form

For the first line: x+21=y−2=z−52=t\frac{x+2}{1}=\frac{y}{-2}=\frac{z-5}{2}=t1x+2​=−2y​=2z−5​=t So, x=t−2,y=−2t,z=2t+5x=t-2,\quad y=-2t,\quad z=2t+5x=t−2,y=−2t,z=2t+5 A point on it is P1(−2,0,5)P_1(-2,0,5)P1​(−2,0,5) and its direction vector is d⃗1=(1,−2,2).\vec d_1=(1,-2,2).d1​=(1,−2,2).

For the second line: x−41=y−12=z+30=s\frac{x-4}{1}=\frac{y-1}{2}=\frac{z+3}{0}=s1x−4​=2y−1​=0z+3​=s So, x=s+4,y=2s+1,z=−3x=s+4,\quad y=2s+1,\quad z=-3x=s+4,y=2s+1,z=−3 A point on it is P2(4,1,−3)P_2(4,1,-3)P2​(4,1,−3) and its direction vector is d⃗2=(1,2,0).\vec d_2=(1,2,0).d2​=(1,2,0).

  1. Formula for shortest distance between two skew lines

If two lines are r⃗=a⃗1+λd⃗1,r⃗=a⃗2+μd⃗2,\vec r=\vec a_1+\lambda \vec d_1,\qquad \vec r=\vec a_2+\mu \vec d_2,r=a1​+λd1​,r=a2​+μd2​, then the shortest distance is D=∣(a⃗2−a⃗1)⋅(d⃗1×d⃗2)∣∣d⃗1×d⃗2∣.D=\frac{|(\vec a_2-\vec a_1)\cdot(\vec d_1\times \vec d_2)|}{|\vec d_1\times \vec d_2|}.D=∣d1​×d2​∣∣(a2​−a1​)⋅(d1​×d2​)∣​.

Here, a⃗2−a⃗1=(4−(−2), 1−0, −3−5)=(6,1,−8).\vec a_2-\vec a_1=(4-(-2),\,1-0,\,-3-5)=(6,1,-8).a2​−a1​=(4−(−2),1−0,−3−5)=(6,1,−8).

  1. Compute the cross product
\begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & -2 & 2\\ 1 & 2 & 0 \end{vmatrix}$$ $$=\hat i((-2)(0)-2\cdot 2)-\hat j(1\cdot 0-2\cdot 1)+\hat k(1\cdot 2-(-2)\cdot 1)$$ $$=(-4,2,4).$$ Its magnitude is $$|\vec d_1\times \vec d_2|=\sqrt{(-4)^2+2^2+4^2}= \sqrt{16+4+16}=\sqrt{36}=6.$$ --- 4. **Compute the scalar triple product** $$ (\vec a_2-\vec a_1)\cdot(\vec d_1\times \vec d_2) =(6,1,-8)\cdot(-4,2,4) $$ $$=6(-4)+1(2)+(-8)(4)=-24+2-32=-54.$$ So, $$\left|(\vec a_2-\vec a_1)\cdot(\vec d_1\times \vec d_2)\right|=54.$$ --- 5. **Shortest distance** $$D=\frac{54}{6}=9.$$ Thus, the shortest distance is $$\boxed{9}.$$ So the correct option is **D**.
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