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3D Geometry question

2020 · 8 Jan · Shift 1 · Q24
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  5. /2020 · 8 Jan · Shift 1 · Q24

3D Geometry question

2020 · 8 Jan · Shift 1 · Q24

JEE MainMathematics3D GeometryMCQ+4 / −1
The shortest distance between the lines x−33=y−8−1=z−31{{x - 3} \over 3} = {{y - 8} \over { - 1}} = {{z - 3} \over 1}3x−3​=−1y−8​=1z−3​ and x+3−3=y+72=z−64{{x + 3} \over { - 3}} = {{y + 7} \over 2} = {{z - 6} \over 4}−3x+3​=2y+7​=4z−6​ is :
  1. A
    3
  2. B
    7230{7 \over 2}\sqrt {30}27​30​
  3. C
    3303\sqrt {30}330​
  4. D
    2302\sqrt {30}230​
View written solutionFree

Correct answer: C

  1. Write the lines in vector form

The given lines are

x−33=y−8−1=z−31\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}3x−3​=−1y−8​=1z−3​ and x+3−3=y+72=z−64.\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}.−3x+3​=2y+7​=4z−6​.

So their parametric forms are:

  • Line L1L_1L1​ passes through A(3,8,3)A(3,8,3)A(3,8,3) with direction vector d⃗1=(3,−1,1).\vec d_1=(3,-1,1).d1​=(3,−1,1).

  • Line L2L_2L2​ passes through B(−3,−7,6)B(-3,-7,6)B(−3,−7,6) with direction vector d⃗2=(−3,2,4).\vec d_2=(-3,2,4).d2​=(−3,2,4).

  1. Formula for shortest distance between two skew lines

For lines through points A,BA, BA,B with direction vectors d⃗1,d⃗2\vec d_1, \vec d_2d1​,d2​, 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=(−3−3,−7−8,6−3)=(−6,−15,3).\overrightarrow{AB}=B-A=(-3-3,-7-8,6-3)=(-6,-15,3).AB=B−A=(−3−3,−7−8,6−3)=(−6,−15,3).

  1. Compute the cross product d⃗1×d⃗2\vec d_1\times \vec d_2d1​×d2​
\begin{vmatrix} \hat i & \hat j & \hat k\\ 3 & -1 & 1\\ -3 & 2 & 4 \end{vmatrix}.$$ Expanding, $$\vec d_1\times \vec d_2= \hat i((-1)(4)-(1)(2)) -\hat j((3)(4)-(1)(-3)) +\hat k((3)(2)-(-1)(-3)).$$ $$=\hat i(-4-2)-\hat j(12+3)+\hat k(6-3)$$ $$=(-6,-15,3).$$ Interesting: this is exactly the same as $\overrightarrow{AB}$. --- 4. **Compute numerator** $$|(\overrightarrow{AB})\cdot(\vec d_1\times \vec d_2)| =|(-6,-15,3)\cdot(-6,-15,3)|$$ $$=|36+225+9|=270.$$ --- 5. **Compute denominator** $$|\vec d_1\times \vec d_2|=\sqrt{(-6)^2+(-15)^2+3^2} =\sqrt{36+225+9}= \sqrt{270}=3\sqrt{30}.$$ --- 6. **Shortest distance** $$D=\frac{270}{3\sqrt{30}}=\frac{90}{\sqrt{30}}=3\sqrt{30}.$$ --- 7. **Check options** The correct option is $$\boxed{3\sqrt{30}}$$ which is **Option C**. --- 8. **Comparison with stored correct answer** Stored correct answer: **C** Our derived answer: **C** So they agree.
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