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3D Geometry question

2022 · 25 Jul · Shift 2 · Q33
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  5. /2022 · 25 Jul · Shift 2 · Q33

3D Geometry question

2022 · 25 Jul · Shift 2 · Q33

JEE MainMathematics3D GeometryMCQ+4 / −1
The shortest distance between the lines x+7−6=y−67=z\frac{x+7}{-6}=\frac{y-6}{7}=z−6x+7​=7y−6​=z and 7−x2=y−2=z−6\frac{7-x}{2}=y-2=z-627−x​=y−2=z−6 is :
  1. A
    2292 \sqrt{29}229​
  2. B
    1
  3. C
    3729\sqrt{\frac{37}{29}}2937​​
  4. D
    292\frac{\sqrt{29}}{2}229​​
View written solutionFree

Correct answer: A

  1. Write the lines in symmetric/parametric form

Given x+7−6=y−67=z\frac{x+7}{-6}=\frac{y-6}{7}=z−6x+7​=7y−6​=z Let the common parameter be λ\lambdaλ. Then x=−6λ−7,y=7λ+6,z=λx=-6\lambda-7,\quad y=7\lambda+6,\quad z=\lambdax=−6λ−7,y=7λ+6,z=λ So line L1L_1L1​ passes through A(−7,6,0)A(-7,6,0)A(−7,6,0) and has direction vector d⃗1=(−6,7,1).\vec d_1=(-6,7,1).d1​=(−6,7,1).

Now for the second line: 7−x2=y−2=z−6\frac{7-x}{2}=y-2=z-627−x​=y−2=z−6 Let the common parameter be μ\muμ. Then 7−x2=μ,y−2=μ,z−6=μ\frac{7-x}{2}=\mu,\quad y-2=\mu,\quad z-6=\mu27−x​=μ,y−2=μ,z−6=μ Hence x=7−2μ,y=μ+2,z=μ+6x=7-2\mu,\quad y=\mu+2,\quad z=\mu+6x=7−2μ,y=μ+2,z=μ+6 So line L2L_2L2​ passes through B(7,2,6)B(7,2,6)B(7,2,6) and has direction vector d⃗2=(−2,1,1).\vec d_2=(-2,1,1).d2​=(−2,1,1).

  1. Use the 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​, Shortest distance=∣AB→⋅(d⃗1×d⃗2)∣∣d⃗1×d⃗2∣.\text{Shortest distance} = \frac{|\overrightarrow{AB}\cdot(\vec d_1\times \vec d_2)|}{|\vec d_1\times \vec d_2|}.Shortest distance=∣d1​×d2​∣∣AB⋅(d1​×d2​)∣​.

First, AB→=B−A=(7−(−7), 2−6, 6−0)=(14,−4,6).\overrightarrow{AB}=B-A=(7-(-7),\,2-6,\,6-0)=(14,-4,6).AB=B−A=(7−(−7),2−6,6−0)=(14,−4,6).

  1. Compute the cross product
\begin{vmatrix} \hat i & \hat j & \hat k\\ -6 & 7 & 1\\ -2 & 1 & 1 \end{vmatrix}$$ $$=\hat i(7\cdot 1-1\cdot 1)-\hat j((-6)\cdot 1-1\cdot(-2))+\hat k((-6)\cdot 1-7\cdot(-2))$$ $$=6\hat i-(-4)\hat j+8\hat k=(6,4,8).$$ Its magnitude is $$|\vec d_1\times \vec d_2|=\sqrt{6^2+4^2+8^2}=\sqrt{36+16+64}=\sqrt{116}=2\sqrt{29}.$$ --- 4. **Compute the scalar triple product** $$\overrightarrow{AB}\cdot(\vec d_1\times \vec d_2)=(14,-4,6)\cdot(6,4,8)$$ $$=14\cdot 6+(-4)\cdot 4+6\cdot 8=84-16+48=116.$$ Therefore, $$\text{Shortest distance} = \frac{116}{2\sqrt{29}}=\frac{58}{\sqrt{29}}=2\sqrt{29}.$$ --- 5. **Match with options** $$2\sqrt{29}$$ is **Option A**. --- 6. **Compare with stored correct answer** Stored correct answer: **A** Our derived answer: **A** So the stored answer is correct.
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