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3D Geometry question

2016 · 9 Apr · Shift 1 · Q36
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3D Geometry question

2016 · 9 Apr · Shift 1 · Q36

JEE MainMathematics3D GeometryMCQ+4 / −1
The shortest distance between the lines x2=y2=z1{x \over 2} = {y \over 2} = {z \over 1}2x​=2y​=1z​ and x+2−1=y−48=z−54{{x + 2} \over { - 1}} = {{y - 4} \over 8} = {{z - 5} \over 4}−1x+2​=8y−4​=4z−5​ lies in the interval :
  1. A
    [0, 1)
  2. B
    [1, 2)
  3. C
    (2, 3]
  4. D
    (3, 4]
View written solutionFree

Correct answer: C

  1. Write the lines in vector form

The first line is x2=y2=z1=t\frac{x}{2}=\frac{y}{2}=\frac{z}{1}=t2x​=2y​=1z​=t So its parametric form is x=2t,y=2t,z=tx=2t,\quad y=2t,\quad z=tx=2t,y=2t,z=t Hence, a point on it is A(0,0,0)A(0,0,0)A(0,0,0) and its direction vector is d⃗1=(2,2,1).\vec d_1=(2,2,1).d1​=(2,2,1).

The second line is x+2−1=y−48=z−54=s\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}=s−1x+2​=8y−4​=4z−5​=s So its parametric form is x=−2−s,y=4+8s,z=5+4sx=-2-s,\quad y=4+8s,\quad z=5+4sx=−2−s,y=4+8s,z=5+4s Hence, a point on it is B(−2,4,5)B(-2,4,5)B(−2,4,5) and its direction vector is d⃗2=(−1,8,4).\vec d_2=(-1,8,4).d2​=(−1,8,4).

  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=(−2,4,5).\overrightarrow{AB}=B-A=(-2,4,5).AB=B−A=(−2,4,5).

  1. Compute the cross product
\begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & 2 & 1\\ -1 & 8 & 4 \end{vmatrix}$$ Expanding, $$\vec d_1\times \vec d_2= \hat i(2\cdot 4-1\cdot 8)-\hat j(2\cdot 4-1\cdot(-1))+\hat k(2\cdot 8-2\cdot(-1)).$$ So, $$\vec d_1\times \vec d_2= \hat i(8-8)-\hat j(8+1)+\hat k(16+2) =(0,-9,18).$$ Therefore, $$|\vec d_1\times \vec d_2|=\sqrt{0^2+(-9)^2+18^2}=sqrt{81+324}=\sqrt{405}=9\sqrt{5}.$$ --- 4. **Compute the scalar triple product** $$\overrightarrow{AB}\cdot(\vec d_1\times \vec d_2)=(-2,4,5)\cdot(0,-9,18).$$ Thus, $$= (-2)(0)+4(-9)+5(18)=0-36+90=54.$$ So, $$D=\frac{|54|}{9\sqrt{5}}=\frac{6}{\sqrt{5}}=\frac{6\sqrt{5}}{5}.$$ Now, $$\frac{6}{\sqrt{5}}\approx 2.68.$$ --- 5. **Identify the interval** Since $$2<2.68\le 3,$$ the shortest distance lies in $$(2,3].$$ So the correct option is **C**. --- 6. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** They agree.
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