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3D Geometry question

2023 · 25 Jan · Shift 2 · Q28
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  5. /2023 · 25 Jan · Shift 2 · Q28

3D Geometry question

2023 · 25 Jan · Shift 2 · Q28

JEE MainMathematics3D GeometryMCQ+4 / −1
The shortest distance between the lines x+1=2y=−12zx+1=2y=-12zx+1=2y=−12z and x=y+2=6z−6x=y+2=6z-6x=y+2=6z−6 is :
  1. A
    3
  2. B
    52\frac{5}{2}25​
  3. C
    32\frac{3}{2}23​
  4. D
    2
View written solutionFree

Correct answer: D

  1. Write each line in parametric form

The first line is x+1=2y=−12zx+1=2y=-12zx+1=2y=−12z Let the common value be ttt. Then x=t−1,y=t2,z=−t12.x=t-1,\quad y=\frac t2,\quad z=-\frac t{12}.x=t−1,y=2t​,z=−12t​. So line L1L_1L1​ can be written as (x,y,z)=(−1,0,0)+t(1,12,−112).(x,y,z)=(-1,0,0)+t\left(1,\frac12,-\frac1{12}\right).(x,y,z)=(−1,0,0)+t(1,21​,−121​). Thus, a point on L1L_1L1​ is A(−1,0,0)A(-1,0,0)A(−1,0,0) and a direction vector is d⃗1=(1,12,−112).\vec d_1=\left(1,\frac12,-\frac1{12}\right).d1​=(1,21​,−121​). To avoid fractions, multiply by 121212: d⃗1=(12,6,−1).\vec d_1=(12,6,-1).d1​=(12,6,−1).

The second line is x=y+2=6z−6x=y+2=6z-6x=y+2=6z−6 Let the common value be sss. Then x=s,y=s−2,z=s+66=s6+1.x=s,\quad y=s-2,\quad z=\frac{s+6}{6}=\frac s6+1.x=s,y=s−2,z=6s+6​=6s​+1. So line L2L_2L2​ can be written as (x,y,z)=(0,−2,1)+s(1,1,16).(x,y,z)=(0,-2,1)+s\left(1,1,\frac16\right).(x,y,z)=(0,−2,1)+s(1,1,61​). Thus, a point on L2L_2L2​ is B(0,−2,1)B(0,-2,1)B(0,−2,1) and a direction vector is d⃗2=(1,1,16).\vec d_2=\left(1,1,\frac16\right).d2​=(1,1,61​). Again multiplying by 666: d⃗2=(6,6,1).\vec d_2=(6,6,1).d2​=(6,6,1).

  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​)∣​.

Here, AB→=B−A=(0−(−1),−2−0,1−0)=(1,−2,1).\overrightarrow{AB}=B-A=(0-(-1),-2-0,1-0)=(1,-2,1).AB=B−A=(0−(−1),−2−0,1−0)=(1,−2,1).

  1. Compute the cross product
\begin{vmatrix} \hat i & \hat j & \hat k\\ 12 & 6 & -1\\ 6 & 6 & 1 \end{vmatrix}$$ $$=\hat i(6\cdot 1-(-1)\cdot 6)-\hat j(12\cdot 1-(-1)\cdot 6)+\hat k(12\cdot 6-6\cdot 6)$$ $$=\hat i(12)-\hat j(18)+\hat k(36)$$ $$=(12,-18,36).$$ Its magnitude is $$|\vec d_1\times \vec d_2|=\sqrt{12^2+(-18)^2+36^2} =\sqrt{144+324+1296} =\sqrt{1764}=42.$$ --- 4. **Compute the scalar triple product** $$(\overrightarrow{AB})\cdot (\vec d_1\times \vec d_2) =(1,-2,1)\cdot (12,-18,36)$$ $$=1\cdot 12+(-2)(-18)+1\cdot 36=12+36+36=84.$$ Therefore, $$\text{Shortest distance}=\frac{|84|}{42}=2.$$ --- 5. **Check options** The shortest distance is $$2.$$ So the correct option is: **D: $2$**
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