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3D Geometry question

2024 · 30 Jan · Shift 1 · Q57
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3D Geometry question

2024 · 30 Jan · Shift 1 · Q57

JEE MainMathematics3D GeometryNumerical+4 / −1
If d1\mathrm{d}_1d1​ is the shortest distance between the lines x+1=2y=−12z,x=y+2=6z−6x+1=2 y=-12 z, x=y+2=6 z-6x+1=2y=−12z,x=y+2=6z−6 and d2\mathrm{d}_2d2​ is the shortest distance between the lines x−12=y+8−7=z−45,x−12=y−21=z−6−3\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}, \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}2x−1​=−7y+8​=5z−4​,2x−1​=1y−2​=−3z−6​, then the value of 323 d1 d2\frac{32 \sqrt{3} \mathrm{~d}_1}{\mathrm{~d}_2} d2​323​ d1​​ is :
Numerical answer
View written solutionFree

Correct answer: 16

  1. Interpret the first pair of lines

The lines are written as x+1=2y=−12z,x=y+2=6z−6.x+1=2y=-12z, \qquad x=y+2=6z-6.x+1=2y=−12z,x=y+2=6z−6.

We convert each into parametric form.

Line L1L_1L1​

Let x+1=2y=−12z=t.x+1=2y=-12z=t.x+1=2y=−12z=t. 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 a point on L1L_1L1​ is A(−1,0,0)A(-1,0,0)A(−1,0,0) and a direction vector is a⃗=(1,12,−112).\vec a=\left(1,\frac12,-\frac1{12}\right).a=(1,21​,−121​). Multiplying by 121212 for convenience, a⃗=(12,6,−1).\vec a=(12,6,-1).a=(12,6,−1).

Line L2L_2L2​

Let x=y+2=6z−6=s.x=y+2=6z-6=s.x=y+2=6z−6=s. 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 a point on L2L_2L2​ is B(0,−2,1)B(0,-2,1)B(0,−2,1) and a direction vector is b⃗=(1,1,16).\vec b=\left(1,1,\frac16\right).b=(1,1,61​). Multiplying by 666, b⃗=(6,6,1).\vec b=(6,6,1).b=(6,6,1).

  1. Shortest distance between two skew lines

For lines through points A,BA,BA,B with direction vectors a⃗,b⃗\vec a,\vec ba,b, d=∣(AB→)⋅(a⃗×b⃗)∣∣a⃗×b⃗∣.d=\frac{|(\overrightarrow{AB})\cdot(\vec a\times \vec b)|}{|\vec a\times \vec b|}.d=∣a×b∣∣(AB)⋅(a×b)∣​.

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

Now compute

\begin{vmatrix} \hat i & \hat j & \hat k\\ 12 & 6 & -1\\ 6 & 6 & 1 \end{vmatrix}.$$ So, $$\vec a\times \vec b= \hat i(6\cdot1-(-1)\cdot6) -\hat j(12\cdot1-(-1)\cdot6) +\hat k(12\cdot6-6\cdot6).$$ $$\vec a\times \vec b=(12,-18,36)=6(2,-3,6).$$ Hence, $$|\vec a\times \vec b|=\sqrt{12^2+(-18)^2+36^2} =\sqrt{144+324+1296} =\sqrt{1764}=42.$$ Now, $$(\overrightarrow{AB})\cdot(\vec a\times \vec b) =(1,-2,1)\cdot(12,-18,36) =12+36+36=84.$$ Therefore, $$d_1=\frac{84}{42}=2.$$ --- 3. **Interpret the second pair of lines** The lines are $$\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}, \qquad \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}.$$ ### Line $M_1$ Let $$\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}=\lambda.$$ Then $$x=1+2\lambda,\quad y=-8-7\lambda,\quad z=4+5\lambda.$$ So a point is $$P(1,-8,4)$$ and direction vector is $$\vec p=(2,-7,5).$$ ### Line $M_2$ Let $$\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}=\mu.$$ Then $$x=1+2\mu,\quad y=2+\mu,\quad z=6-3\mu.$$ So a point is $$Q(1,2,6)$$ and direction vector is $$\vec q=(2,1,-3).$$ --- 4. **Compute $d_2$** Use $$d_2=\frac{|(\overrightarrow{PQ})\cdot(\vec p\times \vec q)|}{|\vec p\times \vec q|}.$$ Here, $$\overrightarrow{PQ}=Q-P=(0,10,2).$$ Now compute cross product: $$\vec p\times \vec q= \begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & -7 & 5\\ 2 & 1 & -3 \end{vmatrix}.$$ Thus, $$\vec p\times \vec q= \hat i((-7)(-3)-5\cdot1) -\hat j(2(-3)-5\cdot2) +\hat k(2\cdot1-(-7)\cdot2).$$ $$\vec p\times \vec q=(16,16,16).$$ Hence, $$|\vec p\times \vec q|=\sqrt{16^2+16^2+16^2}=16\sqrt3.$$ Also, $$(\overrightarrow{PQ})\cdot(\vec p\times \vec q) =(0,10,2)\cdot(16,16,16)=160+32=192.$$ Therefore, $$d_2=\frac{192}{16\sqrt3}=\frac{12}{\sqrt3}=4\sqrt3.$$ --- 5. **Evaluate the required expression** We need $$\frac{32\sqrt3\, d_1}{d_2}.$$ Substitute $d_1=2$ and $d_2=4\sqrt3$: $$\frac{32\sqrt3\cdot 2}{4\sqrt3} =\frac{64\sqrt3}{4\sqrt3}=16.$$ --- 6. **Final answer** $$\boxed{16}$$ The derived answer matches the stored correct answer.
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