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3D Geometry question

2022 · 25 Jun · Shift 2 · Q45
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  5. /2022 · 25 Jun · Shift 2 · Q45

3D Geometry question

2022 · 25 Jun · Shift 2 · Q45

JEE MainMathematics3D GeometryNumerical+4 / −1
Let l1 be the line in xy-plane with x and y intercepts 18{1 \over 8}81​ and 142{1 \over {4\sqrt 2 }}42​1​ respectively, and l2 be the line in zx-plane with x and z intercepts −18- {1 \over 8}−81​ and −163- {1 \over {6\sqrt 3 }}−63​1​ respectively. If d is the shortest distance between the line l1 and l2, then d −-− 2 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 51

  1. Write equations of the two lines
  • Line l1l_1l1​ lies in the xyxyxy-plane, so z=0z=0z=0.
  • Its xxx- and yyy-intercepts are 18\dfrac1881​ and 142\dfrac{1}{4\sqrt2}42​1​.

Using intercept form,

x1/8+y1/(42)=1\frac{x}{1/8}+\frac{y}{1/(4\sqrt2)}=11/8x​+1/(42​)y​=1

so

8x+42 y=1,z=0.8x+4\sqrt2\,y=1, \qquad z=0.8x+42​y=1,z=0.

A convenient point on l1l_1l1​ is

A(18,0,0)A\left(\frac18,0,0\right)A(81​,0,0)

and a direction vector is obtained from the equation 8x+42y=18x+4\sqrt2 y=18x+42​y=1 in the plane z=0z=0z=0:

u⃗1=(1,−2,0).\vec{u}_1=(1,-\sqrt2,0).u1​=(1,−2​,0).
  • Line l2l_2l2​ lies in the zxzxzx-plane, so y=0y=0y=0.
  • Its xxx- and zzz-intercepts are −18-\dfrac18−81​ and −163-\dfrac{1}{6\sqrt3}−63​1​.

Using intercept form,

x−1/8+z−1/(63)=1\frac{x}{-1/8}+\frac{z}{-1/(6\sqrt3)}=1−1/8x​+−1/(63​)z​=1

which gives

−8x−63 z=1-8x-6\sqrt3\,z=1−8x−63​z=1

or equivalently

8x+63 z=−1,y=0.8x+6\sqrt3\,z=-1, \qquad y=0.8x+63​z=−1,y=0.

A convenient point on l2l_2l2​ is

B(−18,0,0)B\left(-\frac18,0,0\right)B(−81​,0,0)

and a direction vector is

u⃗2=(−33,0,4)\vec{u}_2=( -3\sqrt3, 0, 4)u2​=(−33​,0,4)

(since 8x+63z=−18x+6\sqrt3 z=-18x+63​z=−1 has direction satisfying 8 dx+63 dz=08\,dx+6\sqrt3\,dz=08dx+63​dz=0).


  1. Use formula for shortest distance between two skew lines

For lines through points A,BA,BA,B with direction vectors u⃗1,u⃗2\vec{u}_1,\vec{u}_2u1​,u2​,

d=∣(AB→)⋅(u⃗1×u⃗2)∣∣u⃗1×u⃗2∣.d=\frac{|(\overrightarrow{AB})\cdot(\vec{u}_1\times \vec{u}_2)|}{|\vec{u}_1\times \vec{u}_2|}.d=∣u1​×u2​∣∣(AB)⋅(u1​×u2​)∣​.

Here,

AB→=B−A=(−14,0,0).\overrightarrow{AB}=B-A=\left(-\frac14,0,0\right).AB=B−A=(−41​,0,0).

Now compute the cross product:

u⃗1×u⃗2=∣i^j^k^1−20−3304∣\vec{u}_1\times \vec{u}_2= \begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & -\sqrt2 & 0\\ -3\sqrt3 & 0 & 4 \end{vmatrix}u1​×u2​=​i^1−33​​j^​−2​0​k^04​​ =i^(−42)−j^(4)−k^(−36)=(−42,−4,−36).=\hat i(-4\sqrt2)-\hat j(4)-\hat k(-3\sqrt6) =(-4\sqrt2,-4, -3\sqrt6).=i^(−42​)−j^​(4)−k^(−36​)=(−42​,−4,−36​).

Hence,

∣u⃗1×u⃗2∣=(42)2+42+(36)2=32+16+54=102.|\vec{u}_1\times \vec{u}_2| =\sqrt{(4\sqrt2)^2+4^2+(3\sqrt6)^2} =\sqrt{32+16+54} =\sqrt{102}.∣u1​×u2​∣=(42​)2+42+(36​)2​=32+16+54​=102​.

Also,

(AB→)⋅(u⃗1×u⃗2)=(−14,0,0)⋅(−42,−4,−36)=2.(\overrightarrow{AB})\cdot(\vec{u}_1\times \vec{u}_2) =\left(-\frac14,0,0\right)\cdot(-4\sqrt2,-4,-3\sqrt6)=\sqrt2.(AB)⋅(u1​×u2​)=(−41​,0,0)⋅(−42​,−4,−36​)=2​.

Therefore,

d=2102=151=151.d=\frac{\sqrt2}{\sqrt{102}}=\sqrt{\frac1{51}}=\frac1{\sqrt{51}}.d=102​2​​=511​​=51​1​.
  1. Interpret the asked quantity

The computed shortest distance is

d=151.d=\frac1{\sqrt{51}}.d=51​1​.

So

d−2=51.d^{-2}=51.d−2=51.

Since the stored answer is 515151, the intended quantity is clearly d−2d^{-2}d−2 (the problem statement appears to have a formatting issue where it shows d−2d-2d−2 instead of d−2d^{-2}d−2).

Thus the required integer is

51.\boxed{51}.51​.
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