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3D Geometry question

2021 · 1 Sep · Shift 2 · Q32
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  5. /2021 · 1 Sep · Shift 2 · Q32

3D Geometry question

2021 · 1 Sep · Shift 2 · Q32

JEE MainMathematics3D GeometryMCQ+4 / −1
The distance of line 3y−2z−1=0=3x−z+43y - 2z - 1 = 0 = 3x - z + 43y−2z−1=0=3x−z+4 from the point (2, −-− 1, 6) is :
  1. A
    26\sqrt {26}26​
  2. B
    252\sqrt 525​
  3. C
    262\sqrt 626​
  4. D
    424\sqrt 242​
View written solutionFree

Correct answer: C

  1. Write the line as intersection of two planes

The line is given by 3y−2z−1=0,3x−z+4=0.3y-2z-1=0, \qquad 3x-z+4=0.3y−2z−1=0,3x−z+4=0.

So it is the intersection of the planes 3y−2z−1=03y-2z-1=03y−2z−1=0 and 3x−z+4=0.3x-z+4=0.3x−z+4=0.

We need the perpendicular distance from the point P(2,−1,6)P(2,-1,6)P(2,−1,6) to this line.

  1. Find a point on the line

Let us take z=tz=tz=t.

From 3y−2z−1=0  ⟹  3y=2t+1  ⟹  y=2t+13,3y-2z-1=0 \implies 3y=2t+1 \implies y=\frac{2t+1}{3},3y−2z−1=0⟹3y=2t+1⟹y=32t+1​,

and from 3x−z+4=0  ⟹  3x=t−4  ⟹  x=t−43.3x-z+4=0 \implies 3x=t-4 \implies x=\frac{t-4}{3}.3x−z+4=0⟹3x=t−4⟹x=3t−4​.

Choose a convenient value of ttt, say t=4t=4t=4. Then x=0,y=8+13=3,z=4.x=0, \qquad y=\frac{8+1}{3}=3, \qquad z=4.x=0,y=38+1​=3,z=4.

Hence a point on the line is A(0,3,4).A(0,3,4).A(0,3,4).

  1. Find the direction vector of the line

The normals of the two planes are n⃗1=(0,3,−2),n⃗2=(3,0,−1).\vec n_1=(0,3,-2), \qquad \vec n_2=(3,0,-1).n1​=(0,3,−2),n2​=(3,0,−1).

Therefore, the direction vector of their intersection line is d⃗=n⃗1×n⃗2.\vec d=\vec n_1\times \vec n_2.d=n1​×n2​.

Compute:

d⃗=∣i^j^k^03−230−1∣=i^(3⋅(−1)−(−2)⋅0)−j^(0⋅(−1)−(−2)⋅3)+k^(0⋅0−3⋅3).\vec d= \begin{vmatrix} \hat i & \hat j & \hat k \\ 0 & 3 & -2 \\ 3 & 0 & -1 \end{vmatrix} =\hat i(3\cdot(-1)-(-2)\cdot0)-\hat j(0\cdot(-1)-(-2)\cdot3)+\hat k(0\cdot0-3\cdot3).d=​i^03​j^​30​k^−2−1​​=i^(3⋅(−1)−(−2)⋅0)−j^​(0⋅(−1)−(−2)⋅3)+k^(0⋅0−3⋅3).

So, d⃗=(−3,−6,−9).\vec d=(-3,-6,-9).d=(−3,−6,−9).

This is proportional to d⃗=(1,2,3).\vec d=(1,2,3).d=(1,2,3).

  1. Use distance formula from a point to a line in 3D

If AAA is a point on the line and d⃗\vec dd is the direction vector, then distance from point PPP to the line is

Distance=∣AP→×d⃗∣∣d⃗∣.\text{Distance}=\frac{|\overrightarrow{AP}\times \vec d|}{|\vec d|}.Distance=∣d∣∣AP×d∣​.

Now,

AP→=P−A=(2−0,−1−3,6−4)=(2,−4,2).\overrightarrow{AP}=P-A=(2-0,-1-3,6-4)=(2,-4,2).AP=P−A=(2−0,−1−3,6−4)=(2,−4,2).

Take d⃗=(1,2,3).\vec d=(1,2,3).d=(1,2,3).

Then

AP→×d⃗=∣i^j^k^2−42123∣.\overrightarrow{AP}\times \vec d= \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & -4 & 2 \\ 1 & 2 & 3 \end{vmatrix}.AP×d=​i^21​j^​−42​k^23​​.

Compute:

=i^((−4)(3)−2(2))−j^(2⋅3−2⋅1)+k^(2⋅2−(−4)⋅1)=\hat i((-4)(3)-2(2)) - \hat j(2\cdot 3-2\cdot 1) + \hat k(2\cdot 2-(-4)\cdot 1)=i^((−4)(3)−2(2))−j^​(2⋅3−2⋅1)+k^(2⋅2−(−4)⋅1) =i^(−12−4)−j^(6−2)+k^(4+4)=\hat i(-12-4)-\hat j(6-2)+\hat k(4+4)=i^(−12−4)−j^​(6−2)+k^(4+4) =(−16,−4,8).=(-16,-4,8).=(−16,−4,8).

Its magnitude is

∣AP→×d⃗∣=(−16)2+(−4)2+82=256+16+64=336=421.|\overrightarrow{AP}\times \vec d|=\sqrt{(-16)^2+(-4)^2+8^2} =\sqrt{256+16+64} =\sqrt{336}=4\sqrt{21}.∣AP×d∣=(−16)2+(−4)2+82​=256+16+64​=336​=421​.

Also,

∣d⃗∣=12+22+32=14.|\vec d|=\sqrt{1^2+2^2+3^2}=\sqrt{14}.∣d∣=12+22+32​=14​.

Hence,

Distance=42114=42114=432=26.\text{Distance}=\frac{4\sqrt{21}}{\sqrt{14}}=4\sqrt{\frac{21}{14}}=4\sqrt{\frac32}=2\sqrt{6}.Distance=14​421​​=41421​​=423​​=26​.
  1. Match with the options

262\sqrt{6}26​ corresponds to Option C.

  1. Compare with stored correct answer

Stored correct answer: C

Our derived answer: C

So the answer agrees with the stored correct answer.

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