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3D Geometry question

2022 · 26 Jul · Shift 1 · Q35
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  5. /2022 · 26 Jul · Shift 1 · Q35

3D Geometry question

2022 · 26 Jul · Shift 1 · Q35

JEE MainMathematics3D GeometryMCQ+4 / −1
The length of the perpendicular from the point (1,−2,5)(1,-2,5)(1,−2,5) on the line passing through (1,2,4)(1,2,4)(1,2,4) and parallel to the line x+y−z=0=x−2y+3z−5x+y-z=0=x-2 y+3 z-5x+y−z=0=x−2y+3z−5 is :
  1. A
    212\sqrt{\frac{21}{2}}221​​
  2. B
    92\sqrt{\frac{9}{2}}29​​
  3. C
    732\sqrt{\frac{73}{2}}273​​
  4. D
    1
View written solutionFree

Correct answer: A

  1. Interpret the given line

The line is given as the intersection of the planes x+y−z=0x+y-z=0x+y−z=0 and x−2y+3z−5=0.x-2y+3z-5=0.x−2y+3z−5=0.

A line parallel to this intersection line will have the same direction vector as the intersection line.

  1. Find the direction vector of the given line

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

So 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^11−11−23∣=i^(1⋅3−(−1)(−2))−j^(1⋅3−(−1)(1))+k^(1⋅(−2)−1⋅1).\vec d= \begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & 1 & -1\\ 1 & -2 & 3 \end{vmatrix} =\hat i(1\cdot 3-(-1)(-2)) - \hat j(1\cdot 3-(-1)(1)) + \hat k(1\cdot (-2)-1\cdot 1).d=​i^11​j^​1−2​k^−13​​=i^(1⋅3−(−1)(−2))−j^​(1⋅3−(−1)(1))+k^(1⋅(−2)−1⋅1).

Thus,

d⃗=i^(3−2)−j^(3+1)+k^(−2−1)=(1,−4,−3).\vec d=\hat i(3-2)-\hat j(3+1)+\hat k(-2-1)=(1,-4,-3).d=i^(3−2)−j^​(3+1)+k^(−2−1)=(1,−4,−3).

So the required line passes through (1,2,4)(1,2,4)(1,2,4) and has direction vector d⃗=(1,−4,−3).\vec d=(1,-4,-3).d=(1,−4,−3).

  1. Equation of the required line

A point on the line is A=(1,2,4),A=(1,2,4),A=(1,2,4), and the external point is P=(1,−2,5).P=(1,-2,5).P=(1,−2,5).

The perpendicular distance from point PPP to the line through AAA in direction d⃗\vec dd is

Distance=∣AP→×d⃗∣∣d⃗∣.\text{Distance} = \frac{\left|\overrightarrow{AP}\times \vec d\right|}{|\vec d|}.Distance=∣d∣​AP×d​​.
  1. Compute AP→\overrightarrow{AP}AP
AP→=P−A=(1−1,−2−2,5−4)=(0,−4,1).\overrightarrow{AP}=P-A=(1-1,-2-2,5-4)=(0,-4,1).AP=P−A=(1−1,−2−2,5−4)=(0,−4,1).
  1. Compute the cross product
AP→×d⃗=∣i^j^k^0−411−4−3∣\overrightarrow{AP}\times \vec d= \begin{vmatrix} \hat i & \hat j & \hat k\\ 0 & -4 & 1\\ 1 & -4 & -3 \end{vmatrix}AP×d=​i^01​j^​−4−4​k^1−3​​ =i^[(−4)(−3)−1(−4)]−j^[0(−3)−1(1)]+k^[0(−4)−(−4)(1)]=\hat i[(-4)(-3)-1(-4)]-\hat j[0(-3)-1(1)]+\hat k[0(-4)-(-4)(1)]=i^[(−4)(−3)−1(−4)]−j^​[0(−3)−1(1)]+k^[0(−4)−(−4)(1)] =i^(12+4)−j^(0−1)+k^(0+4)=(16,1,4).=\hat i(12+4)-\hat j(0-1)+\hat k(0+4) =(16,1,4).=i^(12+4)−j^​(0−1)+k^(0+4)=(16,1,4).

So,

∣AP→×d⃗∣=162+12+42=256+1+16=273.\left|\overrightarrow{AP}\times \vec d\right|=\sqrt{16^2+1^2+4^2}= \sqrt{256+1+16}=\sqrt{273}.​AP×d​=162+12+42​=256+1+16​=273​.
  1. Compute ∣d⃗∣|\vec d|∣d∣
∣d⃗∣=12+(−4)2+(−3)2=1+16+9=26.|\vec d|=\sqrt{1^2+(-4)^2+(-3)^2}= \sqrt{1+16+9}=\sqrt{26}.∣d∣=12+(−4)2+(−3)2​=1+16+9​=26​.

Hence the distance is

27326=27326=212.\frac{\sqrt{273}}{\sqrt{26}}=\sqrt{\frac{273}{26}}= \sqrt{\frac{21}{2}}.26​273​​=26273​​=221​​.
  1. Match with options

212\sqrt{\frac{21}{2}}221​​ corresponds to Option A.

Therefore, the correct answer is A.

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