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JEE Main 2019 · 12 Apr · Shift II

JEE Main 2019 Mathematics question on Three Dimensional Geometry

The length of the perpendicular drawn from the point (2,1,4)(2, 1, 4) to the plane containing the lines r⃗=(i^+j^)+λ(i^+2j^−k^)\vec{r} = (\hat{i} + \hat{j}) + \lambda(\hat{i} + 2\hat{j} - \hat{k}) and r⃗=(i^+j^)+μ(−i^+j^−2k^)\vec{r} = (\hat{i} + \hat{j}) + \mu(-\hat{i} + \hat{j} - 2\hat{k}) is
  1. A.
    3
  2. B.
    13\dfrac{1}{3}
  3. C.
    3\sqrt{3}
  4. D.
    13\dfrac{1}{\sqrt{3}}
Show answer and solution

Correct option: C

Solution

The normal to the plane is the cross product of the two direction vectors (1,2,-1) and (-1,1,-2):

n = (2·(-2) - (-1)·1, (-1)·(-1) - 1·(-2), 1·1 - 2·(-1)) = (-4+1, 1+2, 1+2) = (-3, 3, 3), simplified to (1,-1,-1).

The plane passes through (1,1,0), so its equation is 1(x-1) - 1(y-1) - 1(z-0) = 0 → x - y - z = 0.

Distance from (2,1,4) = |2 - 1 - 4| / √(1+1+1) = |-3|/√3 = 3/√3 = √3.

[JEE Main 2019 · 12 Apr · Shift II]

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