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JEE Main 2016

JEE Main 2016 Mathematics question on Three Dimensional Geometry

The distance of the point (1,−5,9)(1, -5, 9) from the plane x−y+z=5x - y + z = 5 measured along the line x=y=zx = y = z is
  1. A.
    3103\sqrt{10}
  2. B.
    10310\sqrt{3}
  3. C.
    103\dfrac{10}{\sqrt{3}}
  4. D.
    203\dfrac{20}{3}
Show answer and solution

Correct option: B

Solution

The line through (1, -5, 9) parallel to x = y = z has direction (1,1,1), parametrized as (1+t, -5+t, 9+t). Substituting into the plane x - y + z = 5: (1+t) - (-5+t) + (9+t) = 5 → 15 + t = 5 → t = -10. The point of intersection is (1-10, -5-10, 9-10) = (-9, -15, -1). Distance = √((1-(-9))² + (-5-(-15))² + (9-(-1))²) = √(100+100+100) = 10√3.

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