JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Three Dimensional Geometry

If the distance of the point

(1, - 2, 3)

from the plane

x + 2 y - 3 z + 10 = 0

measured parallel to the line,

\frac{x - 1}{3} = \frac{2 - y}{m} = \frac{z + 3}{1}

\frac{7}{2}

, then the value of

∣ m ∣

is equal to......

Answer: 2

Solution

The line through

(1, - 2, 3)

parallel to the given line has direction

(3, - m, 1)

, so its points are

(1 + 3 t, - 2 - m t, 3 + t)

. Substituting in the plane,

(1 + 3 t) + 2 (- 2 - m t) - 3 (3 + t) + 10 = - 2 - 2 m t = 0 \Rightarrow t = - \frac{1}{m}

The distance measured along this line is

∣ t ∣

times the length of the direction vector:

\frac{1}{∣ m ∣} 9 + m^{2} + 1 = \frac{7}{2} \Rightarrow \frac{10 + m ^{2}}{m ^{2}} = \frac{7}{2} \Rightarrow 5 m^{2} = 20

∣ m ∣ = 2

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Integer answer

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