JEE Main · 2004 · 24 Apr 2004 · Offline · Three Dimensional Geometry

Distance between two parallel planes

2 x + y + 2 z = 8

and

4 x + 2 y + 4 z + 5 = 0

(A) $\frac{3}{2}$
(B) $\frac{5}{2}$
(C) $\frac{7}{2}$ ✓ Correct
(D) $\frac{9}{2}$

Answer: (C)

Solution

Scale the second plane

4 x + 2 y + 4 z + 5 = 0

down by

2

to match normals:

2 x + y + 2 z + \frac{5}{2} = 0

, i.e.

2 x + y + 2 z = - \frac{5}{2}

, while the first is

2 x + y + 2 z = 8

. The gap between parallel planes is

\frac{∣ 8 - ( - \frac{5}{2} ) ∣}{2 ^{2} + 1 ^{2} + 2 ^{2}} = \frac{\frac{21}{2}}{3} = \frac{7}{2}

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option

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