JEE Main · 2013 · 07 Apr 2013 · 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) $\frac{5}{2}$
(B) $\frac{7}{2}$ ✓ Correct
(C) $\frac{9}{2}$
(D) $\frac{3}{2}$

Answer: (B)

Solution

Scale the first plane so the normals match:

2 x + y + 2 z = 8

becomes

4 x + 2 y + 4 z = 16

, alongside

4 x + 2 y + 4 z = - 5

. The distance between parallel planes is

\frac{∣ d _{1} - d _{2} ∣}{a ^{2} + b ^{2} + c ^{2}} = \frac{∣16 - ( - 5 ) ∣}{16 + 4 + 16} = \frac{21}{6} = \frac{7}{2}

JEE Main · 2013 · 07 Apr 2013 · Offline · Single correct option

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