Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is

Question

Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is (A)  (3/2) (B)  (5/2) (C)  (7/2) (D)  (9/2)

Tardigrade Admin · Accepted Answer

Given planes are 2x + y + 2z â 8 = 0, 4x + 2y + 4z + 5 = 0 â 2x + y + 2z + 5/2 = 0 Distance between planes = (|d1-d2|/√a2+b2+c2) = (|-8-5/2|/√22+12+22) = (7/2).

Q. Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is

$\frac{3}{2}$

$\frac{5}{2}$

$\frac{7}{2}$

$\frac{9}{2}$

Given planes are
$2 x + y + 2 z - 8 = 0, 4 x + 2 y + 4 z + 5 = 0 \Rightarrow 2 x + y + 2 z + 5/2 = 0$
Distance between planes $= \frac{∣ d _{1} - d _{2} ∣}{a ^{2} + b ^{2} + c ^{2}} = \frac{∣ - 8 - 5/2 ∣}{2 ^{2} + 1 ^{2} + 2 ^{2}} = \frac{7}{2} .$