Solveeit Logo

Question

Mathematics Question on Three Dimensional Geometry

Distance between two parallel planes 2x+y+2z=82x+ y + 2z = 8 and 4x+2y+4z+5=04x + 2y + 4z + 5 = 0 is

A

32\frac{3}{2}

B

52\frac{5}{2}

C

72\frac{7}{2}

D

92\frac{9}{2}

Answer

72\frac{7}{2}

Explanation

Solution

Given planes are
4x+2y+4z+5=04x + 2y + 4z +5 = 0 ....(1)
4x+2y+4z16=04x + 2y + 4z -16 = 0 ....(2)
[d=c1c216+4+116]d=5+1616+4+16\left[d=\frac{\left|c_{1}-c_{2}\right|}{\sqrt{16+4+116}}\right]\Rightarrow d = \left|\frac{5+16}{\sqrt{16+4+16}}\right|
=216=72= \frac{21}{6} = \frac{7}{2}