Question

What is the distance(in units) between the two planes
3x + 5y + 7z = 3 and 9x + 15y + 21z = 9 ?
(a) 0 (b) 3 (c) 6/√83 (d) 6

Answer 1

The distance(in units) between the two planes

3x + 5y + 7z = 3 and 9x + 15y + 21z = 9 is "(a) zero(0)".

Explanation:

Given,

The two panes are

3x + 5y + 7z = 3    ......(1)

and 9x + 15y + 21z = 9

⇒ 3x + 5y + 7z = 3    ......(2)

Here, both plane are parallel.

We know that,

The distance(in units) between the two paralel planes

= 0

Because, D_2 - D_1= 3 - 3 = 0

Hence, The distance(in units) between the two planes

3x + 5y + 7z = 3 and 9x + 15y + 21z = 9 is "(a) zero(0)".