Question

15.
The distance between parallel planes 2x + y - 2z – 6 = 0 and
4x + 2y – 4z = 0 is

units.​

Answer 1

Given : equations of parallel planes are,

$2x+y-2z=6 =c_1\hfill (1)$

$4x+2y-4z=0=c_2 \hfill (2)$

Now, multiply (1) by 2 we will get,

$4x+2y-4z=12= c_1=ax+by+cz$

To find : distance between (1) and (2).

Step-by-step explanation :

We know that distance between two parallel planes,

$|\frac{c_2-c_1}{\sqrt{a^2+b^2+c^2}}|$

$=|\frac{12-8}{\sqrt{4^2+6^2+8^2}}|$

$=\frac{4}{\sqrt{116}}$

Distance between parallel planes is $\frac{4}{\sqrt{116}}$ units.

Answer 2

Answer:

2 units

Step-by-step explanation:

P1  ⇒ 2x + y - 2z – 6 = 0 ---( 1 )

P2 ⇒ 4x + 2y – 4z = 0    ---( 2 )

( 1 ) x 2 ⇒ 4x+2y-4z-12=0

d1 = 12

d2 = 0

therefore, distance between 2 parallel planes are :

d = $|\frac{d1-d2}{ \sqrt{a^{2} +b^{2}+c^{2}}}|$

  = $|\frac{12-0}{ \sqrt{(4)^{2} +(2)^{2}+(4)^{2}}}|$

  = $|\frac{12}{ \sqrt{16 + 4 + 16}}|$

  = $|\frac{12}{ \sqrt{36}}|$

  = $|\frac{12}{6} |$

  = 2 units