Question

Find the distance between the planes $2x+3y+4z=7$ and $2x+3y+4z=17$

Answer 1

We have two parallel planes

$2x+3y+4z=7$

and $2x+3y+4z=17$

Let's find out a point on the plane $2x+3y+4z=7$

When $x=y=0$, in the above plane then $z=\frac{7}{4}$

$\therefore$ Let's find out the distance between point$(0,0,0)$ & plane $2x+3y+4z=17$

$\therefore D=\frac{\mid a{x}_1+b{y}_1+c{z}_1+d\mid }{\sqrt{a^2+b^2+c^2}}$

$=\frac{\mid 2x_0+3x_0+4]\times \frac{7}{4}-17\mid }{\sqrt{2^2+3^2+4^2}}$
$=\frac{\mid 7-17\mid }{\sqrt{16+4+9}}$

$=\frac{10}{\sqrt{2}9}$