Question

Find the distance between the point $(7,2,4)$ and the plane determined by the points $A(2,5,-3),B(-2,-3,5)$ and $C(5,3,-3).$

Answer 1

The general equation of a plane passing through $A(2,5,-3)$ is $a(x-2)+b(y-5)+c(z+3)=0$--- (1)

It will pass through $B(-2,-3,5)$ and $C(5,3,-3)$ if

$a(-2-2)+b(-3-5)+c(5+3)=0\Rightarrow -4a-8b+8c=0$

$=a+2b-2c=0$--- (2)

and $a(5-2)+b(3-5)+c(-3+3)=0\Rightarrow 3a-2b=0$--- (3)

Solving (2) and (3) by cross multiplying method, we get

$\frac{a}{0-4}=\frac{b}{-6-0}=\frac{c}{-2-6}\Rightarrow \frac{a}{-4}=\frac{b}{-6}=\frac{c}{-8}$

$\Rightarrow \frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\lambda$

So, $a=2\lambda ,b=3\lambda ,c=4\lambda$

Put these values of a, b and c in (1), we get

$2\lambda (x-2)+3\lambda (y-5)+4\lambda (z+3)=0$

$\Rightarrow 2x-4+3y-15+4z+12=0$

$\Rightarrow 2x+3y+4z-7=0$

We know that the distance of a point $(x_1,y_1,z_1)$ from plane $ax+by+cz+d=0$ is given as

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

$=\frac{\left|2\right(7)+3(2)+4(4)-7|}{\sqrt{4+9+16}}$

$=\frac{|14+6+16-7|}{\sqrt{29}}\Rightarrow \frac{29}{\sqrt{29}}\Rightarrow \sqrt{29}.$