Question

Find the equation of the plane which passes through the point $(3,2,0)$ and the line $\frac{x-4}{1}=\frac{y-6}{5}=\frac{z-4}{4}$ is?

Answer 1

The equation of a plane passing through a point $(3,2,0)$ is $a(x-3)+b(y-2)+c(z-0)=0$ .........$\left(1\right)$

Since the above plane contains the line

$\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}$

So, the point $(3,6,4)$ satisfies the equation of plane $\left(1\right)$

$\therefore a(3-3)+b(6-2)+c(4-0)=0$

$\Rightarrow 4b+4c=0$

$\Rightarrow b+c=0$ or $b=-c$ .........$\left(2\right)$

As the plane contains the line, therefore normal to the plane is perpendicular to the line

$a_1{a}_2+b_1{b}_2+c_1{c}_2=0$ where $a_1=a,b_1=b,c_1=c$ and $a_2=1,b_2=5,c_2=4$

$\therefore a\times 1+b\times 5+c\times 4=0$

or $a+5b+4c=0$

or $a-5b+4c=0$ from $\left(2\right)$

$\Rightarrow a=c$

On putting the values of $a,b$ and $c$ in eqn$\left(1\right)$, we get

$c(x-3)-c(y-2)+c(z-0)=0$

or $c(x-3-y+2+z-0)=0$

or $x-y+z=1$ is the required equation of the plane.