Question

The equation of the plane passing through the points $(-1,2,0)(2,2,-1)$ and perpendicular the line $\frac{x-1}{1}=\frac{2y+1}{2}=\frac{z-1}{-1}$

Answer 1

Cartesian equation of a plane passing through point $(x_1,y_1,z_1)$ having direction ratios $a,b,c$ for its normal is $a(x-x_1)+b(y-y_1)+c(z-z_1)=0$

It is given that points $(-1,2,0)$ & $(2,2,-1)$ lie on the plane.

Let the equation of plane through $(-1,2,0)$ be

$a(x+1)+b(y-2)+c(z-0)=0$------------(1)

Now, if it passes through $(2,2,-1)$ the substitute for $(x,y,z)$.

$a(2+1)+b(2-2)+c(-1-0)=0$

$\Rightarrow 3a-c=0$----------------(2)

$\Rightarrow c=3a$---------------(3)

Given the plane is perpendicular to the line

$\frac{x-1}{1}=\frac{2y+1}{2}=\frac{z-1}{-1}$

$\therefore$ a\times 1+2\times b-c\times 1=0$

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

Using Equation(3),

$\Rightarrow 2b-2a=0$

$\Rightarrow b=a$---------------(5)

$\therefore$ Equation of plane is

$a(x+1)+b(y-2)+c\left(z\right)=0$

$\Rightarrow a(x+1)+a(y-2)+3az=0$

$\Rightarrow x+1+y-2+3z=0$

$\Rightarrow x+y+3z=1$