Question

Find the equation of the plane passing through the point $(1,1-1)$ and perpendicular to the plnes $x+2y+3z-7=0$ and $2x-3y+4z=0$.

Answer 1

Let the DR's of the normal to the required plane be $a,b,c$.

Since the required plane passes through the point $(1,1,-1)$.

So, its equation is :

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

Also, plane 1 is perpendicular to each of the given planes. Therefore,

$a+2b+3c=0$ .......(2)

and $2a-3b+4c=0$ .........(3)

On solving equation 2 and 3, we get,

$\frac{a}{17}=\frac{b}{2}=\frac{c}{-7}=\lambda$

$a=17\lambda ,b=2\lambda ,c=-7\lambda$

On putting the values of a, b, c in equation 1, we get,

$17\lambda (x-1)+2\lambda (y-1)+(-7\lambda )(z+1)=0$

$\lambda [17x+2y-7z-26]=0$

$17x+2y-7z=26$