Question

Find the equation of the plane through the points (2,1,-1), (-1,3,4) and perpendicular to the plane x-2y+4z=10.

Answer 1

The equation of the plane passing through (2,1,-1) is
a(x-2)+b(y-1)+c(z+1)=0 ...(i)
Since, this passes through (-1,3,4).
$\therefore a(-1-2)+b(3-1)+c(4+1)=0$
$\Rightarrow -3+2b+5c=0$ ...(ii)

Since, the plane (i) is perpendicular to the plane x-2y+4z=10.
$\therefore 1.a-2.b+4.c=0$
$\Rightarrow a-2b+4c=0$ ...(iii)
On solving Eqs. (ii) and (iii), we get ...(iii)
$\frac{a}{8+10}=\frac{-b}{-17}=\frac{c}{4}=\lambda$
$\Rightarrow a=18\lambda ,b=17\lambda ,c=4\lambda$

From Eq.(i).
$18\lambda (x-2)+17\lambda (y-1)+4\lambda (z+1)=0\Rightarrow 18x-36+17y-17+4z+4=0\Rightarrow 18x+17y+4z-49=0\therefore 18x+17y+4z=49$