Question

Find the equation of the plane passing through the intersection of the planes $3x+2y-z+1=0$ and $x+y+z-2=0$ and the point $(2,2,1)$ .

Answer 1

Let us take $z=0,$ then

$x-y=1$

$3x=3$

$⟹x=1,y=1$

Thus the point is $(1,1,0)$

when $x=0,$ then

$2y-z=-1$

$y-z=2$

Multiplying by 3, we get

$3y-3z=6$

$-y+3z=1$

$2y=7$

$⟹y=\frac{7}{2},z=\frac{3}{2}$

Thus the point is $(0,\frac{7}{2},\frac{3}{2})$

Let a be the vector from $(-1,2,1)$ to$(1,1,0)$

$a=⟨1-(-1),1-2,0-1⟩=⟨2,-1,-1⟩$

Let b be the vector from $(-1,2,1)$ to $(0,7/2,3/2):$

$b=⟨0-(-1),\frac{7}{2}-2,\frac{3}{2}-1⟩=⟨1,\frac{3}{2},\frac{1}{2}⟩$

The normal to the plane is:

$n=a\times b=\left|\begin{array}{ccc}i& j& k\\ 2& -1& -1\\ 1& \frac{3}{2}& \frac{1}{2}\end{array}\right|=1i-2j+4k$

Hence, the equation of plane is given\quad by

$n\ast ⟨x-x_0,y-y_0,z-z_0⟩=0$

$⟹⟨1,-2,4⟩\ast ⟨x-(-1),y-2,z-1⟩=0$

$⟹1(1+x)-2(y-2)+4(-1)=0$

$⟹x-2y+4=-1$ which is the required equation of plane.