Question

Find the equation of the plane passing through the intersection of the planes
2x-3y+z+4=0 and x-y+z+1=0 and perpendicular to the plane x+2y-3z+6=0.
Find the equation of the plane passing through the intersection of the planes​

Answer 1

Our plane passes through the intersection of the planes,

• $2x-3y+z+4=0$

• $x-y+z+1=0$

where the intersection is a straight line. First we are finding a direction ratio of that straight line.

It is clear that the vector $\left<2,\ -3,\ 1\right>$ and $\left<1,\ -1,\ 1\right>$ are perpendicular to the first and the second planes respectively. So these two vectors are perpendicular to their intersection (the straight line).

Hence the cross product of the two vectors will be a direction ratio of the line, i.e.,

$\longrightarrow\vec{b_1}=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\1&-1&1\\2&-3&1\end{array}\right|$

$\longrightarrow\vec{b_1}=\left<2,\ 1,\ -1\right>$

This is a direction ratio of the intersection line, which lies on our plane.

Let $(a,\ b,\ c)$ be a point on the line. As this point lies on both those planes, we have,

• $2a-3b+c+4=0$

• $a-b+c+1=0$

We find some possible point $(a,\ b,\ c)$ that lies on the intersection line. Here we assume a particular real number value to any coordinate of the point.

Here I assume $b=0,$ and so the plane equations become,

• $2a+c+4=0$

• $a+c+1=0$

Solving these two, we get $(a,\ b,\ c)=(-3,\ 0,\ 2).$

Let $(x,\ y,\ z)$ be a point on our plane, so that the vector $\left<x+3,\ y,\ z-2\right>$ lies on our plane.

Given that our plane is perpendicular to the plane $x+2y-3z+6=0,$ to which it is clear that the vector $\left<1,\ 2,\ -3\right>$ is perpendicular. That means this vector is parallel to our plane.

Now we have two vectors parallel to our plane:

• $\vec{b_1}=\left<2,\ 1,\ -1\right>$

• $\vec{b_2}=\left<1,\ 2,\ -3\right>$

Let us find the normal vector to our plane.

$\longrightarrow\vec{n}=\vec{b_1}\times\vec{b_2}$

$\longrightarrow\vec{b_1}=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\2&1&-1\\1&2&-3\end{array}\right|$

$\longrightarrow\vec{n}=\left<-1,\ 5,\ 3\right>$

As this normal vector is perpendicular to the vector $\left<x+3,\ y,\ z-2\right>,$ their dot product is zero, i.e.,

$\longrightarrow\left<x+3,\ y,\ z-2\right>\cdot\left<-1,\ 5,\ 3\right>=0$

$\longrightarrow-(x+3)+5y+3(z-2)=0$

$\longrightarrow\underline{\underline{x-5y-3z+9=0}}$

This is the equation of our plane.