Equation of plane given a point and contain line of intersection of 2 plane

Question

What is the equation of the plane that passes through the point $(−1, 2, 1)$ and contains the line of intersection of the planes $x + y − z = 2$ and $2x − y + 3z = 1$?

How can i find the equation of the plane?

Answer 1

We must first find two other points on the plane. When $z = 0:$

$$x + y = 2$$ $$2x - y = 1$$

$$3x = 3$$

$$ x = 1, y = 1$$

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

When $x = 0:$ $$y - z = 2$$ $$-y+3z=1$$ $$y=\frac72,z=\frac32$$ Thus the point is $( 0, \frac72, \frac32 ).$

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 $( 0, \frac72, \frac32):$

$$b = < 0 - (-1), 7/2 - 2, 3/2 - 1 > = < 1, 3/2, 1/ 2 >$$

The normal to the plane is:

$n=a$ x $b$ $$=\begin{vmatrix} i & j & k \\ 2 & -1 & -1 \\ 1 & \frac32 & \frac12 \end{vmatrix}$$ $$=i\begin{vmatrix} -1 & -1 \\ \frac32 & \frac12 \end{vmatrix}-j\begin{vmatrix} 2 & -1 \\ 1 & \frac12 \end{vmatrix}+k\begin{vmatrix} 2 & -1\\ 1 & \frac32 \end{vmatrix}$$ $$=1i-2j+4k$$

The general equation for a plane is:

$$n.<x-x_0,y-y_0,z-z_0>=0$$ $$<1,-2,4>.<x-(-1),y-2,z-1>=0$$ $$1(x+1)-2(y-2)+4(z-1)=0$$

$$x-2y+4z=-1$$

Answer 2

First, find the line of intersection. You know that the line of intersection is parallel to the plane. Then, find the vector orthogonal to this line. Use this orthogonal vector and the point given to construct the equation of the plane.

Answer 3

Hint:

Consider the pencil of planes that have this line in common:

$$\lambda(x+y-z-2)+(1-\lambda)(2x-y+3z-1)=0.$$

By plugging the coordinates of the given point, you will determine $\lambda$.

Answer 4

We know that the equation of plane through line of intersection of the two other planes is given by

$$(x+y-z-2)+k(2x-y+3z-1)=0$$ or $$x(2k+1)+y(1-k)+z(3k-1)-(2+k)=0\qquad\qquad\qquad(1)$$

Now given that it passes through (-1,2,1), so $(1)$ must be satisfied by substituting that point. Solving the equation after substitution we obtain $k=-1$ where by inserting in (1) yields to:$$x-2y+4z+1=0$$