Chapter 11, Problem 66RE

To determine

To find: The general form of the equation of the plane with the given characteristics.

Answer to Problem 66RE

The required equation of the plane is $2x-y+z-6=0$ .

Explanation of Solution

Given information: The plane passes through the points $(0,0,6)$ and $(2,1,3)$ and is perpendicular to the plane $2x+3y-z=5$ .

Calculation:

To find the equation of the plane, a point in the plane is needed and a vector that is normal to the plane.

Let $u$ be the vector from $(0,0,6)$ to $(2,1,3)$ .

Then $u=\langle 2-0,1-0,3-6\rangle =\langle 2,1,-3\rangle$ .

As this vector is perpendicular to the plane, so the normal vector can be found by taking the cross product of $u$and the vector $v$ , where $v=(2,3,-1)$ is the vector formed by taking the direction numbers of the given plane.

So the normal vector is:

  $n=u\times v=\left|\begin{array}{l}ijk\\ 21-3\\ 23-1\end{array}\right|=8i-4j+4k$

Using the direction numbers for n, i.e., $n=\langle a,b,c\rangle =\langle 8,-4,4\rangle =4\langle 2,-1,1\rangle$ and the initial point $(x_1,y_1,z_1)=(0,0,6)$ in the standard form of the equation of the plane,

  $\begin{array}{l}a(x-x_1)+b(y-y_1)+c(z-z_1)=0\\ \Rightarrow 2(x-0)+(-1)(y-0)+1(z-6)=0\\ \Rightarrow 2x-y+z-6=0\end{array}$

Thus, the general form of the equation of the plane passing through the points $(0,0,6)$ and $(2,1,3)$ and perpendicular to the plane $2x+3y-z=5$ is $2x-y+z-6=0$ .