Chapter 10.4, Problem 27E

To determine

To find: the general form of the equation of the plane.

Answer to Problem 27E

The general form of the equation of the plane is $-2x-4y+3z+4=0$ .

Explanation of Solution

Given information:

Passing points are $\left(3,1,2\right),\left(2,3,4\right),\left(-3,4,2\right)$

Calculation: to find the equation of the plane, first find the vector that is normal to the plane.

So find the normal vector,

  $\begin{array}{l}u=\langle 2-3,3-1,4-2\rangle \\ u=\langle -1,2,2\rangle \text{ [subtract]}\\ v=\langle -3-3,4-1,2-2\rangle \\ v=\langle -6,3,0\rangle \text{ [subtract]}\end{array}$

So the normal vector to the plane is,

  $\begin{array}{l}n=u\times v\\ n=\begin{array}{ccc}i& j& k\\ -1& 2& 2\\ -6& 3& 0\end{array}\\ n=\left(0-6\right)i-j\left(0+12\right)+k\left(-3+12\right)\text{ [simplify]}\\ n=-6i-12j+9k\end{array}$

Since perpendicular vector is $n=-6i-12j+9k$ , so direction numbers for $n$ are $a=-6,b=-12,c=9$ .

And the point is $\left(3,1,2\right)$ ,

So the general equation can be defined as:

  $\begin{array}{l}a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0\\ -6\left(x-3\right)-12\left(y-1\right)+\left(9\right)\left(z-2\right)=0\text{ [substitute]}\\ -6x+18-12y+12+9z-18=0\text{ [multiply]}\\ -6x-12y+9z+12=0\text{ [simplify]}\\ -2x-4y+3z+4=0\text{ [divide by 3]}\end{array}$

Thus, the general equation is $-2x-4y+3z+4=0$ .