Chapter 11.4, Problem 31E

To determine

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

Answer to Problem 31E

  $7x+y-11z-5=0$

Explanation of Solution

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

Calculation:

Let $\langle a,b,c\rangle$ be the direction ratios of the normal to the plane , then equation of plane is

  $a\left(x-2\right)+b\left(y-2\right)+c\left(z-1\right)=0$

The point (-1,1,-1) lies on it .

  $\begin{array}{l}a\left(-1-2\right)+b\left(1-2\right)+c\left(-1-1\right)=0\\ -3a-b-2c=0\\ 3a+b+2c=0\mathrm{........}\left(1\right)\end{array}$

We know that if $\langle a_1,b_1,c_1\rangle$ and $\langle a_2,b_2,c_2\rangle$ are direction ratios of the normal to the two perpendicular planes, then

  $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

Now, the plane is perpendicular to plane $2x-3y+z=3$ and direction ratios of normal to the $2x-3y+z=3$ plane are $\langle 2,-3,1\rangle$ .

Therefore,

  $\begin{array}{l}a\left(2\right)+b\left(-3\right)+c\left(1\right)=0\\ 2a-3b+c=0\mathrm{........}\left(2\right)\end{array}$

On solving the equations (1) and (2), we get

  $\begin{array}{l}\frac{a}{1+6}=\frac{b}{4-3}=\frac{c}{-9-2}\\ \frac{a}{7}=\frac{b}{1}=\frac{c}{-11}\\ a=7k,b=k,c=-11k\end{array}$

where k is any non zero real number.

Therefore, equation of plane is

  $\begin{array}{l}7k\left(x-2\right)+k\left(y-2\right)-11k\left(z-1\right)=0\\ 7\left(x-2\right)+\left(y-2\right)-11\left(z-1\right)=0\\ 7x+y-11z-5=0\end{array}$