Chapter 10, Problem 62RE

To determine

To find: The general form of the equation of the plane that passes through $\left(6,-3,4\right)$ .

Answer to Problem 62RE

The general form of equation of plane that passes through $\left(6,-3,4\right)$ is $-x+y-2z+12=0$ .

Explanation of Solution

Given information:

A plane passes through the point $\left(6,-3,4\right)$ .The plane is perpendicular to $x=1-t$ , $y=2+t$ and $z=4-2t$ .

Calculation:

Consider the equations.

  $x=1-t$.........(1)

  $y=2+t$.........(2)

  $z=4-2t$.........(3)

Substitute $t=0$ in equation (1) to obtain the first point.

  $x=1$

Substitute $t=0$ in equation (2) to obtain the first point.

  $y=2$

Substitute $t=0$ in equation (3) to obtain the first point.

  $z=4$

Substitute $t=1$ in equation (1) to obtain the second point.

  $\begin{array}{c}x=1-1\\ =0\end{array}$

Substitute $t=1$ in equation (2) to obtain the second point.

  $\begin{array}{c}y=2+1\\ =3\end{array}$

Substitute $t=1$ in equation (3) to obtain the second point.

  $\begin{array}{c}z=4-2\left(1\right)\\ =2\end{array}$

So, the first point is $P\left(1,2,4\right)$ and the second point is $Q\left(0,3,2\right)$ .

Calculate the coordinates of vector $PQ$ .

  $\begin{array}{c}\overrightarrow{PQ}=\left(0-1,3-2,2-4\right)\\ =\left(-1,1,-2\right)\end{array}$

The general equation of the plane is:

  $ax+by+cz+d=0$

Substitute $-1$ for $a$ , $1$ for $b$ and $-2$ for $c$ in the above equation.

  $-x+y-2z+d=0$.........(4)

Substitute the points given in the above equation.

  $\begin{array}{c}-1+1-2\left(6\right)+d=0\\ d=12\end{array}$

Substitute $d=12$ in the equation (4).

  $-x+y-2z+12=0$

Therefore, the general form of the equation of the plane is $-x+y-2z+12=0$ .