Chapter 9, Problem 54RE

To determine

To find : The equation of plane.

Explanation of Solution

Given: The lines are $x=2+2t$ , $y=4t$ , $z=-6$ and the point $P\left(5,3,0\right)$ .

Consider vector $u=\langle 2,4,0\rangle$ is parallel to the line. Another vector parallel is required to the desired plane. Choose a line where $t=$ 0 and draw a vector from there to $P$ .

$\begin{array}{c}v=P-Q\\ =\left(5,3,0\right)-\left(2,0,-6\right)\\ =\langle 3,3,6\rangle \end{array}$

The vector normal to the plane is calculated as,

$\begin{array}{c}n=\left|\begin{array}{ccc}i& j& k\\ 2& 4& 0\\ 3& 3& 6\end{array}\right|\\ =\left(24-0\right)i+\left(0-12\right)j+\left(6-12\right)k\\ =24i-12j-6k\\ =4i-2j-1k\end{array}$

The equation normal to the plane is,

$\begin{array}{c}4\left(x-5\right)-2\left(y-3\right)-z=0\\ 4x-2y-z=14\end{array}$

Therefore, the equation normal to the plane is $4x-2y-z=14$ .