Chapter 9.6, Problem 35E

(a)

To determine

To find: The point on the line intersection the plane.

Answer to Problem 35E

The point on the line intersection the plane is $5x-2y-2z=1.$

Explanation of Solution

Given information:

The parametric equation are,

  $\begin{array}{l}x=2+t\\ y=3t\\ z=5-t\end{array}$

And the equation of plane is $5x-2y-2z=1.$

Calculation:

Calculate a line intersection the plane, then the values of $x$ , $y$ and $z$ will also satisfy the equation of the plane.

Therefore, putting the values of $x=2+t$ , $y=3t$ and $z=5-t$ in the plane $5x-2y-2z=1.$

  $5\left(2+t\right)-2\left(3t\right)-2\left(5-t\right)=1.$

Simplify the above equation,

  $\begin{array}{c}10+5t-6t-10+2t=1\\ t=1\end{array}$

Therefore, for $t=1$ the line with parametric equation $x=2+t$ , $y=3t$ and $z=5-t$ intersects the plane $5x-2y-2z=1.$

(b)

To determine

To find: The point on the line intersection the plane.

Answer to Problem 35E

The point on the line intersection the plane is $\left(3,3,4\right)$ .

Explanation of Solution

Given information:

The parametric equation are,

  $\begin{array}{l}x=2+t\\ y=3t\\ z=5-t\end{array}$

And the plane has equation $5x-2y-2z=1.$

Calculation:

Calculate the point at which the line with parametric equation $x=2+t$ , $y=3t$ and $z=5-t$ intersects the plane $5x-2y-2z=1.$

Since , in the $\left(a\right)$ , we have calculate that for $t=1$ the line with parametric equations $x=2+t$ , $y=3t$ and $z=5-t$ intersects the plane $5x-2y-2z=1.$

Thus, the point at which the given line intersects the given plane can be found by putting $t=1$ in the line with parametric equations $x=2+t$ , $y=3t$ and $z=5-t$ .

Putting the value of $t=1$ in the given line

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

  $\begin{array}{c}y=3\left(1\right)\\ =3,\end{array}$

And

  $\begin{array}{c}z=5-1\\ =4.\end{array}$

Therefore, the point at which the line with parametric equation $x=2+t$ , $y=3t$ and $z=5-t$ intersects the plane $5x-2y-2z=1.$ is $\left(3,3,4\right)$ .