Chapter 9.6, Problem 37E

(a)

To determine

To find: The coordinate of parametric equation.

Answer to Problem 37E

The coordinate of parametric equation is $\left(0,3,-1\right)$

Explanation of Solution

Given information:

The parametric equation are,

Line 1: $x=1-t,$

  $y=3t,$

  $z=-5+2t$

Line 2: $x=-1+2t,$

  $y=6-6t,$

  $z=4-10t$

Calculation:

Calculate two points that lie on line1

  $x=1-t,$

  $y=3t,$

  $z=-5+2t$

By sitting $t=0$ in line 1, we get

  $x=-1+2t,$

  $y=6-6t,$

  $z=4-10t$

Putting $t=0$ in line 1 ,

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

  $\begin{array}{c}y=3\times \left(0\right)\\ =0,\end{array}$

  $\begin{array}{c}z=-6+5\times 0\\ =-6,\end{array}$

The point that lies on line 1 by setting $t=0$ is $\left(1,0,-6\right)$

Putting $t=1$ in line 1,

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

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

  $\begin{array}{c}z=-6+5\left(1\right)\\ =-1,\end{array}$

Thus, the point that lie on the 1 by setting $t=1$ is $\left(0,3,-1\right)$

To check whether point $\left(1,0,-6\right)$ and $\left(0,3,-1\right)$ lie on line 2, we need to substitute these point one by one and if the value of $t$ same for both points , then both will lie on the line 2.

Putting the point $\left(1,0,-6\right)$ in line 2 ,we get

  $\begin{array}{l}1=-1+2t\\ t=1,\end{array}$

  $\begin{array}{l}0=6-6t\\ t=1,\end{array}$

  $\begin{array}{c}-6=4-10t\\ t=1,\end{array}$

Since , value of t are same for $x=1,$

  $y=0$ and $z=-6$ .

Therefore, the point $\left(1,0,-6\right)$ lies on line 2: $x=-1+2t,$

  $y=6-6t,$

  $z=4-10t$.

Putting the point $\left(0,3,-1\right)$ in line2,

  $\begin{array}{l}0=-1+2t\\ t=\frac{1}{2},\end{array}$

  $\begin{array}{l}3=6-6t\\ t=\frac{1}{2},\end{array}$

  $\begin{array}{l}-1=4-10t\\ t=\frac{1}{2}\end{array}$

Therefore, the value of $t$ are same for $x=0,$

  $y=3$ and $z=-1.$

Therefore, the point $\left(0,3,-1\right)$ lies on line 2.

(b)

To determine

To find: The coordinate of parametric equation.

Answer to Problem 37E

The coordinate of line 3 and line 4 are not same.

Explanation of Solution

Given information:

Line3: $x=4t,$

  $y=3-6t,$

  $z=-5+2t$

Line 4: $x=8-2t$

  $y=-9+3t,$

  $z=6-t$

Calculation:

We have to calculate two points that lie on line 3:

  $x=4t,$

  $y=3-6t,$

  $z=-5+2t$

And the line 4:

  $x=8-2t$

  $y=-9+3t,$

  $z=6-t$

Are not same by calculating a point on a line3 and the point does not lie on line4

Let $t=0$

Putting $t=0$ in line 3 ,

  $x=0,$

  $y=3,$

  $z=-5$

Hence the point on line 3 at $t=0$ is $\left(0,3,-5\right)$ ’

The point $\left(0,3,-5\right)$ will on line 4 , if value of t are same for $x=0,$

  $y=3,$

  $z=-5$ in line4.

Putting the point $\left(0,3,-5\right)$ in line 4 ,we get

  $\begin{array}{l}0=8-2t\\ t=4,\end{array}$

  $\begin{array}{l}3=-9+3t\\ t=4,\end{array}$

  $\begin{array}{l}-5=6-1\\ t=11.\end{array}$

Therefore, the line 3 and 4 are not same by finding a point on line 3 and does not lie on line 4.