Question

In Exercises 35–38, find an equation of the line of intersection of the two given planes.

$x-2z=7\text{and}x-3y-4z=0$

Short answer

The equation is$\mathbf{r}\left(t\right)=⟨2-3t,-3+3t,5+2t⟩.$

Step-by-step solution

Step 1:Given information

$\text{Consider the pcints}P(2,-3,5),Q(3,1,0)\text{and}R(-1,0,7)$

Step 2:Calculation

$\text{Consider the points}P(2,-3,5),Q(3,1,0)\text{and}R(-1,0,7)\text{.}$

$\text{The equation of the line containing the points}A\left(x_0,y_0,z_0\right)\text{and}B\left(x_1,y_1,z_1\right)\text{is given by:}$

Where,

$\text{The equation of line containing the points}P(2,-3,5)\text{and}R(-1,0,7)\text{is:}$

$\mathbf{r}\left(t\right)=⟨2+t(-1-2),-3+t(0+3),5+t(7-5)⟩$

$=⟨2-3t,-3+3t,5+2t⟩$

$\text{Therefore, the equation of line the points}P(2,-3,5)\text{and}R(-1,0,7)\text{is}$

$\mathbf{r}\left(t\right)=⟨2-3t,-3+3t,5+2t⟩.$