Find the point of intersection given their parametric equations, then find the equation of the plane they exist on.

The given is

Line $$(1):$$ $$x= 1+2t, \space y=4t, \space z= 3-3t$$ Line $$(2):$$ $$x= 2+t , y=-1-t , z=-3t$$

First I changed the variable in line $$2$$ to $$s$$ as to find the point of intersection of the $$y$$ coord. $$y=4t=-1-s$$ $$s=-1-4t$$ Now we set $$t=0$$ to solve for $$x$$ and $$z$$ $$x=2+(-1-4(0))$$ $$x=1$$ $$z=-3(-1-4(0))$$ $$z=3$$ So our point of intersection is $$(1,0,3)$$ The $$y$$ coord being zero means that the intersection is parallel to the $$xy$$ and $$zy$$ planes. So now we extrapolate out two vector from our given parametric equations $$\vec V_1= \langle 2,4,-3\rangle$$ $$\vec V_2= \langle 1,-1,-3\rangle$$ $$\vec V_1 X \vec V_2=\langle-15,3,-6\rangle$$ So the equation of the line is $$-15(x-1)+3(y)-6(z-3)=0$$ Am I correct?

• How did you arrive at $V_2=(1,-1,-1)$? Sep 23, 2020 at 20:14
• Your right it should have been$(1,-1,-3)$ Sep 23, 2020 at 20:19
• In the question, you write $y = -2 - t$ which seems a typo because that is not possible. In the working, you have used $-1 - t$ which seems fine. Sep 23, 2020 at 20:21
• your cross product seems wrong Sep 23, 2020 at 20:23
• Ok, I fixed the errors. How's the cross product looking? Sep 23, 2020 at 20:40