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?

  • $\begingroup$ How did you arrive at $V_2=(1,-1,-1)$? $\endgroup$ Sep 23, 2020 at 20:14
  • $\begingroup$ Your right it should have been$(1,-1,-3)$ $\endgroup$ Sep 23, 2020 at 20:19
  • $\begingroup$ 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. $\endgroup$
    – Math Lover
    Sep 23, 2020 at 20:21
  • $\begingroup$ your cross product seems wrong $\endgroup$
    – Math Lover
    Sep 23, 2020 at 20:23
  • $\begingroup$ Ok, I fixed the errors. How's the cross product looking? $\endgroup$ Sep 23, 2020 at 20:40


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