Let L be the line through (1, 2, 3) and (3, 1, 2) and let L' be the line through (1, −1, 1) and (0, 2, 1). a) find Find the equation of a plane π containing L, and parallel to a plane containing L'. b)Find a point p ∈L and a point p'∈L' such that p − p' is orthogonal to both L and L'

A) so for A what I did was find the equation of both lines (parametric equation) L=(x,y,z) = (1,2,3) +t(2,-1,-1) L'=(x,y,z) = (1,-1,1) + t (-1,3,0)

Next thing I did was find a normal vector for these planes that have to be parallel to eachother, buy doing the crossproduct of the direction vectors of the two lines, and got n=[3,1,5]

Now I have no clue how to write the equation of these two planes, because dont I need a 3rd point to write one an equation of a plane?

b) have no clue where to start

Thank for your help!

  • $\begingroup$ I think the line $L$ should be given by $(x,y,z)=(1,2,3)+t(2,-1,-1)$ not $(1,2,3)+t(2,-1,1)$. $\endgroup$ – Dom Sep 29 '14 at 13:22
  • $\begingroup$ yes you're right I made a typo, ill fix that $\endgroup$ – user2877301 Sep 29 '14 at 13:23

You just need one vector $n$ that is orthogonal to both lines (you can take the cross product of the direction vectors of the lines as you did). Then if you take a point $p_1$ on the first line the plane is given by all points $v$ so that $(v-p_1)\cdot n=0$ where $\cdot$ is the dot product. You can do the same thing for the second plane (use the same normal vector $n$).

If you expand this out, you should get that the plane is given by \begin{align*}0&=((x,y,z)-(1,2,3))\cdot(3,1,5) \\ &=(x-1,y-2,z-3)\cdot(3,1,5) \\ &=3x-3+y-2+5z-15 \\ &=3x+y+5z-20\end{align*} as the equation for the first plane. Just do the exact same working for the second plane, by subtitute $(1,-1,1)$ for $(1,2,3)$.

  • $\begingroup$ So I use any point P1 (ex: for line L point (1,2,3)) and the normal , and V is an arbitrary (x,y,z) and have to solve for it? $\endgroup$ – user2877301 Sep 29 '14 at 13:31
  • $\begingroup$ Yep, the point is, when you think about it geometrically, that the vector between any two points on the plane must be orthogonal to the normal vector, so you can shoose any point as the "base" point for the formula. Of course, if the plane goes through the origin you don't need one, but I didn't want to bother working out if it did. $\endgroup$ – Dom Sep 29 '14 at 13:42
  • $\begingroup$ Umm, Dom, See I initially got that answer, but I threw it away because when I subbed the line back in x=1+2t, y=2-t, z=3+t I get 10t = 0, for the line to be on the plane when you sub x, y, z into the equation of the plane, is is supposed to be 0=0 or any constant (let c be the constant) ct=0, so t must be zero $\endgroup$ – user2877301 Sep 29 '14 at 13:42
  • $\begingroup$ When you sub in the line you do get 0, $3(1+2t)+(2-t)+5(3-t)-20=3+6t+2-t+15-5t-20=0$. You've sub in the wrong this for $z$, it should be $3-t$. $\endgroup$ – Dom Sep 29 '14 at 13:48
  • $\begingroup$ omg, I'm dumb I was using the one where I made a mistake on that you initially commented on, where it should be negative.... Okay thank you So much Dom! I'd like ur post, but it says I cant until I get 15 rep -.- $\endgroup$ – user2877301 Sep 29 '14 at 13:50

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