# Find two planes which are parallel to each other, given that each plane contains a line (just a summary of the question)

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!

• I think the line $L$ should be given by $(x,y,z)=(1,2,3)+t(2,-1,-1)$ not $(1,2,3)+t(2,-1,1)$. – Dom Sep 29 '14 at 13:22
• yes you're right I made a typo, ill fix that – 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 $(1,2,3)$.
• 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 $3-t$. – Dom Sep 29 '14 at 13:48