# Plane geometry problem

1. Sep 2, 2011

### chmate

1. The problem statement, all variables and given/known data

Find the equation of plane which contains the line $l:\left\{\begin{array}{l} x=t+2 \\y=2t-1\\z=3t+3 \end{array}\right.$, and makes the angle of $\frac{2\pi}{3}$ with the plane $\pi:x+3y-z+8=0$.

3. The attempt at a solution

My attempt was to find the normal vector of plane which contains the line by the cross product of vector $\vec{a}$ of line and some vector created between points $P(2,-1,3)$ of line and point $Q(x,y,z)$, than by using the formula to find angles, but this leads me to complications.

I think there's another way to solve this, which I don't know.

Thank you

2. Sep 2, 2011

### HallsofIvy

Staff Emeritus
If two planes make a given angle, $\theta$, with each other, then their normal vectors make the same angle. The dot product of two vectors, u and v, is given by $u\cdot v= |u||v|cos(\theta)$. So if we take <a, b, c> to be a unit vector perpendicular to the desired plane, we must have $<a, b, c>\cdot<1, 3, -1>= \sqrt{1+ 9+ 1} cos(2\pi/3)$ or $a+ 3b- c= -.5\sqrt{11}$. That, together with $a^2+ b^2+ c^2= 1$ gives two equations to solve for a, b, and c.