Let P be the plane in 3-space that passes through the point A = (−3, −2, 1) and is perpendicular to the line L with vector equation t<1,2,1>.

Question

I've been working on this (not-for-marks) problem for a couple hours now and am completely lost. It's review for our midterm. Can anyone walk me through it?

Answer 1

The equation for $P$ is: $1x+2y+1z = d$, where $n = (1,2,1)$ is the normal vector to $P$. The plane passes through $(-3,-2,1)$ means that $1(-3)+2(-2)+1(1) = d\implies d = -6\implies P: x+2y+z = -6$. The intersection $Q$ of $L$ and $P$ is found by plugging in $x = t, y = 2t, z = t$ into the equation of $P$. Thus: $t+2(2t)+t = -6 \implies 6t=-6\implies t = -1\implies Q=(-1,-2,-1)$.