# Equation of a plane perpendicular to a line

I've read this but can't seem to reason it around for the reverse case: finding equation of a plane given a perpendicular line.

I'm looking for an equation for a plane perpendicular to the line $$L(t) =(4+3t, 2t, 4t-1)$$ that passes through $$(5, -4, 2)$$.

I tried to find two vectors $$u, v$$ whose cross product equals $$L(t)$$ or a vector $$u$$ whose dot product to $$L(t)$$ is zero and passes through $$(5,-4,2)$$ but couldn't (maybe the t is throwing me off). How should I approach this?

• Do you know how to write the equation of a plane, given a normal vector? Hint: by looking at the $t$-coefficients, you can see that the vector $v = (3,2,4)$ is a normal vector to the plane.
– Nick
Commented Oct 13, 2019 at 22:24
• I know it as $0=i(x-x_0)+j(y-y_0)+k(z-z_0)$ for the normal $n=(i,j,k)$. Working on understanding the rest of your hint now... Commented Oct 13, 2019 at 22:29
• Correct. In this case $i=3$, $j=2$, $k=4$, and $x_0=5$, $y_0=-4$, $z_0=2$.
– Nick
Commented Oct 13, 2019 at 22:31
• Omg it's so simple that way...I was too thrown off by the fact that there was a point I think somehow...So to recap the overall approach was to realize they gave us a line with direction vector $(3,2,4)$ (we don't care about it's $(4,0,-1)$ point) and that since it's the normal to our desired plane i just plug it into the "plane equation"? And from my understanding any point in a plane can go into the $x_0, y_0, z_0$ spots so we just use the 5,-4,2 it gave us. EDIT: I can't +1 comments but bless u for answering in a quick and clear way...post ur comment as an answer so I can vote it! Commented Oct 13, 2019 at 22:40
• Correct! All that matters is the direction vector of the line, and any point on the plane will work.
– Nick
Commented Oct 13, 2019 at 22:41

Since $$L(t) = (4,0,-1) + (3,2,4)t$$, the vector $$v = (3,2,4)$$ is a normal vector to the plane. Then using the given point $$(5,-4,2)$$, the equation of the plane is
$$0 = 3(x-5) + 2(y+4) + 4(z-2)$$