Multivariable Calculus Vectors

Question

Find an equation for the plane that is perpendicular to the line $l(t)=(5,0,2)t+(3,-1,1)$ and passes through $(5,-1,0)$

Answer 1

From the equation of the line, you can determine the normal vector of the plane; it's (5,0,2). Do you see why?

Now you have a normal vector and a point for a plane. Let's call the normal vector $\hat{n}$, and the point $p_0$. We know that any vector in the plane is perpendicular to its normal vector. This can be written as follows: $ \hat{n} \cdot (p-p_0) = 0 $. And this is the equation of the plane.

The point $p$ can be rewritten componentwise as $p=(x,y,z)$. Then our equation becomes $(5,0,2) \cdot ( (x,y,z) - (5,-1,0) ) = 0$. This can be expanded out further if desired.

Answer 2

First find two linear independent vectors perpendicular to vector $(5,0,2)$, for example $(0,1,0)$ and $(-2,0,5)$. Next you can write quation of the plane which you find $\{(0,1,0)t+(-2,0,5)s+(5,-1,0) : s,t \in \mathbb{R}\}$ (if you put $s=t=0$ you see that plane passes through point $(5,-1,0)$).