I need to find the equations for the line which passes through the point $(-5,3)$ and is perpendicular to the line $<3+t,-5-2t>$.
I know I can get the direction vector $<1,-2>$ out of the line equation. If I needed a parallel line I would just use the direction vector and the given point to construct a vector parametric equation for the line.
But, I am not sure what it means for two lines to be perpendicular with respect to their direction vector. My textbook (Stewart's Calculus 5th edition) does not define this, and I don't have any other calculus references for the course.
My intuition is to find the vector that is orthogonal to the direction vector, but I'm not sure if this is correct.
I set my direction vector as $\vec{a}$ and said $\vec{a}\cdot\vec{b} = \vec{0}$, and solving this using the dot production definition gives me $\vec{b} = <-1,2>$. Is this the vector I use to define the perpendicular line?
Best Answer
You are correct that you should find a vector that is orthogonal to $<1, -2>$, however the vector you calculated as $\vec{b}$ does not seem to satisfy that requirement:
$$ \vec{a}\cdot\vec{b} = (1)(-1) + (-2)(2) = -5 \neq 0 $$
One $\vec{b}$ that satisfies the requirement would be $<1, 1/2>$, since we would have
$$ \vec{a}\cdot\vec{b} = (1)(1) + (-2)(1/2) = 1 - 1 = 0 $$ as desired. You can then use that vector (or any scalar multiple of it) as the direction vector for your line.