# What makes two lines in 3-space perpendicular?

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?

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.

• So, if I my knowledge of high school plane geometry is correct, two lines are perpendicular if their slopes are negative reciprocals. So, in 3D, I see that principle still applies: I can take the negative reciprocal of each component in the direction vector and get the perpendicular direction vector? Jan 30, 2019 at 1:31
• For one thing, it's important to note that we're still talking about 2D here: the line you gave above is only on a plane. To answer your question, not quite. If you try that here, you'd get $<-1, 1/2>$ which is not orthogonal to $<1, -2>$. However, you can switch the entries and flip the sign on one: $<2, 1>$ is orthogonal to $<1, -2>$.
– Alex
Jan 30, 2019 at 1:37
• So, if I find any one orthogonal vector, will all of the scalar multiples of that vector be orthogonal to the original direction as well? I notice that $<2,1>$ is equal to $2*<1,1/2>$. I think I've constructed a rough proof that this is true but I wanted to double check in case I made a mathematical mistake. Jan 30, 2019 at 1:51
• Yes, that's true. Think about it geometrically: two vectors on the plane are orthogonal if there is a right angle between them. Making any vector $c$ times longer (i.e. multiplying it by scalar $c$) does not change its direction, so it won't change the angle.
– Alex
Jan 30, 2019 at 16:19