Question

How do you know if two lines are perpendicular in 3-D space?

Answer 1

Hint: We use the concept of vectors and their dot products to find the lines being perpendicular in 3-D space. We use the equation in their parametric form and find their intersecting point. Then using the dot product, we find the multiplication as 0.

Complete step-by-step solution:
We can solve the condition of two lines being perpendicular in 3-D space.
We can use the concept of multiple of their slopes being $-1$ or we take their vector form to use the dot product on them.
Depending on what form the lines' equations take on, you need to find an $\left( x,y,z \right)$ point that's on both lines. If the lines are in parametric form $x=a+bt,y=c+dt,z=e+ft$ for line 1 and $x=g+hu,y=i+ju,z=k+mu$ for line 2, then set \[x=a+bt=g+hu\] etc, and you have 3 equations in 2 unknowns, t and u. These may or may not have a solution; if they do then the lines intersect at $\left( x,y,z \right)$.
The next step is to find two vectors starting from the point of intersection: let $\left( p,q,r \right)$ be the intersection point, and on line 1 use the equation to find any other point, say $\left( a,c,e \right)$ with $t=0$.
Then $v=\left\langle a-p,c-q,e-r \right\rangle $ is a vector in line 1 direction; do the same for line 2 to get vector w. Then v is perpendicular to w if and only if $v.w=0$. You can do the dot product coordinate-wise, using the formula $\left\langle a,b,c \right\rangle .\left\langle d,e,f \right\rangle =ad+be+cf$.

Note: Now the test for perpendicular is that the dot product of the direction vectors of the 2 lines has to be 0. Remember if the dot product of two vectors is 0 they're perpendicular. So, if the direction vector is, if 2 lines are perpendicular then the lines are perpendicular.