So, I'm given a line with parametric equations
- $x = 2t + 3$
- $y = 1 – t$
- $z = 5t$
and I'm supposed to find symmetric and parametric equations of another line containing the origin and is perpendicular to the given line
i think a way to answer this would be to find a plane containing the given line because the normal vector can be used as a parallel vector of the missing line but i don't know how to get that normal vector. tips and help would be highly appreciated 🙂
Best Answer
Let's call the given line $L$, and let's take a line $M$ that passes through the origin and the point $(2t+3,1-t,5t)$ on $L$. We're going to adjust the value of $t$ to make $M$ perpendicular to $L$.
The direction vector of $L$ is $(2,-1,5)$, and the direction vector of $M$ is $(2t+3,1-t,5t)$. So, in order for $M$ to be perpendicular to $L$, we need $$ (2t+3,1-t,5t) \cdot (2,-1,5) = 0 $$ This gives $30t+5=0$, so $t=-\tfrac16$. Putting $t=-\tfrac16$ in the equation of $L$, we get the point $P=\tfrac16(16,7,-5)$. The desired line $M$ passes through the origin and $P$, so its equation is $M(s) = \tfrac{s}{6}(16,7,-5)$.
You can confirm that $L$ and $M$ are perpendicular, because $$ (2,-1,5) \cdot (16,7,-5) = 0 $$