I'm trying to find parametric equations for a line that passes through the point (0,1,2) and is perpendicular to the line:
x = 1 + t
y = 1 - t
z = 2t
Here's what I know: I have a directional vector, v, given by <1,-1,2> and another vector, r0, given by <1,1,0>. I can also make another vector for the point, P, <0,1,2>.
Can I use any of that information to solve my question? How?
Best Answer
Find a vector perpendicular to the line $\underbrace{(1,1,0)}_{\vec{x}_0} + t\underbrace{(1,-1,2)}_{\vec{v}}$ and through $(0,1,2)$ i.e. a vector in the plane containing the line and the point and perpendicular to the line.
To do this, consider the vector connecting $(1,1,0)$ and $(0,1,2)$ i.e. $\vec{u} = (-1,0,2)$. Subtract out the projection of this vector $\vec{u}$ onto $\vec{v}$. $$\dfrac{\vec{u} \cdot \vec{v}}{\Vert \vec{v} \Vert} = \dfrac{-1+4}{\sqrt{6}} = \dfrac{3}{\sqrt6} = \dfrac{\sqrt6}2$$ Hence, $$\vec{n} = \vec{u} - \dfrac{\sqrt6}2 \dfrac{\vec{v}}{\Vert \vec{v} \Vert}$$ Hence, the equation of the line perpendicular to the given line passing through $(0,1,2)$ is $$(0,1,2) + s \vec{n}$$ where $s \in \mathbb{R}$.