# Find parametric equations for a line

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?

• $t$ is supposed to be a number, not a vector. Are the original line and then one you want to intersect? If not then the answer is not unique. In fact the whole plane through $P_0$ and perpendicular to the given line will be the answer. Jan 27, 2013 at 2:33
• The original line, given by the parametric equations above, is supposed to be perpendicular to the answer line. And the answer line passes through the point (0,1,2). Jan 27, 2013 at 2:35
• Let $R=(1+t,1-t,2t)$ be a point on line given. Let $r_0=(0,1,2)$ be the given point. Let $u=(1,-1,2)$ be direction of the given line then you want $(R-r_0)\cdot u=0$. Solve that for $t$. Jan 27, 2013 at 2:40
• 1/4? And then what? Jan 27, 2013 at 2:48
• Now use $t=1/4$ in equation of given line to get coordinate of the end of line segment that start at $r_0$, say $r_1$. Now find the equation of line that goes through $r_0$ and $r_1$, that is the line that goes through $r_0$ and hits the given line at $r_1$ in a perpendicular fashion. Jan 27, 2013 at 3:06

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 $s \in \mathbb{R}$.