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?

  • $\begingroup$ $t$ 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. $\endgroup$
    – Maesumi
    Jan 27, 2013 at 2:33
  • $\begingroup$ 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). $\endgroup$ Jan 27, 2013 at 2:35
  • $\begingroup$ 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 $t$. $\endgroup$
    – Maesumi
    Jan 27, 2013 at 2:40
  • $\begingroup$ 1/4? And then what? $\endgroup$ Jan 27, 2013 at 2:48
  • $\begingroup$ 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. $\endgroup$
    – Maesumi
    Jan 27, 2013 at 3:06

1 Answer 1


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 $s \in \mathbb{R}$.


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