Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Find the equation of the line passing through a point $B$, with position vector $ \vec b$ relative to an origin $O$, which is perpendicular to and intersects the line $\vec r= a+ \lambda \cdot c$, with $c \neq 0$, given that $B$ is not a point of the line.

Any help with this would be appreciated.


share|cite|improve this question
up vote 0 down vote accepted

The question feels to me like it should be in 3 dimensions, since it says "perpendicular to and intersects" and the "intersects" part would be trivial in 2d (and I think Bill Cook's answer covers the 2d case).

As in that answer, the direction of your given line is the vector $c$ and you need to find a direction vector $d$ for the new line, which should be perpendicular to $c$. The direction vector $d$ also has to lie in the plane determined by the known line and the known point. We can find the normal vector $n$ to that plane by taking the cross-product of the direction vector $c$ of the line and a vector from $B$ to any point on the line. Once we have $n$, $d$ must satisfy two conditions: $$d\cdot c=0$$ so that our new line is perpendicular to the existing line and $$d\cdot n=0$$ so that our new line lies in the same plane as the existing line and the given point. Having two equations is sufficient to solve for a 3d vector up to the length of that vector—that is, any $d$ that satisfies those two equations will be in the same direction as the line we want, and an equation for that line will be $\vec{r}=b+\lambda d$.

share|cite|improve this answer
So how can this be solved? – Euden Feb 13 '12 at 16:14
@Euden: If you let $d=\langle d_x,d_y,d_z\rangle$, the two dot-product equations become two equations in three unknowns, which you can reduce to a parameterized solution by various methods, including Gaussian elimination (equivalently, row-reduction on the adjoined coefficient matrix). – Isaac Feb 13 '12 at 22:48

The direction of your given line is the vector $c$, say $c=(c_1,c_2)$. You need to find a direction vector perpendicular to $c$, lets call such a vector $d=(d_1,d_2)$. There are a number of ways to find such a vector.

  • $d {\;\bf\cdot\;} c=0$ (their dot product must be zero so they're perpendicular). Thus you need to solve $c_1d_1+c_2d_2=0$. One solution is $d_1=c_2$ and $d_2=-c_1$. Thus $d=(c_2,-c_1)$ is the desired direction vector.

  • You could compute something like a 2-dimensional "cross-product" $$\begin{vmatrix} {\bf i} & {\bf j} \\ c_1 & c_2 \end{vmatrix} = c_2{\bf i}-c_1{\bf j} = (c_2,-c_1)=d$$

Then the line you're looking for is $\vec{r}(\lambda) = b+\lambda d = b+\lambda(c_2,-c_1)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.