# Perpendicular Vectors

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.

Thanks

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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 $\vec{r}=b+\lambda d$.

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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 $\vec{r}(\lambda) = b+\lambda d = b+\lambda(c_2,-c_1)$.

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