# Finding vector equation of a line

Show that the equation of a straight line passing through the point with position vector $$\vec{b}$$ and perpendicular to the line $$\vec{r}=\vec{a}+\mu \vec{c}$$ is of the form $$\vec{r}=\vec{b}+\beta \vec{c}×\{(\vec{a}-\vec{b})×\vec{c}\}$$.

How to derive the vector parallel to the required line? I get that this vector must be perpendicular to $$\vec{c}$$ but I can't derive the $$\vec{c}×\{(\vec{a}-\vec{b})×\vec{c}\}$$ form.

In terms of unknown, it may be abstract. Let's recall in our first multivariable class, with numerical example. Like find a line passing through $$b=(\pi,1,2)$$ such that it is perpendicular to the line $$\ell:(1,2,3)t+(1,1,1)$$ How do you solve it? You first identify that the set of vectors perpendicular to $$\ell$$ must satisfy $$v\cdot(1,2,3)=0$$, so this gives a plane with $$c:=(1,2,3)$$ as normal vector (equivalently $$x+2y+3z=0$$).

Now we want a line passing through $$b$$, so we should adjust,parallelly, to $$\Pi:x+2y+3z=D$$ so that $$b$$ lies on $$\Pi$$. Now we need to choose one line that passing through $$b$$ and lie on $$\Pi$$. How can you choose? Possibly, you identify that $$a:=(1,1,1)$$ is on $$\ell$$, but $$a$$ may not lie on $$\Pi$$, so is $$a-b$$. Hence we need to consider the projection of $$a-b$$ on $$\Pi$$. Then this gives the directional vector of the line we need.

Okay finish numerical example. Now can you construct your proof based on this idea?

Let the required line be $$\vec{r} = \vec{b} + \beta \vec{l}$$. Now as per the diagram below, we can say that vector $$\vec{p} = (\vec{a}-\vec{b}) \times \vec{c}$$ is perpendicular to the plane containing these two lines.

With this, we can say that $$\vec{l}$$ is perpendicular to $$\vec{c}$$ and $$\vec{p}$$.

So we can write $$\vec{l}$$ is parallel to $$\vec{c} \times \vec{p}$$. So the line equation becomes

$$\vec{r} = \vec{b} + \beta \vec{c} \times ((\vec{a}-\vec{b}) \times \vec{c})$$

## EDIT

After the comments, I am adding more details here. Assume that these two lines lie in plane $$M$$.

Now $$\vec{a} - \vec{b}$$ and $$\vec{c}$$ lie in plane $$M$$. So the vector perpendicular to the plane is $$\vec{p}$$. As $$\vec{l}$$ also lies in plane $$M$$, $$\vec{l}$$ is perpendicular to $$\vec{p}$$.

• What is the need for involving $\vec{a}-\vec{b}$ and why is it that $\vec{l}$ is perpendicular to both $\vec{c}$ and $\vec{p}$. It isn't clear from your figure. Mar 25 at 5:46
• I have added another figure which is somewhat better than previous one. please check it. The need to involve $\vec{a} - \vec{b}$ is to find another vector which is in a plane with $\vec{c}$. Another question that why $\vec{l}$ is perpendicular to $\vec{p}$ ? I suppose the new figure will answer your question Mar 25 at 6:04

Let $$H$$ be the orthogonal projection of the point $$B$$ on the line $$A+\Bbb R\vec c$$.

$$H=A+k\vec c\text{ and }\overrightarrow{BH}\perp\vec c$$

hence $$0=\overrightarrow{BH}\cdot\vec c=(\overrightarrow{BA}+k\vec c)\cdot\vec c$$, i.e. $$k\|\vec c\|^2=-\overrightarrow{BA}\cdot\vec c.$$

A vector `parallel to the line' $$(BH)$$ is therefore: \begin{align}\|\vec c\|^2\overrightarrow{BH}&=\|\vec c\|^2(\overrightarrow{BA}+k\vec c)\\&=(\vec c\cdot\vec c)\overrightarrow{BA}-(\vec c\cdot\overrightarrow{BA})\vec c\\&=\vec c\times(\overrightarrow{BA}\times\vec c). \end{align}