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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.

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3 Answers 3

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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?

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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 $\vec{p}$.

rough-sketch

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 $\vec{p}$.

rough-sketch-v2

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    $\begingroup$ 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. $\endgroup$
    – a_i_r
    Mar 25 at 5:46
  • $\begingroup$ 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 $\endgroup$ Mar 25 at 6:04
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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: $$\begin{align}\|\vec c\|^2\overrightarrow{BH}&=\|\vec c\|^2(\overrightarrow{BA}+k\vec c)\\&=(\vec c\cdot\vec c)\overrightarrow{BA}-(\vec c\cdot\overrightarrow{BA})\vec c\\&=\vec c\times(\overrightarrow{BA}\times\vec c). \end{align}$$

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