0
$\begingroup$

At Line $L_{1}$ has equation $r = \begin{pmatrix} -5\\ -3\\ 2 \end{pmatrix} + \lambda \begin{pmatrix} -1\\ 2\\ 2 \end{pmatrix}$

A line $L_{2}$ passing through the origin intersects $L_{1}$ and perpendicular to $L_{1}$. Find the vector equation of $L_{2}$!

I find the vector for $L_{2}$ using cross product $\begin{pmatrix} i & j & k\\ -5 & -3 & 2\\ -1 & 2 & 2 \end{pmatrix}$, and I don't have any idea to do with that vector....

$\endgroup$
4
  • $\begingroup$ You need to solve $r\cdot\begin{pmatrix}-1\\ 2\\ 2 \end{pmatrix}=0$ for $\lambda$, it gives the foot of the perpendicular. $\endgroup$ Jun 3, 2015 at 1:51
  • $\begingroup$ What is $r$ on that dot product? I know that $\begin{pmatrix}-1\\ 2\\ 2 \end{pmatrix}$ is the direction vector. $\endgroup$
    – Johny
    Jun 3, 2015 at 2:12
  • $\begingroup$ It's $r$ for $L_1$ from above $\endgroup$ Jun 3, 2015 at 2:13
  • $\begingroup$ So $r\cdot\begin{pmatrix}-1\\ 2\\ 2 \end{pmatrix}=0$ is for searching the lambda. After do the dot product, I get : $\begin{pmatrix} a\\ b\\ c \end{pmatrix}\cdot \begin{pmatrix} -1\\ 2\\ 2 \end{pmatrix}=0$ and become : $-a+2b+2c=0$ $\endgroup$
    – Johny
    Jun 3, 2015 at 2:18

2 Answers 2

1
$\begingroup$

You had the right idea that the cross product will give you an orthogonal vector, but taking $(-5,-3,2)\times(-1,2,2)$ will give you a vector orthogonal to the direction of the line and a point on the line, not necessarily the direction perpendicular to the line that passes through the origin and $L_1$.

We know that the equation of $L_2$ will be of the form $r = r_0 + v\mu$. We also know that $r_0$ will be 0 since the line passes through the origin.

To have $L_2$ be orthogonal to $L_1$ and intersect $L_1$ we need the following to hold:

$$ v\cdot\begin{pmatrix}-1\\2\\2\end{pmatrix} = 0 \\ \text{and } v\mu = \begin{pmatrix}-5\\-3\\2\end{pmatrix} + \lambda \begin{pmatrix}-1\\2\\2\end{pmatrix} \text{ for some $\mu$ and $\lambda$}$$

Substituting in $v$ as $\begin{pmatrix}x\\y\\z\end{pmatrix}$ we get:

$$ -x + 2y + 2z = 0 $$ $$ x\mu = -5-\lambda $$ $$ y\mu = -3 + 2\lambda $$ $$ z\mu = 2 + 2\lambda $$

Rearranging the last three equations for $x,y,z$ and substituting them into the first gives:

$$ -\frac{-5-\lambda}{\mu} + 2\frac{-3+2\lambda}{\mu} + 2\frac{2+2\lambda}{\mu} = 0 $$

Solving for $\lambda$ gives $\lambda = -\frac{1}{3}$. This gives the following equation for $L_2$:

$$ v\mu = \begin{pmatrix}-5\\-3\\2\end{pmatrix} + -\frac{1}{3}\begin{pmatrix}-1\\2\\2\end{pmatrix} $$ $$ v\mu = \frac{1}{3}\begin{pmatrix}-14\\-11\\4\end{pmatrix} $$

Therefore our line $L_2$ is parallel to \begin{pmatrix}-14\\-11\\4\end{pmatrix} and the equation of the line is given by:

$$ L_2: r = \mu\begin{pmatrix}-14\\-11\\4\end{pmatrix} $$

$\endgroup$
4
  • $\begingroup$ Thanks for the solution... $\endgroup$
    – Johny
    Jun 3, 2015 at 2:36
  • $\begingroup$ If, I already get $L_{1}$ and $L_{2}$ equation, Can I get the shortest distance? Can I just $L_{2}$ vector - $L_{1}$ and then count the distance using (($L_{2}$ vector - $L_{1}$ vector)^2)^1/2 $\endgroup$
    – Johny
    Jun 3, 2015 at 3:27
  • $\begingroup$ For the two lines in the question the distance is $0$ since they are intersecting lines. In general though, for lines $L_1: r=r_1 + \lambda v_1$ and $L_2: r=r_2 + \mu v_2$ the shortest distance is given by $D = (r_2-r_1)\cdot\frac{v_1\times v_2}{||v_1\times v_2||}$. This can be thought of as the projection of the distance between the points $r_1$ and $r_2$ onto a direction orthogonal to both lines (given by $\frac{v_1\times v_2}{||v_1\times v_2||}$). $\endgroup$ Jun 3, 2015 at 4:32
  • $\begingroup$ Ok. I will try to get the information from your suggestion.... Thanks $\endgroup$
    – Johny
    Jun 3, 2015 at 4:59
0
$\begingroup$

Sloving $\left(\begin{pmatrix} -5\\ -3\\ 2 \end{pmatrix} + \lambda \begin{pmatrix} -1\\ 2\\ 2 \end{pmatrix}\right)\cdot \begin{pmatrix} -1\\ 2\\ 2 \end{pmatrix}=0$ for $\lambda$ we obtain a point $X$ on $L_1$, for which $OX\perp L_1$ so $r=O+tX$ will be the desired line equation.

$\endgroup$
2
  • $\begingroup$ Finally, I understand that equation. We must find a vector that perpendicular with the same direction to $L_{1}$. Is it right? $\endgroup$
    – Johny
    Jun 3, 2015 at 2:38
  • $\begingroup$ Sure.${}{}{}{}$ $\endgroup$ Jun 3, 2015 at 2:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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