# Another Line Equation Case

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....

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

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:

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

• Thanks for the solution... Jun 3, 2015 at 2:36
• 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 Jun 3, 2015 at 3:27
• 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 $\frac{v_1\times v_2}{||v_1\times v_2||}$). Jun 3, 2015 at 4:32
• Ok. I will try to get the information from your suggestion.... Thanks Jun 3, 2015 at 4:59

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.

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