1. ## Find symmetric equation

Find the symmetric equation of the line passing through the point (-5,-3,2) and perpendicular to both lines:

(x-1)/3 = (4-y)/1 = z/3 and x/-2 = (y-1)/4, z = 2

For the first line would the direction vector be (3,1,3) and for the second line would it be (-2,4,1) ?

2. Originally Posted by Random-Hero-
Find the symmetric equation of the line passing through the point (-5,-3,2) and perpendicular to both lines:

(x-1)/3 = (4-y)/1 = z/3 and x/-2 = (y-1)/4, z = 2

For the first line would the direction vector be (3,1,3) and for the second line would it be (-2,4,1) ?
1. Re-write the equations of the lines:

$\displaystyle \begin{array}{l}x=1+3t \\ y=4-t \\ z=0+3t\end{array}$ ...... and ...... $\displaystyle \begin{array}{l}x=0-2t \\ y=1+4t \\ z=2+0t\end{array}$

Therefore the direction vectors are $\displaystyle \vec u=(3, -1, 3)$ and $\displaystyle \vec v = (-2, 4, 0)$

2. The vector perpendicular to both direction is:

$\displaystyle \vec n = \vec u \times \vec v = (-12, -6, 10)$

3. Now use the point-direction-form of the equation of a line:

$\displaystyle \vec r = (-5, -3, 2) + s \cdot (-12, -6, 10)$

4. Re-write:

$\displaystyle \begin{array}{l}x=-5-12s\\y=-3-6s\\z=2+10s\end{array}~\implies~ \begin{array}{l}\dfrac{x+5}{-12}=s\\ \dfrac{y+3}{-6}=s\\ \dfrac{z-2}{10}=s\end{array}$

5. Symmetric equation:

$\displaystyle \dfrac{x+5}{-12}= \dfrac{y+3}{-6}= \dfrac{z-2}{10}$

3. Wow thanks earboth!!! I totally spaced on the cross product! You're the man! *high fives*