# How do you find the parametric equation of a line in R3 that passes through P(1,1,1) and is orthogonal to both the lines x=3-t, y= -1+4t, z = -2t and x=2+2t, y=1+t, z=-2+3t where t is any real number?

Dec 28, 2016

The answer is x=1+14s ; y=1-s ;z=1-9s ; $s \in \mathbb{R}$

#### Explanation:

To find a line orthogonal to 2 other lines, we must perfom a cross product

The first line $L 1$ is

$x = 3 - t$

$y = - 1 + 4 t$

$z = - 2 t$

The vector parallel to line $L 1$ is vecL_1=〈-1,4,-2〉

The second line $L 2$ is

$x = 2 + 2 t$

$y = 1 + t$

$z = - 2 + 3 t$

The vector parallel to line $L 2$ is vecL_2=〈2,1,3〉

The vector orthogonal to ${\vec{L}}_{1}$ and ${\vec{L}}_{2} =$ is given by the cross product

$| \left(\hat{i} , \hat{j} , \hat{k}\right) , \left(- 1 , 4 , - 2\right) , \left(2 , 1 , 3\right) |$

$= \hat{i} | \left(4 , - 2\right) , \left(1 , 3\right) | - \hat{j} | \left(- 1 , - 2\right) , \left(2 , 3\right) | + \hat{k} | \left(- 1 , 4\right) , \left(2 , 1\right) |$

$= \hat{i} \left(12 + 2\right) - \hat{j} \left(- 3 + 4\right) + \hat{k} \left(- 1 - 8\right)$

=〈14,-1,-9〉

Verification by doing a dot product

〈14,-1,-9〉.〈-1,4,-2〉=-14-4+18=0

〈14,-1,-9〉.〈2,1,3〉=28-1-27=0

The equation of the line orthogonal to ${L}_{1}$ and ${L}_{2}$ and passing through $P \left(1 , 1 , 1\right)$ is

$x = 1 + 14 s$

$y = 1 - s$

$z = 1 - 9 s$