# Prove the following lines are perpendicular.

I am not understanding how to follow through on this exercise, so if someone could explain it to me I would appreciate it.

r:

x + y - 3z - 1 = 0

2x - y - 9z - 2 = 0

is perpendicular to

s:

2x + y + z + 5 = 0

2x - 2y - z + 2 = 0

and the second part requires me to prove

r:

2x + y − 4z + 2 = 0

4x − y − 5z + 4 = 0

is perpendicular to

s:

x = 1 + 2λ

y = −2 + 3λ

z = 1 − 6λ

I apologize for the bad formatting, I wasn't sure how else I could arrange the equations. Thanks for any help.

• Do you require that the lines intersect for them to be perpendicular? – amd Apr 28 '18 at 0:42
• See the help center for information about how to format equations. – Simply Beautiful Art Apr 28 '18 at 12:49

The direction vector of $\mathbb{r}$ is given by $(\vec{i}+\vec{j}-3\vec{k})\times(2\vec{i}-\vec{j}-9\vec{k})$. You can find the direction vector of $\mathbb{s}$ by cross product in a similar way.
You can find the direction vector of $\mathbb{r}$ by cross product. The direction vector of $\mathbb{s}$ is $2\vec{i}+3\vec{j}-6\vec{k}$.