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.
Part 1
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.
The dot product of the two direction vectors should be zero if they are perpendicular to each other.
Part 2
You can find the direction vector of $\mathbb{r}$ by cross product. The direction vector of $\mathbb{s}$ is $2\vec{i}+3\vec{j}-6\vec{k}$.
Find the dot product of the two direction vectors.