text: # 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 $2\vec{i}+3\vec{j}-6\vec{k}$.