# Perpendicular Vector Lines

• Jan 9th 2009, 06:50 AM
Simplicity
Perpendicular Vector Lines
How can I show that these two lines are perpendicular?

$\b{r} = \left(2\b{i}+1\b{j}+1\b{k}\right)+\lambda \left(-2\b{i}+4\b{j}+2\b{k}\right)$

$\b{s} = \left(5\b{i}-4\b{j}+2\b{k}\right)+\lambda \left(-1\b{i}+1\b{j}+3\b{k}\right)$
• Jan 9th 2009, 07:05 AM
Simplicity
I realised that I can show it by direction cosines but is there an alternative method?
• Jan 9th 2009, 09:10 AM
vincisonfire
Usually you would take the dot product. If it is equal to 0 then the line are perpendicular.
• Jan 9th 2009, 10:14 AM
HallsofIvy
I agree with vincisonfire. Take the dot products of the direction vectors-the vectors multiplied by $\lambda$.

(Actually, I see a distinct problem with "showing these two lines are perpendicular"- they are not! Are you sure you haven't lost a sign somewhere?)
• Jan 9th 2009, 10:49 AM
Simplicity
Quote:

Originally Posted by HallsofIvy
I agree with vincisonfire. Take the dot products of the direction vectors-the vectors multiplied by $\lambda$.

(Actually, I see a distinct problem with "showing these two lines are perpendicular"- they are not! Are you sure you haven't lost a sign somewhere?)

Yes, I realised. There was a minus. It should have been:

$\b{s} = \left(5\b{i}-4\b{j}+2\b{k}\right)+\lambda \left(-1\b{i}+1\b{j}-3\b{k}\right)$

Then I considered the direction vectors which work to give angle between them as $90^o$.
• Jan 9th 2009, 12:07 PM
Plato
Quote:

Originally Posted by Air
Yes, I realised. There was a minus. It should have been:
$\b{s} = \left(5\b{i}-4\b{j}+2\b{k}\right)+\lambda \left(-1\b{i}+1\b{j}-3\b{k}\right)$

Now the two direction vectors are perpendicular.
However, in order for the lines to be perpendicular they must intersect.
Do they intersect?
• Jan 10th 2009, 03:23 AM
Simplicity
Quote:

Originally Posted by Plato
Now the two direction vectors are perpendicular.
However, in order for the lines to be perpendicular they must intersect.
Do they intersect?

By equating the lines and finding $\lambda$ then substituting that back into one of the equation to get a vector point. Am I correct?
• Jan 10th 2009, 08:36 AM
Plato
Quote:

Originally Posted by Air
By equating the lines and finding $\lambda$ then substituting that back into one of the equation to get a vector point. Am I correct?

No, you must write the two lines with different parameters.
$r(\lambda ) = \left\{ \begin{gathered} x = 2 - 2\lambda \hfill \\ y = 1 + 4\lambda \hfill \\ z = 1 + 2\lambda \hfill \\ \end{gathered} \right.\,\& \,s(\gamma ) = \left\{ \begin{gathered} x = 5 - \gamma \hfill \\ y = - 4 + \gamma \hfill \\ z = 2 - 3\gamma \hfill \\ \end{gathered} \right.$

The set two of them equal. Say $\begin{gathered} 2 - 2\lambda = 5 - \gamma \hfill \\ 1 + 4\lambda = - 4 + \gamma \hfill \\ \end{gathered}$.

Now solve, making sure the solution is consistent in the z position.