# Finding Perpendicular Vector, Help!!!

• Apr 27th 2008, 08:43 AM
Stylis10
Finding Perpendicular Vector, Help!!!
I really need help on vectors were you have to find a perpendicular vector. I have 2 questons that are really confusing me:

1. Find a vector which is perpendicular to both a and b
a = $(i+3j-k)$
b = $(3i-j-k)$

2. Show that lines L1 and L2 are perpendicular to each other.

L1: $r = \left(\begin{array}{c}5\\3\\0\end{array}\right) + \lambda \left(\begin{array}{c}-7\\-7\\7\end{array}\right)$

L2: $r = \left(\begin{array}{c}1\\-3\\-4\end{array}\right) + \mu \left(\begin{array}{c}1\\2\\3\end{array}\right)$

Thank You, this part of vectors is really confusing me
• Apr 27th 2008, 08:56 AM
elizsimca
Does it help to remember that if the dot product of two vectors is zero, then they are orthogonal (perpendicular).

Sorry, I'm in the same boat you are it seems, we just started vectors and I'm wading my way through the material.
• Apr 27th 2008, 09:15 AM
Soroban
Hello, Stylis10!

Quote:

1. Find a vector which is perpendicular to both $\vec{a}\text{ and }\vec{b}$

. . $\begin{array}{ccc}\vec{a} &= & {\bf i}+3{\bf j}-{\bf k} \\
\vec{b} &= & 3{\bf i}-{\bf j}-{\bf k} \end{array}$

A vector perpendicular to two vectors is their cross product.

. . $\vec{a} \times \vec{b} \;=\;\begin{vmatrix}{\bf i} & {\bf j} & {\bf k} \\ 1 & 3 & \text{-}1 \\ 3 & \text{-}1 & \text{-}1 \end{vmatrix} \;=\;-4{\bf i} - 2{\bf j} - 10{\bf k}\;\;\text{ or }\;\;\boxed{2{\bf i} + {\bf j} + 5{\bf k}}$

Quote:

2. Show that lines $L_1\text{ and }L_2$ are perpendicular.

$\begin{array}{cccc}L_1\!: & r &= & \left(\begin{array}{c}5\\3\\0\end{array}\right) + \lambda \left(\begin{array}{c}\text{-}7\\\text{-}7\\7\end{array}\right) \\ \\[-3mm]
L_2\!: & r & = & \left(\begin{array}{c}1\\\text{-}3\\\text{-}4\end{array}\right) + \mu \left(\begin{array}{c}1\\2\\3\end{array}\right) \end{array}$

Two vectors are perpendicular if their dot product is zero.

We have: . $(-7{\bf i} - 7{\bf j} + 7{\bf k})\cdot({\bf i} + 2{\bf j} + 3{\bf k}) \;=\;-7 - 14 + 21 \;=\;{\color{blue}0}$

. . Therefore: . $L_1 \perp L_2$

• Apr 27th 2008, 10:28 AM
Stylis10
Thank You both, but Soro could u please explain the method for question 1 in a bit more detail to me please.
• Apr 27th 2008, 11:25 AM
o_O
One of the properties of the cross product of two vectors $\vec{a}$ and $\vec{b}$ is that the resultant vector $\vec{a} \times \vec{b}$ is orthogonal (perpendicular) to both $\vec{a}$ and $\vec{b}$.

So, you are given two vectors. Their cross product will yield the vector perpendicular to both of them.
• Apr 27th 2008, 12:04 PM
elizsimca
Quote:

Originally Posted by Soroban

A vector perpendicular to two vectors is their cross product.

. .

I got told!!! hehe