# Help with vectors

• September 3rd 2006, 01:48 PM
Ranger SVO
Help with vectors
I was doing good until I got to the last two problems.

14. Give a unit vector that points in the opposite direction to the vector
-i + 2k

16. Find two nonparallel vectors that are perpendicular to i - j + k

I would like an explination alot more than an answer
• September 3rd 2006, 02:39 PM
ThePerfectHacker
Quote:

Originally Posted by Ranger SVO

14. Give a unit vector that points in the opposite direction to the vector
-i + 2k

To find the unit vector divide by the norm (magnitude) which is,
$||\bold{v}||=\sqrt{1^2+2^2}=\sqrt{5}$
Thus,
$-\frac{1}{\sqrt{5}}\bold{i}+\frac{2}{\sqrt{5}}\bold {k}$
In the opposite direction means that is a negative of the vector.
Multiply by (-1) to get,
$\frac{1}{\sqrt{5}}\bold{i}-\frac{2}{\sqrt{5}}\bold{k}$.
• September 3rd 2006, 02:48 PM
ThePerfectHacker
Quote:

Originally Posted by Ranger SVO
16. Find two nonparallel vectors that are perpendicular to i - j + k

Any vector perpendicular (orthogonal) to another has its dot porduct zero.
Thus, you want to find a vector perpendicular to the given one.
If the components of the first vector are $v_1,v_2,v_3$
Then, you need that, (dot product)
$v_1-2v_2+v_3=0$

The second vector has components of $w_1,w_2,w_3$
Then you need that, (dot product)
$w_1-2w_2+w_3=0$

And they are not parrallet meaning $v_k$ and $w_k$ are not (scalar) multiples of each other.

There a infinitely many solutions to the first and second equation.
One such solution is,
$v_1=1,v_2=0,v_3=-1$
$v_2=2,v_1=1,w_3=0$
Note they are not multiples.
Thus,
$\bold{v}=\bold{i}-\bold{k}$
$\bold{w}=2\bold{i}-2\bold{k}$
• September 3rd 2006, 03:10 PM
Ranger SVO
Quote:

Originally Posted by ThePerfectHacker
To find the unit vector divide by the norm (magnitude) which is,
$||\bold{v}||=\sqrt{1^2+2^2}=\sqrt{5}$
Thus,
$-\frac{1}{\sqrt{5}}\bold{i}+\frac{2}{\sqrt{5}}\bold {k}$
In the opposite direction means that is a negative of the vector.
Multiply by (-1) to get,
$\frac{1}{\sqrt{5}}\bold{i}-\frac{2}{\sqrt{5}}\bold{k}$.

I should have seen that, thank you very much for your time.
• September 3rd 2006, 03:15 PM
CaptainBlack
Quote:

Originally Posted by ImPerfectHacker
...
There a infinitely many solutions to the first and second equation.
One such solution is,
$v_1=1,v_2=0,v_3=-1$
$v_2=2,v_1=1,w_3=0$
Note they are not multiples.
Thus,
$\bold{v}=\bold{i}-\bold{k}$
$\bold{w}=2\bold{i}-2\bold{k}$

Typos here, presumably we should have:

One such solution is,
$v_1=1,v_2=0,v_3=-1$
$w_1=2,w_2=1,w_3=0$
Note they are not multiples.
Thus,
$\bold{v}=\bold{i}-\bold{k}$
$\bold{w}=2\bold{i}+1\bold{j}$?

RonL