Question

Determine whether the vectors u and v are parallel, orthogonal, or neither
u=<6,-2>; v=<2,6>
Please explain this concept please?

Answer 1

To check of they are parallel: They must be scalar multiples of each other, i.e.
If
\[ \vec{u}=<a,b>\] then
\[ \vec{v}=k\vec{u}=<ka,kb>\], for some \(k\) being a real number.

To check if they are orthogonal:
You can use the dot product:
\[\vec{u}\cdot \vec{v}=|\vec{u}||\vec{v}|\cos\theta \], and you know they are perpendicular if the angle is 90°. So \(\cos 90°=0\), so you should expect the dot product to be 0 if they are orthogonal. (Recall, you can also calculate the dot product the following way:
If \(\vec{u}=<x_1,y_1>\) and \(\vec{v}=<x_2,y_2>\), then \[ \vec{u}\cdot \vec{v}=x_1x_2+y_1y_2\]

Answer 2

Thank you, but I don't understand that explanation. Can we walk thru the problem bit by bit please?

Answer 3

So the parallel one: Say we have 2 vectors like
u = <2, 4> and v = <6, 12>

These are parallel since you can multiply vector u by the scalar 3 to get v
3u = <3(2), 3(4)> = <6, 12> = v

So if you can divide both components and they give the same constant, they are parallel (i.e. I can divide 6/2 = 3
and 12/4 = 3... you get 3 in both cases when you divide each component respectively).

Does this one make sense?

Answer 4

Somewhat
So using that concept, my original problem wouldn't be parallel since it can't become <2,6>? Right?

Answer 5

yup exactly.

Answer 6

For the orthogonal one:
2 vectors are orthogonal if they form a right angle:


So we can use the dot product to help us. One way to write the dot product is:
\(\vec{u}\cdot\vec{v}=|\vec{u}||\vec{v}|\cos \theta\)
So if the vectors form a right angle, that means the angle is 90°

So by the dot product formula, you get:
\(\vec{u}\cdot\vec{v}=|\vec{u}||\vec{v}|\cos 90° \), but \(\cos 90°=0\), so:
\(\vec{u}\cdot\vec{v}=|\vec{u}||\vec{v}|(0)\\
\vec{u}\cdot\vec{v}=0\)

So the dot product will be 0 if the vectors are orthogonal.

But there is another equivalent formula for the dot product: If \(\vec{u}=<x_1, y_1>\) and \(\vec{v}=<x_2,y_2>\), then
\[ \vec{u}\cdot\vec{v}=x_1x_2+y_1y_2\] so with your vectors, you should calculate:
\(\vec{u}\cdot\vec{v}=(6)(2)+(-2)(6) \)
If that gives 0, then the vectors are orthogonal

Answer 7

OOOoooooooooOoOooOoo I understand the 2nd formula so much better! So it is Orthogonal! :) Correct?

Answer 8

yes :)

Answer 9

Thank you! I submitted it in and got it right :) Though on the entire exam I got a 6/10, I hope I did well enough on the essay question to boost me... :)
Again, thank you <3

Answer 10

Hopefully! :) and your welcome!