Question

Determine whether the vectors u and v are parallel, orthogonal, or neither.

u = <10, 6>, v = <9, 5>

Answer 1

Use the dot product:
\[u \cdot v = |u||v|\cos \theta\]
If they are parallel, \(\theta = 0\) or \(\theta = \pi\), and then
\[u \cdot v = |u||v|\cos 0 = |u||v|\]
or
\[u \cdot v = |u||v|\cos \pi = -|u||v|\]

If they are orthogonal, \(\large \theta = \frac{\pi}{2}\) and
\[u \cdot v = |u||v|\cos \frac{\pi}{2} = 0\]

So if \(u \cdot v\) equals \(\pm|u||v|\) they are parallel, and if \(u \cdot v = 0\) they are parallel. If it's not one of those they are neither

Answer 2

thankyou very much! would you mind helping me on one more?

Answer 3

Sure

Answer 4

Find the angle between the given vectors to the nearest tenth of a degree.

u = <2, -4>, v = <3, -8>

Answer 5

Again you can use the dot product, but this time we'll use it a bit differently
\[u \cdot v = |u||v|\cos \theta\]
Now we want to find the angle \(\theta\), so we isolate \(\cos \theta\)
\[\cos \theta = \frac{u \cdot v}{|u||v|}\]
The angle is simply the inverse cosine
\[\theta = \cos^{-1}\left(\frac{u \cdot v}{|u||v|}\right)\]

Answer 6

so how do i plug the given coordinates in?

Answer 7

The dot product for two two-dimensional vectors can be calculated by using
\[<u_1,u_2> \cdot <v_1, v_2> = u_1v_1+u_2v_2\]

And |u| and |v| mean the magnitude of the vectors, which can be calculated by using
\[|<u_1, u_2>| = \sqrt{u_1^2 + u_2^2} \]
notice that this is basically just the Pythagorean theorem :p

Answer 8

3.0°

6.0°

-7.0°

16.0
are the only answers given.
i really still don't understand it im sorry

Answer 9

We want to know the angle between the 2 vectors u = <2, -4> and v = <3, -8>
The first step is calculating the dot product using the method I listed above
u \(\cdot\) v = <2, -4> \(\cdot\) <3, -8> = (2 * 3) + (-4 * -8) = 38

The dot product is also equal to the product of the magnitudes, times the cosine of the angle between the two vectors (u \(\cdot\) v = |u||v|cos \(\theta\))
u \(\cdot\) v = \(\sqrt{2^2 + (-4)^2}\sqrt{(3)^2 + (-8)^2}\cos \theta = \sqrt{20}\sqrt{73}\cos \theta\)

So setting those equal we get
\[38 = \sqrt{20}\sqrt{73} \cos \theta\]
\[\cos \theta = \frac{38}{\sqrt{20}\sqrt{73}}\]
\[\theta = \cos^{-1} \frac{38}{\sqrt{20}\sqrt{73}}\]

Which is around 6 degrees.

Sorry for the late reply by the way!

Answer 10

THANKYOU THANKYOU THANKYOU you're a life saver :)