Author Topic: Vectors help (Read 1298 times) Tweet Share

Stick · « **on:** August 03, 2012, 08:07:15 pm »

Although I'm only in GMA at the moment, my teacher has set us some homework questions from the Year 12 textbook to practice using the dot product. Anyway, I'm stuck on the following multiple choice questions:

1. If $(\mathbf{u}-\mathbf{v}).(\mathbf{u}+\mathbf{v})=0$ then:

A: $\mathbf{u}$ is parallel to $\mathbf{v}$
B: $\mathbf{u}$ and $\mathbf{v}$ have equal magnitudes
C: $\mathbf{u}$ is perpendicular to $\mathbf{v}$
D: $\mathbf{u}$ is a multiple of $\mathbf{v}$
E: None are true.

2. If $(\mathbf{u}-\mathbf{v}).(\mathbf{u}+\mathbf{v})=|\mathbf{v}|^2$ then:

A: $\mathbf{u}=\mathbf{v}$
B: $\mathbf{u}$ must be equal to the zero vector, $0$
C: $\mathbf{u}$ is perpendicular to $\mathbf{v}$
D: $\mathbf{u}$ is a multiple of $\mathbf{v}$
E: None are true.

I managed to work it out, but it was more of a 'look at the answer and see how we got there' approach. Perhaps someone could show me how to do these properly in case they appeared in a test. Thanks.

Lasercookie · « **Reply #1 on:** August 03, 2012, 08:37:34 pm »

didn't bother with bold face tildes etc.

1. You can expand it out and get $u^2 - v^2 = |u|^2 - |v|^2 = 0 \implies |u| = |v|$ , hence B

2. Continuing from before, since the magnitudes are equal, we can say
$(u-v) \dot (u+v) = u^2 - v^2 = |u|^2 - |v|^2$

How do we get this equal to $|v|^2$ ?

We know that: $|u|^2 - |v|^2 = |v|^2 \implies |u|^2 = 2|v|^2$

$|u| = \sqrt{2}|v|$

If $u = v$ , then the magnitudes are also equal. We know this isn't the case, otherwise the dot product would be equal to zero, as in Question 1.

We can also see that if u was the zero vector, and hence |u| = 0, we get $-|v|^2$ . So not that.

If $v = k \cdot u$ , then what does this imply about the magnitudes?

Lets say $v = xi + yi$ and that means $u = kxi + kyi$

$|v| = \sqrt{x^2 + y^2}$

$|u| = \sqrt{(kx)^2 + (ky)^2} = \sqrt{k^2x^2 + k^2y^2} = \sqrt{k^2(x^2 + y^2)} = k\sqrt{x^2+y^2} = k \cdot |v|$

So this is equivalent to what we had shown before, that $|u| = \sqrt{2}|v|$

So D

kamil9876 · « **Reply #2 on:** August 03, 2012, 08:47:24 pm »

The condition $(u-v).(u+v)=|v|^2$ is EQUIVALENT to $|u|=\sqrt{2}|v|$ (i.e if you have one then you have the other). But clearly you can construct vectors such that none of A,B,C,D are true but $|u|=\sqrt{2}|v|$ is true(check this yourself, choose some "bad" pair of vectors, one of length 1 and one of length sqrt2). Which I guess makes the correct answer E to 2, (where we interpret is as "none of the A,B,C,D have to necessarily hold")

Stick · « **Reply #3 on:** August 03, 2012, 09:08:52 pm »

I'm confused as to why you went from $u^2-v^2=0$ to $|u|^2-|v|^2=0$ . Instead, I did $u^2=v^2$ which led me to $u=v$ . What have I done wrong? :S

Lasercookie · « **Reply #4 on:** August 03, 2012, 09:14:00 pm »

Quote from: Stick on August 03, 2012, 09:08:52 pm

I'm confused as to why you went from $u^2-v^2=0$ to $|u|^2-|v|^2=0$ . Instead, I did $u^2=v^2$ which led me to $u=v$ . What have I done wrong? :S

Sorry, it's probably better to write u.u rather than u^2. So $\mathbf{u}.\mathbf{u} = |\mathbf{u}|^2$ The dot product of a vector by itself is just it's magnitude squared. Try evaluating the dot product of $\mathbf{a}.\mathbf{a}$ if you want to see one way of working out that result.

$(\mathbf{u}-\mathbf{v})(\mathbf{u}+\mathbf{v}) = \mathbf{u}.\mathbf{u} - \mathbf{v}.\mathbf{u} + \mathbf{v}.\mathbf{u} - \mathbf{v}.\mathbf{v} = \mathbf{u}.\mathbf{u} - \mathbf{v}.\mathbf{v} = |\mathbf{u}|^2 - |\mathbf{v}|^2$

(or just use difference of two squares when expanding it out)

edit: that bold font looks yuck

Oh and also on $\mathbf{u}^2 = \mathbf{v}^2$ , what does it mean to square a vector? If I'm not mistaken, multiplication of a vector and a vector isn't defined.

Stick · « **Reply #5 on:** August 03, 2012, 09:23:42 pm »

THE LIGHT BULB MOMENT JUST HAPPENED TO ME!

I didn't realise that it is better to use u.u instead of u^2 for vectors.

Now for my final question. For question 2, I did $u-v=v$ giving $u=2v$ and also $u+v=v$ giving $u=0$ . How is this wrong? I'm assuming it's along the lines to my misunderstanding in question 1.

TrueTears · « **Reply #6 on:** August 03, 2012, 09:32:07 pm »

Quote from: Stick on August 03, 2012, 09:23:42 pm

THE LIGHT BULB MOMENT JUST HAPPENED TO ME! I didn't realise that it is better to use u.u instead of u^2 for vectors.

No it's not that it's "better" to use u.u instead of u^2, rather that u^2 is an undefined expression, as laseredd pointed out, what definition does u^2 follow? (Assuming u here is a vector), if you defined u to be the vector then u to be the magnitude of u, then u.u = u^2 are interchangeable

Stick · « **Reply #7 on:** August 03, 2012, 09:41:26 pm »

I feel so dumb. I just realised I've been using the null factor law incorrectly for 3 years. Wow. -.- Thanks for all your help and putting up with my stupid questions!

Phy124 · « **Reply #8 on:** August 03, 2012, 09:44:06 pm »

Quote from: Stick on August 03, 2012, 09:41:26 pm

I feel so dumb. I just realised I've been using the null factor law incorrectly for 3 years. Wow. -.- Thanks for all your help and putting up with my stupid questions!

Better to find out now than in an exam

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Author Topic: Vectors help (Read 1298 times) Tweet Share

Stick

Vectors help

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