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Hello I am not sure of the answers I gave. Can I have your opinion on them, please?

Q1: Three vectors $\vec u$, $\vec v$ and $\vec w$ of length 5, 12 and 15. If their sum gives $\vec 0$ what's the angle between the vectors $\vec u$ and $\vec v$ ?

Answer 1: 90 degrees

Q2: Two vectors $\vec u$, $\vec v$ of length 3 and 5, respectively, are added, and the result vector is perpendicular to $\vec u$. Give the angle that separates $\vec u$ and $\vec v$.

Answer 2 : 90 degrees

Thanks in advance for your help.

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  • $\begingroup$ Did you mean $5, 12, $ and $13$? That would be $90^\circ$ (cf. Pythagorean Theorem) $\endgroup$ – J. W. Tanner Nov 24 '19 at 1:24
  • $\begingroup$ No, it's actually written 5, 12, 15 $\endgroup$ – aita.kane Nov 24 '19 at 1:25
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    $\begingroup$ Then it's not $90^\circ$ $\endgroup$ – J. W. Tanner Nov 24 '19 at 1:26
  • $\begingroup$ and for Q2, the angle that separates $\vec u$ and $\vec u + \vec v$ is $90^\circ$ $\endgroup$ – J. W. Tanner Nov 24 '19 at 1:28
  • $\begingroup$ Try $\vec u \cdot \vec v=|u| |v| \cos\theta$ $\endgroup$ – J. W. Tanner Nov 24 '19 at 1:30
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a)
That $\mathbb u + \mathbb v + \mathbb w = 0$ means that if we place the vectors head to tail to head to tail, they form a triangle.

By the law of cosines

$15^2 = 5^2 + 12^2 - 2(5)(12)\cos \theta\\ \frac {225 - 144 - 25}{120} = -\cos\theta\\ \frac {56}{120} = \cos\theta\\ \theta = \arccos -\frac {7}{15}$

But that is the angle between $u,v$ when they are head to tail. When they are tail to tail you will get the supplement of that angle.

$\arccos \frac 7{15}$

You could also do this with $\|\mathbb u + \mathbb v\| = \|\mathbb w\|$

b)
$(\mathbb u+\mathbb v)\cdot \mathbb u = 0\\ \mathbb u\cdot \mathbb u + \mathbb u\cdot \mathbb v = 0\\ \|\mathbb u\|^2 + \|\mathbb u\|\|\mathbb v\|\cos\theta = 0\\ 9 + 15\cos\theta = 0\\ \theta = \arccos-\frac 35$

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Your answers are wrong.

For question $1$, answer would be $90^\circ$, if the question were $5, 12, $ and $1\color{red}3$.

For question $2$, the answer is not $90^\circ$ nor $180^\circ.$

If $\vec u\cdot (\vec u+\vec v)=0,$ then $\vec u\cdot \vec u=-\vec u\cdot \vec v $,

so $|\vec u|^2=-|\vec u||\vec v|\cos\theta,$ and you are given $|\vec u|$ and $|\vec v|$;

you should be able to find $\theta$ from here.

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An approach could be as follows (I'm assuming you are working with real spaces). You know that: $$u\cdot v = ||u||||v||\cos\theta$$ and you want to find out the value of $\theta$. For this, you need to calculate $u\cdot v$. But note that, because $u+v+w = 0$, we have: $$ 0 = (u+v-w)\cdot (u+v+w) = u\cdot u + u\cdot v + u\cdot w + v\cdot u + v\cdot v +v\cdot w -w\cdot u -w\cdot v -w\cdot w = ||u||^{2}+||v||^{2}-||w||^{2}+2 u\cdot v$$ So, you know that $2 u\cdot v = ||w||^{2}-||u||^{2}-||v||^{2}$.

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1) Start with $v+u+w=0$ and square,

$$v^2+u^2+w^2 +2(u\cdot v-w^2)=0$$

Thus, the angle is

$$\cos\theta = \frac{u\cdot v }{|u||v|} = -\frac{ v^2+u^2-w^2 }{2|u||v|}=\frac7{15}$$

2) $(u+v)\cdot u= 0$ leads to

$$\cos\theta = \frac{u\cdot v }{|u||v|}= -\frac{u^2}{|u||v|}= -\frac{|u|}{|v|}=-\frac35 $$

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