# Find the angle between the vectors

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 $$\vec v$$ ?

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 $$\vec v$$.

• Did you mean $5, 12,$ and $13$? That would be $90^\circ$ (cf. Pythagorean Theorem) – J. W. Tanner Nov 24 '19 at 1:24
• No, it's actually written 5, 12, 15 – aita.kane Nov 24 '19 at 1:25
• Then it's not $90^\circ$ – J. W. Tanner Nov 24 '19 at 1:26
• and for Q2, the angle that separates $\vec u$ and $\vec u + \vec v$ is $90^\circ$ – J. W. Tanner Nov 24 '19 at 1:28
• Try $\vec u \cdot \vec v=|u| |v| \cos\theta$ – J. W. Tanner Nov 24 '19 at 1:30

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

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.

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 $$2 u\cdot v = ||w||^{2}-||u||^{2}-||v||^{2}$$.

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

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