Problem with calculation of dot product of two vectors

Question

I have such a question from the exercises given by the professor

$\vec{u}$ and $\vec{v}$ are two vectors on the plane and $\vec{w}=\vec{u}+\vec{v}$. We suppose that $||\vec{u}|| = 3$ and $||\vec{v}|| = 5$ and $\vec{w} \vec{u} = 0$. Please calculate $\vec{u} \vec{v}$

According to the answer key, the procedure is as following

• $\vec{u} \vec{v}$ = $\vec{u}(\vec{w}-\vec{u})$
• =$\vec{u}\vec{w}-\vec{u}\vec{u}$
• =$0-||\vec{u}||^2$
• =-9

However, from my understand, the formula of dot product should be $\vec{u} \vec{v} = ||\vec{u}||||\vec{v}||cos\theta$

So why the $cos\theta$ is not calculated here?

Answer 1

You are given $\vec w \cdot \vec u =0$.

Note $\vec w \cdot \vec u = (\vec u + \vec v) \cdot \vec u$

$= \vec u \cdot \vec u + \vec v \cdot \vec u $

(distributivity of the dot product)

$=\vec u \cdot \vec u + \vec u \cdot \vec v $

(commutativity of the dot product)

$= | \vec u|^2+\vec u \cdot \vec v$

So, $| \vec u|^2+\vec u \cdot \vec v=0$

$3^2+\vec u \cdot \vec v =0$

$\therefore \vec u \cdot \vec v = -9$