Computation with scalar product

Question

Let $\vec{a}$ and $\vec{b}$ be vectors from $V_3$. Suppose, that $|\vec{a}| = 1$, $|\vec{b}|=2$ and the angle between $\vec{a},\vec{b}$ is $\frac{\pi}{3}$. Use the properties of scalar product and compute $|\vec{a}-2\vec{b}|$.

I'm totaly lost in that.

Answer 1

Hint: $|\vec{a}-2\vec{b}|$ is the square root of $(\vec{a}-2\vec{b}) \cdot (\vec{a}-2\vec{b})$.

Answer 2

$\lvert \vec{a} - 2 \vec{b} \rvert^2 = \langle \vec{a} – 2\vec{b},\vec{a} – 2\vec{b} \rangle = \langle \vec{a}, \vec{a} \rangle - 4 \langle \vec{a}, \vec{b} \rangle + 4 \langle \vec{b},\vec{b} \rangle$.

Now, can you compute $\langle \vec{a},\vec{b} \rangle$, knowing $\angle (\vec{a},\vec{b}) = \tfrac{\pi}{3}$?