Computing the length of a vector

Question

The task gives length of two vectors, $|u| = 3$, $|v| = 4$, and the angle between them is $60°$. The task is to find the length of a vector $z = 2u - \frac 32v$. How do I find the length of $z$?

Answer 1

Hint:

\begin{align} |z|^2 &= z \cdot z = \left(2u - \frac{3}{2}v\right) \cdot \left(2u - \frac{3}{2}v\right)\\ &= 4 (u \cdot u) -6 (u\cdot v) + \frac{9}{4} (v \cdot v)\\ &= 4|u|^2-6(u\cdot v)+\frac{9}{4}|v|^2. \end{align} You already know $|u|$ and $|v|$, and you can compute $u \cdot v$ from $$ u \cdot v = |u||v|\cos\alpha, $$ where $\alpha$ is the angle between $u$ and $v$.

Answer 2

$$|z|=\sqrt{(2u)^2+(\frac{-3v}{2})^2+2(2|u|)(\frac{-3|v|}{2})cos(\pi/3)}$$ $$|z|=\sqrt{4|u|^2+\frac{9|v|^2}{4}-3|u||v|}$$ $$|z|=\sqrt{4*9+36-3*3*4}=\sqrt{36}=6$$

Answer 3

In defence of my draw a diagram suggestion.

The angle between $u$ and $v$ is $60^{\circ}$. Then note that the vector $2u$ has length $6$ as does $\frac 32v$. The vector $2u-\frac 32v$ joins the ends of these two vectors and therefore makes an equilateral triangle.

If you see this, there is essentially no computation to do.