# Calculate two vectors given their norms and angle

For two vectors $$\mathbf{u,v}$$ in $$\mathbb{R}^n$$ euclidean space, given:

• $$\|\mathbf{u}\| = 3$$
• $$\|\mathbf{v}\| = 5$$
• $$\angle (\mathbf{u,v})=\frac{2\pi}{3}$$

Calculate the length of the vectors

• $$4\mathbf{u}-\mathbf{v}$$
• $$2\mathbf{u}-7\mathbf{v}$$

I'm not sure how to approach this with the given information

With the formula for the angle between the two vectors being
$$\cos \theta=\frac{\mathbf{u\cdot v}}{\|\mathbf{u}\|\cdot\|\mathbf{v}\|}$$

I already have the denominator, but how do I get the point product of u and v in this case?

A point to start would be most appreciated

• From the given information, you know the angle and the lengths, so you know $\mathbf u\cdot\mathbf v$. Now remember that for any vector $\mathbf w$, $\|\mathbf w\|^2 = \mathbf w\cdot\mathbf w$. May 15, 2013 at 15:26
• Also bear in mind that the (point) inner product of two vectors is additive in both variables, i.e. $$(u+v)\cdot(u+v)=u\cdot u+u\cdot v+v\cdot u+v\cdot v.$$ May 15, 2013 at 15:29

$(\mathbf{u}.\mathbf{v})=||\mathbf{u}||||\mathbf{v}||cos(\frac{2\pi}{3})=\frac{-1}{2}||\mathbf{u}||||\mathbf{v}||$
$(4\mathbf{u}-\mathbf{v}).(4\mathbf{u}-\mathbf{v})=16\mathbf{(u.u)-4(\mathbf{u}.\mathbf{v})}-4(\mathbf{v}.\mathbf{u})+(\mathbf{v}.\mathbf{v})=16||\mathbf{u}||^2-8(\mathbf{u}.\mathbf{v})+||\mathbf{v}||^2=16*9-8*\frac{-15}{2}+25=229$ $|(4\mathbf{u}-\mathbf{v})|=\sqrt{229}$
Hint: You should be able to use the formula you quote to calculate $\mathbf{u\cdot v}$. Then use the linearity of the dot product to expand out $\mathbf{(4u-v)\cdot (2u-7v)}$
You have the angle so it's trivial to rearrange for $\mathbf {u\cdot v}$, by multiplying by the denominator. Then you know all combinations of dot products of the two vectors.
Using the linearity (distributivity) of the dot product allows you to now calculate $$\lVert a\mathbf u + b\mathbf v\rVert^2 = ( a\mathbf u + b\mathbf v)\cdot ( a\mathbf u + b\mathbf v) = a^2 \mathbf {u\cdot u} + \cdots$$