Calculate two vectors given their norms and angle

Question

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

Answer 1

$(\mathbf{u}.\mathbf{v})=||\mathbf{u}||||\mathbf{v}||cos(\frac{2\pi}{3})=\frac{-1}{2}||\mathbf{u}||||\mathbf{v}||$

$(4\mathbf{u}-\mathbf{v}).(4\mathbf{u}-\mathbf{v})=16\mathbf{(u.u)-4(\mathbf{u}.\mathbf{v})}-4(\mathbf{v}.\mathbf{u})+(\mathbf{v}.\mathbf{v})=16||\mathbf{u}||^2-8(\mathbf{u}.\mathbf{v})+||\mathbf{v}||^2=16*9-8*\frac{-15}{2}+25=229$ $|(4\mathbf{u}-\mathbf{v})|=\sqrt{229}$

Answer 2

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 $\mathbf{(4u-v)\cdot (2u-7v)}$

Answer 3

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 $$ \lVert a\mathbf u + b\mathbf v\rVert^2 = ( a\mathbf u + b\mathbf v)\cdot ( a\mathbf u + b\mathbf v) = a^2 \mathbf {u\cdot u} + \cdots$$