2
$\begingroup$

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

$\endgroup$
2
  • 1
    $\begingroup$ 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 $\mathbf w$, $\|\mathbf w\|^2 = \mathbf w\cdot\mathbf w$. $\endgroup$ May 15, 2013 at 15:26
  • 1
    $\begingroup$ Also bear in mind that the (point) inner product of two vectors is additive in both variables, i.e. $$ (u+v)\cdot(u+v)=u\cdot u+u\cdot v+v\cdot u+v\cdot v. $$ $\endgroup$
    – Ian Coley
    May 15, 2013 at 15:29

3 Answers 3

5
$\begingroup$

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

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

$\endgroup$
2
  • $\begingroup$ isn't cos(2pi / 3) something like 0.99something? $\endgroup$
    – deemel
    May 15, 2013 at 17:19
  • $\begingroup$ Nice one, somaye (+1) $\endgroup$
    – Mikasa
    May 15, 2013 at 18:54
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.