0
$\begingroup$

I was doing a linear algebra assignment on vectors but I came across with some doubts. One of the problems says Find the module of a vector $\vec{w}$ if $\vec{c} =(-2;3)$ is perpendicular to $\vec{a}$ , $\langle \vec{a} +2 \vec{w}$ , $\vec{c}\rangle =2 \sqrt{13}$ and the angle formed by $\vec{w}$ and $\vec{c}$ is $60^\circ$

I don't know how to do it because it seems like there are many unknowns. I know that the product between two perpendicular vectors is equal to $0$. But how can I solve it?

Thanks

$\endgroup$
  • 1
    $\begingroup$ If you put $\$$ signs around your Mathjax then your vectors will display properly. For example, putting dollar signs around \vec{w} will give $\vec{w}$. $\endgroup$ – John Douma Apr 5 at 3:31
  • $\begingroup$ Oh thanks! I didn't know how to do it $\endgroup$ – AaronTBM Apr 5 at 3:34
  • 1
    $\begingroup$ You should put dollar signs around the whole mathematical expression, not around each individual symbol. $\endgroup$ – Lord Shark the Unknown Apr 5 at 3:36
  • $\begingroup$ Oh, I didn't know that. Thank you! $\endgroup$ – AaronTBM Apr 5 at 3:42
  • $\begingroup$ Did you get the solution? $\endgroup$ – amitava Apr 5 at 15:52
0
$\begingroup$

As you mentioned, the geometric angle between vectors (in Euclidean space, at least) can be related to their inner product.

The gist of the solution is to use the identities

\begin{align} \langle\vec{u}, \vec{v}\rangle &= uv\cos(\theta) \\ \langle\vec{u}, \vec{v}\rangle &= u_1 v_1 + u_2 v_2 \\ \langle\vec{u} + \vec{v}, \vec{w}\rangle &= \langle\vec{u},\vec{w}\rangle + \langle\vec{v},\vec{w}\rangle \end{align}

in order to turn the given information

\begin{align} \langle \vec{a},\vec{c}\rangle &= 0 \\ \langle \vec{a} + 2\vec{w}, \vec{c} \rangle &= 2\sqrt{13} \\ \langle \vec{w}, \vec{c} \rangle &= wc\cos(60^{\circ}) \end{align}

into an expression for $\vec{c}$ in terms of your basis for $\mathbb{R}^2$.

$\endgroup$
  • $\begingroup$ But if I use $\langle \vec{a} , \vec{c} \rangle =0$ I get $a_1 c_1 + a_2 c_2 = 0$ and operating I get $-2 a_1 + 3 a_2 =0$. And then if I have $ wc \cos 60 =-2 w_1 +3 w_2 $ I can't find the unknowns (the components of $\vec{a}$ and the components of $\vec{w}$ ) because they're four unknowns. $\endgroup$ – AaronTBM Apr 5 at 16:14
  • $\begingroup$ I edited my answer to include the property of inner products being linear in the first argument. This should give you the added constraint you need. This should help you in using the second equation, though you never mentioned how you were using it before. $\endgroup$ – greenbagels Apr 5 at 16:43
  • $\begingroup$ Using $\langle \vec{a} , \vec{c} \rangle$ I get $-2 a_1 + 3 a_2 =0$. Then, using $\langle \vec{a} +2 \vec{a} , \vec{c} \rangle =2 \sqrt{13}$ and the property of inner products I get $-4 w_1 +6 w_2 =2 \sqrt{13}$. And using $\langle \vec{w} , \vec{c} \rangle =wc \cos 60°$ I get $-2 w_1 +3 w_2 =0,5w \sqrt{13}$. But what do I have to do now? I haven't been able to find an equation that allows me to find the components of $\vec{w}$ and $\vec{a}$ $\endgroup$ – AaronTBM Apr 6 at 3:12

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.