0
$\begingroup$

The question is as follows: enter image description here

I started to approach this by thinking of the vectors $v$ and $w$ as of the form $a \hat i+b \hat j$ and $c \hat i+d \hat j$. Based on this, I created three equations:

$\sqrt{a^2+b^2} = 3$ and $\sqrt{c^2+d^2} = \sqrt{2}$ from the vector magnitudes that are given, and $ac+bd=3$ from the vector angle equation $\hat u \cdot \hat v = |\hat u||\hat v|cos( \theta)$.

However, as there are four variables, and only three equations, I am obviously missing something here in order to solve the problem. What vector equation am I missing to be able to determine the vectors, and therefore this arbitrary magnitude which the question asks for?

$\endgroup$
2
  • 2
    $\begingroup$ You don't need to know all the variables to solve the magnitude. Just expand the sum $|v +2w|^2 = (v+2w) \cdot (v+2w)$. The info you're given should be enough to solve it. $\endgroup$ Commented Aug 22, 2021 at 18:36
  • $\begingroup$ Another method: if you draw the two vectors $ \ v \ $ and $ \ 2w \ $ head-to-tail and then draw in their sum, you will form a triangle with the included angle between the two given vectors being $ \ \pi - \frac{\pi}{4} \ = \ \frac{\ 3 \pi}{4} \ \ . $ You will be able to get the length-squared of the sum-vector $ \ v + 2w \ $ from the Law of Cosines. $\endgroup$
    – user882145
    Commented Aug 22, 2021 at 20:12

1 Answer 1

0
$\begingroup$

You have\begin{align}|\mathbf v+2\mathbf w|^2&=|\mathbf v|^2+4|\mathbf w|^2+4\langle\mathbf v,\mathbf w\rangle\\&=17+4\langle\mathbf v,\mathbf w\rangle.\end{align}Furthermore, since the angle between $\mathbf v$ and $\mathbf w$ is $\frac\pi4$,$$\langle\mathbf v,\mathbf w\rangle=|\mathbf v||\mathbf w|\cos\left(\frac\pi4\right)=3.$$Therefore, $|\mathbf v+2\mathbf w|^2=29$.

$\endgroup$
2
  • $\begingroup$ Thanks! Shouldn't $|v|^2+4|w|^2$ be 17 though, and the answer would be 29? $\endgroup$
    – figbar
    Commented Aug 22, 2021 at 18:46
  • $\begingroup$ @figbar Indeed. I've edited my answer. $\endgroup$ Commented Aug 22, 2021 at 18:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .