Question

|u| = 3, |v| = 2, and the angle between |u| and |v| (when placed tail-to-tail) is 45 degrees. Find |2u+3v|.

The answer is 11.09 but could somebody show me how it can be solved?

Answer 1

Now, I think here's what you have to do; I'm not entirely certain about it though.

From the given information, you can set up a system of equations. Assuming \(u\) and \(v\) are vectors in \(\mathbb{R}^2\), you have \(u=\langle u_1,u_2\rangle\) and \(v=\langle v_1,v_2\rangle\).

You know that \(|u|=\sqrt{u_1^2+u_2^2}=3\) and \(|v|=\sqrt{v_1^2+v_2^2}=2\).

Also, you're given the angle between the vectors is \(45^\circ\). Using the dot product formula,
\[a\cdot b=|a||b|\cos\theta\]
you can find \(u\cdot v\).

Another form of the dot product is
\[a\cdot b=a_1b_1+a_2b_2\]
so you have the following system of equations:
\[\begin{cases}\sqrt{u_1^2+u_2^2}=3\\\sqrt{v_1^2+v_2^2}=2\\u_1v_1+u_2v_2=(2)(3)\cos(45^\circ)\end{cases}\]

Answer 2

what would u2 and v2 be in this case?

Answer 3

i've tried approaching it this way but i can't seem to find a solution

Answer 4

I'm thinking you might have to choose any \(u\) that satisfies \(|u|=3\), then find \(v\) such that it forms a 45° angle with \(u\).

So one way would be to assume \(u=\langle 3,0\rangle\). Indeed, \(|u|=\sqrt{3^2+0^2}=\sqrt9=3\). Now you find \(v\) that satisfies the remaining equations:
\[\begin{cases}\sqrt{v_1^2+v_2^2}=2\\3v_1=(2)(3)\cos(45^\circ)\end{cases}\]
Then you immediately find that \(v_1=\sqrt2\), so then you have
\[\sqrt{(\sqrt2)^2+v_2^2}=2\\
v_2^2+2=4\\
v_2=\pm\sqrt2\]

Answer 5

So then you can find \(|2u+3v|\), given that you use \(u=\langle3,0\rangle\) and \(v=\langle \sqrt2,\sqrt2\rangle\).

Answer 6

thank you so much!