# How do you find the max value of a length of a vector?

I have a vector $v = 7j$ and a vector $u$ with a length of 5 that starts at the origin and rotates in the $xy$-plane.

How am I supposed to find the max value of the length of the vector $|u \times v|$ ?

I figured the equations is $|u \times v| = |u|\cdot|v| \sin(t)$ but don't know where I am supposed to go from there.

• What is the maximum value of sin(t)? Feb 9, 2014 at 20:44
• What do you mean by '$u$ rotates in the xy-plane'? Anyway, put $u:=5j$, that will minimize $|u\times v|$. Feb 9, 2014 at 20:47
• @Berci the cross-product of parallel vectors is zero.
– MJD
Feb 9, 2014 at 20:55
• Yes. It was written min value for the first case. Feb 9, 2014 at 23:18

You want to find $u$ that makes $|u\times v|$ as large as possible, and you correctly observed that this value is equal to $|u|\cdot|v|\cdot\sin t$, where $t$ is the angle between $u$ and $v$.
You know $|u| = 5$ and $|v| = 7$ so those are constant. To make $|u|\cdot|v|\cdot\sin t$ as large as possible, you need to make $\sin t$ as large as possible. This occurs when $t = \frac\pi2$, so you want $u$ and $v$ to be perpendicular.
What vector $u$ lies in the $xy$-plane, has length 5, and is perpendicular to $v=7j$?