Question

Linear Algebra, fun and games finding \(\theta\)

For the vectors u and v with magnitude |u|=4 and |v|=5, find the angle theta between u and v which makes \(|proj_{\vec{u}}\vec{v}|=3\) (Which I feel means the magnitude of the orthogonal projection of v onto u)

Answer 1

From this I got the following:
\[\cos\theta = \frac{\vec{v}\cdot \vec{u}}{||\vec{v}||\,||\vec{u}||}\implies \cos\theta = \frac{\vec{v}\cdot \vec{u}}{20}\implies 20\cos\theta = \vec{v}\cdot \vec{u}\]and
\[|proj_{\vec{u}}\vec{v}|=3 \implies
\frac{\vec{v}\cdot \vec{u}}{||\vec{u}||^2}||\vec{u}||=3\implies
\frac{\vec{v}\cdot \vec{u}}{12}4=3\implies
\vec{v}\cdot \vec{u}=12\]

Answer 2

\[\therefore \theta=\cos^{-1}\left(\frac{3}{5}\right) \]That right?

Answer 3

It seems fine to me.

Answer 4

Ooh... we are back finally! made one small typo while putting it up... the \(\frac{\vec{v}\cdot \vec{u}}{12}\) is supposed to be \(\frac{\vec{v}\cdot \vec{u}}{16}\) but that was just a mistake here and not in the math for the answer.

Thanks for the check eliassaab