oopsteekwe

2022-10-16

The vector $A=5i+6j$ is rotated through an $\mathrm{\angle }45$ about the Z axis in the anticlockwise direction. What is the resultant vector?
My attempt: I tried to calculate the resultant vector by using the equation, $R=\sqrt{{A}^{2}+{B}^{2}+2ABCos\theta }$
since it is rotated in anticlockwise direction its direction changes. Any hint will be appreciated.

HINT
First approach
You can solve it by considering the rotation matrix, where $\left({x}^{\prime },{y}^{\prime }\right)$ are the new coordinates after the rotation:
$\begin{array}{r}\left[\begin{array}{c}{x}^{\prime }\\ {y}^{\prime }\end{array}\right]=\left[\begin{array}{cc}\mathrm{cos}\left(\frac{\pi }{4}\right)& -\mathrm{sin}\left(\frac{\pi }{4}\right)\\ \mathrm{sin}\left(\frac{\pi }{4}\right)& \mathrm{cos}\left(\frac{\pi }{4}\right)\end{array}\right]\left[\begin{array}{c}5\\ 6\end{array}\right]\end{array}$
Second approach
Since $A=5i+6j=\left(5,6\right)$, you can multiply it by $\mathrm{exp}\left(\frac{\pi i}{4}\right)$

Paloma Sanford

1)$\stackrel{\to }{i}\to \left(1/\surd 2\right)\left(\stackrel{\to }{i}+\stackrel{\to }{j}\right)$
2)$\stackrel{\to }{j}\to \left(1/\surd 2\right)\left(\stackrel{\to }{-i}+\stackrel{\to }{j}\right)$
3)$5\stackrel{\to }{i}+6\stackrel{\to }{j}\to$
$\left(1/\surd 2\right)\left(-1\right)\stackrel{\to }{i}+\left(1/\surd 2\right)\left(11\right)\stackrel{\to }{j}$

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