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The vector $A=5i+6j$ is rotated through an $\angle 45$ 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.

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    $\begingroup$ What tools do you have available to you? This should be trivial using matrices. $\endgroup$
    – JMoravitz
    May 18, 2020 at 16:08
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    $\begingroup$ This is a question for entrance exam in our university. Will you please help me with a hint? $\endgroup$ May 18, 2020 at 16:14

2 Answers 2

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HINT

First approach

You can solve it by considering the rotation matrix, where $(x',y')$ are the new coordinates after the rotation: \begin{align*} \begin{bmatrix} x'\\ y' \end{bmatrix} = \begin{bmatrix} \cos\left(\frac{\pi}{4}\right) & -\sin\left(\frac{\pi}{4}\right)\\ \sin\left(\frac{\pi}{4}\right) & \cos\left(\frac{\pi}{4}\right) \end{bmatrix} \begin{bmatrix} 5\\ 6 \end{bmatrix} \end{align*}

Second approach

Since $A = 5i + 6j = (5,6)$, you can multiply it by $\exp\left(\frac{\pi i}{4}\right)$.

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    $\begingroup$ It was really helpful $\endgroup$ May 18, 2020 at 16:27
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1) $\vec {i} \rightarrow (1/√2)(\vec {i} +\vec {j})$;

2) $\vec {j} \rightarrow (1/√2)(\vec {-i}+\vec {j})$;

3) $5 \vec {i} +6 \vec {j} \rightarrow $

$(1/√2)(-1)\vec {i} +(1/√2)(11)\vec {j}$.

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