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.

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.

Pradellalo

Beginner2022-10-17Added 16 answers

HINT

First approach

You can solve it by considering the rotation matrix, where $({x}^{\prime},{y}^{\prime})$ 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 $A=5i+6j=(5,6)$, you can multiply it by $\mathrm{exp}\left(\frac{\pi i}{4}\right)$

First approach

You can solve it by considering the rotation matrix, where $({x}^{\prime},{y}^{\prime})$ 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 $A=5i+6j=(5,6)$, you can multiply it by $\mathrm{exp}\left(\frac{\pi i}{4}\right)$

Paloma Sanford

Beginner2022-10-18Added 3 answers

1)$\overrightarrow{i}\to (1/\surd 2)(\overrightarrow{i}+\overrightarrow{j})$

2)$\overrightarrow{j}\to (1/\surd 2)(\overrightarrow{-i}+\overrightarrow{j})$

3)$5\overrightarrow{i}+6\overrightarrow{j}\to $

$(1/\surd 2)(-1)\overrightarrow{i}+(1/\surd 2)(11)\overrightarrow{j}$

2)$\overrightarrow{j}\to (1/\surd 2)(\overrightarrow{-i}+\overrightarrow{j})$

3)$5\overrightarrow{i}+6\overrightarrow{j}\to $

$(1/\surd 2)(-1)\overrightarrow{i}+(1/\surd 2)(11)\overrightarrow{j}$

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