# How to calculate the resultant vector

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.

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

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 $$\exp\left(\frac{\pi i}{4}\right)$$.

• It was really helpful May 18, 2020 at 16:27

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

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

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

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