# Rotation of a vector

Is a vector necessarily changed when it is rotated through an angle?

I think a vector always gets changed because its projection will change, and also its inclination with axes will always change. However the direction may remain same. Kindly make things clear to me.

• If the rotation is around an axis defined by the vector itself, then there will be no change. If the rotation is a multiple of $2\pi$ then there will be no change. – Jon Custer Apr 29 '16 at 17:41
• @JonCuster : Probably worth posting as an Answer? – sammy gerbil Apr 29 '16 at 20:30
• With respect to a fixed system of coordinate axes, the position vector of a point in a rotating body changes. – Procyon Apr 29 '16 at 22:23

Rotation of a 3-vector

We'll find an expression for the rotation of a vector $\mathbf{r}=(x_1,x_2,x_3)$ around an axis with unit vector $\mathbf{n}=(n_1,n_2,n_3)$ through an angle $\theta$, as shown in Figure .

The vector $\mathbf{r}$ is analysed in two components $$\mathbf{r}=\mathbf{r}_\|+\mathbf{r}_\bot \tag{01}$$ one parallel and the other normal to axis $\mathbf{n}$ respectively \begin{eqnarray} &\mathbf{r}_\| &=(\mathbf{n}\boldsymbol{\cdot}\mathbf{r})\mathbf{n} \tag{02a}\\ &\mathbf{r}_\bot &=(\mathbf{n}\times\mathbf{r})\times \mathbf{n}= \mathbf{r}-(\mathbf{n}\boldsymbol{\cdot}\mathbf{r})\mathbf{n} \tag{02b} \end{eqnarray} If $\mathbf{r}$ is rotated to $\mathbf{r}^{\prime}$ $$\mathbf{r}^{\prime}=\mathbf{r}^{\prime}_\|+\mathbf{r}^{\prime}_\bot \tag{03}$$ then the parallel component remains unchanged $$\mathbf{r}^{\prime}_\|=\mathbf{r}_\| =(\mathbf{n}\boldsymbol{\cdot}\mathbf{r})\mathbf{n} \tag{04}$$ while the normal component $\mathbf{r}_\bot =(\mathbf{n}\times\mathbf{r})\times \mathbf{n}$ is rotated by the angle $\theta$, so having in mind that this vector is perpendicular to $\mathbf{n}\times\mathbf{r}$ and of equal norm $$\left\|(\mathbf{n}\times\mathbf{r})\times \mathbf{n}\right\|=\left\|\mathbf{n}\times\mathbf{r}\right\| \tag{05}$$ we find the expression, see Figure below \begin{eqnarray} \mathbf{r}^{\prime}_\bot &=& \cos\theta\left[(\mathbf{n}\times\mathbf{r})\times \mathbf{n}\right]+\sin\theta\left[\mathbf{n}\times\mathbf{r}\right]\nonumber\\ &=& \cos\theta\left[\mathbf{r}-(\mathbf{n}\boldsymbol{\cdot}\mathbf{r})\mathbf{n}\right]+\sin\theta\left[\mathbf{n}\times\mathbf{r}\right]\nonumber\\ &=& \cos\theta\;\mathbf{r}-\cos\theta(\mathbf{n}\boldsymbol{\cdot}\mathbf{r})\mathbf{n}+\sin\theta\left[\mathbf{n}\times\mathbf{r}\right] \tag{06} \end{eqnarray}

and so finally the vector expression

$$\bbox[#FFFF88,12px]{\mathbf{r}^{\prime}= \cos\theta \cdot\mathbf{r}+(1-\cos\theta)\cdot(\mathbf{n}\boldsymbol{\cdot}\mathbf{r})\cdot\mathbf{n}+\sin\theta\cdot(\mathbf{n}\times\mathbf{r})} \tag{07}$$

From this the $3\times3$ rotation matrix reads $$\mathbb{A}\left(\mathbf{n}, \theta\right) = \text { 3D-rotation around axis} \:\:\mathbf{n}=\left(n_{1}, n_{2},n_{3}\right)\:\: \text{through angle} \:\:\theta$$ $$= \bbox[#FFFF88,12px]{ \begin{bmatrix} \cos\theta+(1-\cos\theta)n_1^2&(1-\cos\theta)n_1n_2-\sin\theta n_3&(1-\cos\theta)n_1n_3+\sin\theta n_2\\ (1-\cos\theta)n_2n_1+\sin\theta n_3&\cos\theta+(1-\cos\theta)n_2^2&(1-\cos\theta)n_2n_3-\sin\theta n_1\\ (1-\cos\theta)n_3n_1-\sin\theta n_2&(1-\cos\theta)n_3n_2+\sin\theta n_1&\cos\theta+(1-\cos\theta)n_3^2 \end{bmatrix}} \tag{08}$$

In general it changes although the reason is not exactly because its projections changes.

For example. You start with a vector (let us say the electric field of a parallel plate capacitor) on the plane $xy$. Then you rotate the coordinate system by an angle. The components of the vector on the new coordinate system is changed. But the vector did not change at all (you did not move the capacitor). This is called a passive rotation.

On the other hand, if you keep the axis fixed and rotate the vector (rotate the actual capacitor), it changed (unless you rotate by $2\pi$). This is an active rotation.

• Can you elaborate the active rotation? Passive rotation is totally clear to me but in passive rotation how exactly did the vector changed. – Hisenberg Apr 29 '16 at 18:20
• In the passive rotation the vector does not change. Only its components do. Think of a vector as a physical quantity regardless of your definition of coordinate system. It may be an electric field as above mentioned, the velocity of a car, the oriented displacement from Paris to Rome and so on. An active rotation means a rotation of the vector itself (not the coordinate system). In these examples it would mean to rotate the capacitor, to change the direction of the car, to change the oriented displacement from Paris to Rome to Paris to Madrid. – Diracology Apr 29 '16 at 18:30
• sorry I made a typo I want you to explain how exactly did the vector changed in active rotation. Passive rotation is totally clear to me. – Hisenberg Apr 29 '16 at 18:34

Direction of a vector is determined by the components themselves. Now if the components are changed the direction gets changed by the above definition. All this is with respect to one reference frame.

• Not if the components are all scaled by the same amount. The component change but the direction is does not. – ja72 Apr 29 '16 at 18:34
• @ja72 I quite didn't understand what you meant by "components are all scaled by the same amount". – Hisenberg Apr 29 '16 at 18:43
• The direction of $(1,2,3)$ is the same as the direction of $(2,4,6)$ where all the components are scaled by 2. – ja72 Apr 29 '16 at 19:07

Typically there are two kinds of transformation that do not change the outcome of situation. Think of a force vector $\vec{F}$ passing through a point $\vec{r}_A$.

• Any translation along the line of the force, in the direction $\vec{e} = \frac{\vec{F}}{\| \vec{F} \|}$ will not change the outcome.

• Any rotation about the line of the force would also not change the outcome.

• The only things that make a difference is translations perpendicular to the line, and hence the cross product when looking at torques $\vec{\tau} = \vec{r} \times \vec{F}$.

• And rotations perpendicular to the line change things.

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