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.
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 \begin{equation} \mathbf{r}=\mathbf{r}_\|+\mathbf{r}_\bot \tag{01} \end{equation} 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 $\mathbf{r}^{\prime}$ \begin{equation} \mathbf{r}^{\prime}=\mathbf{r}^{\prime}_\|+\mathbf{r}^{\prime}_\bot \tag{03} \end{equation} then the parallel component remains unchanged \begin{equation} \mathbf{r}^{\prime}_\|=\mathbf{r}_\| =(\mathbf{n}\boldsymbol{\cdot}\mathbf{r})\mathbf{n} \tag{04} \end{equation} 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 \begin{equation} \left\|(\mathbf{n}\times\mathbf{r})\times \mathbf{n}\right\|=\left\|\mathbf{n}\times\mathbf{r}\right\| \tag{05} \end{equation} 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
\begin{equation} \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} \end{equation}
From this the $3\times3$ rotation matrix reads \begin{equation} \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 \end{equation} \begin{equation} = \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} \end{equation}
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.
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.
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.
