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If the coordinate system (frame of reference) rotates, the components of a vector, when expressed in that frame, will change. For example, a vector pointing along the $X$-axis in one frame will have both $X$ and $Y$ components in a frame rotated by $45^\circ$. Although the vector itself (its magnitude and direction in absolute space) remains unchanged, its representation (components) in the rotated frame changes. In physics, a change in components is considered a "change in a vector" in the context of its representation.

If the vector itself rotates, its direction changes. A vector is defined by both its magnitude and direction. If either of these changes, the vector itself changes. For example, if a vector pointing east rotates to point north, it is a different vector.

Both the rotation of the frame of reference (leading to a change in components) and the rotation of the vector itself (leading to a change in direction) result in a change in the vector.

The final answer is $\boxed{\text{C}}$.

Let $\vec{R} = \vec{A} + \vec{B}$ be the resultant vector.
Given that $\vec{R}$ is perpendicular to $\vec{A}$, their dot product is zero:

Let $A = |\vec{A}|$ and $B = |\vec{B}|$. Let $\theta$ be the angle between $\vec{A}$ and $\vec{B}$.

Since $A \ne 0$ (otherwise $\vec{R}$ would be $\vec{B}$ and not perpendicular to $\vec{A}$ unless $\vec{B}$ is also zero), we can divide by $A$:

Since $A$ is a magnitude, $A > 0$. This implies that $\cos\theta$ must be negative, so $\theta$ is an obtuse angle.

Given that the magnitude of the resultant is half of $\vec{B}$:

The magnitude of the resultant of two vectors is given by:

Substitute $R = B/2$:

From (1), $A = -B\cos\theta$. Substitute this into (2):

Assuming $B \ne 0$, divide the entire equation by $B^2$:

Rearrange to solve for $\cos^2\theta$:

Take the square root:

From Step 1, we determined that $\cos\theta$ must be negative.

The angle $\theta$ in the range $[0, \pi]$ for which $\cos\theta = -\frac{\sqrt{3}}{2}$ is:

The final answer is $\boxed{\text{D}}$.

We are given the magnitudes $A=5, B=4, C=3$.
Notice that $3^2 + 4^2 = 9 + 16 = 25 = 5^2$. This means $C^2 + B^2 = A^2$.
For vector addition $\vec{A} = \vec{B} + \vec{C}$, the magnitude is given by $A^2 = B^2 + C^2 + 2BC\cos\phi$, where $\phi$ is the angle between $\vec{B}$ and $\vec{C}$.
Substituting the magnitudes:

This implies $\phi = 90^\circ$ or $\frac{\pi}{2}$ radians. Therefore, vector $\vec{B}$ is perpendicular to vector $\vec{C}$.

Let $\alpha$ be the angle between $\vec{A}$ and $\vec{C}$. We can use the dot product definition:

Substitute $\vec{A} = \vec{B} + \vec{C}$:

Since $\vec{B}$ is perpendicular to $\vec{C}$, $\vec{B} \cdot \vec{C} = 0$.
Also, $\vec{C} \cdot \vec{C} = |\vec{C}|^2 = C^2$.

Since $C \ne 0$, we can divide by $C$:

Substitute the given magnitudes $C=3$ and $A=5$:

The final answer is $\boxed{\text{B}}$.