This might be more of a math question. This is a peculiar thing about three-dimensional space. Note that in three dimensions, an area such as a plane is a two dimensional subspace. On a sheet of paper you only need two numbers to unambiguously denote a point.
Now imagine standing on the sheet of paper, the direction your head points to will always be a way to know how this plane is oriented in space. This is called the "normal" vector to this plane, it is at a right angle to the plane.
If you now choose the convention to have the length of this normal vector equal to the area of this surface, you get a complete description of the two dimensional plane, its orientation in three dimensional space (the vector part) and how big this plane is (the length of this vector).
Mathematically, you can express this by the "cross product" $$\vec c=\vec a\times\vec b$$ whose magnitude is defined as $|c| = |a||b|sin\theta$ which is equal to the area of the parallelogram that those two vectors (which really define a plane) span. To steal this picture from wikipedia's article on the cross product:
As I said in the beginning this is a very special thing for three dimensions, in higher dimensions, it doesn't work as neatly for various reasons. If you want to learn more about this topic a keyword would be "exterior algebra"
Update:
As for the physical significance of this concept, prominent examples are vector fields flowing through surfaces. Take a circular wire. This circle can be oriented in various ways in 3D. If you have an external magnetic field, you might know that this can induce an electric current, proportional to the rate of change of the amount flowing through the circle (think of this as how much the arrows perforate the area). If the magnetic field vectors are parallel to the circle (and thus orthogonal to its normal vector) they do not "perforate" the area at all, so the flow through this area is zero. On the other hand, if the field vectors are orthogonal to the plane (i.e. parallel to the normal), the maximally "perforate" this area and the flow is maximal.
if you change the orientation of between those two states you can get electrical current.
- Consider a changing vector $\vec{A}(t)$ on a fixed coordinate system. The rate of change of the vector components are $$\frac{{\rm d}\vec{A}(t)}{{\rm d}t} = \frac{\partial \vec{A}(t)}{\partial t}= \dot{\vec{A}}(t) $$
- Consider a fixed vector $\vec{A}$ on a rotating coordinate system, with angular velocity $\vec\Omega(t)$. The rate of change of the vector components with respect to the attached frame are zero, but with respect to an inertial frame the rate of change is $$\frac{{\rm d}\vec{A}(t)}{{\rm d}t} = \vec{\Omega}(t) \times \vec{A} $$
- Combined the changing $\vec{A}(t)$ vector on a rotating frame $\vec\Omega(t)$ is
$$\boxed{ \frac{{\rm d}\vec{A}(t)}{{\rm d}t} = \frac{\partial \vec{A}(t)}{\partial t} + \vec{\Omega}(t) \times \vec{A}(t)}$$
The first part is the intrinsic change along the rotating frame, and the second part the change due to the rotation of the coordinates.
Example
Two bodies are connected with a slider joint, defined by an axis $\vec{e}$ fixed to the first body which is rotating by $\Omega(t)$. If the joint distance is $\chi(t)$ then the position of the 2nd body is defined relative to the first body as $$\vec{r}_2(t) = \vec{r}_1(t) + \vec{e}\, \chi(t)$$
The position vectors are differentiated to derive the velocity kinematics as:
$$ \vec{v}_1 = \frac{{\rm d}\vec{r}_1(t)}{{\rm d}t} \\
\vec{v}_2 = \frac{{\rm d}\vec{r}_2(t)}{{\rm d}t} $$
$$\begin{align} \vec{v}_2 & = \vec{v}_1 + \frac{{\rm d}(\vec{e}\,\chi(t))}{{\rm d}t} \\
& = \vec{v}_1 + \frac{\partial \vec{e} \chi(t)}{\partial t} + \vec{\Omega}(t) \times \vec{e} \,\chi(t) \\
& = \vec{v}_1 + \vec{e} \dot\chi (t) + \vec{\Omega}(t) \times \left( \vec{r}_2(t) - \vec{r}_1(t) \right)
\end{align}$$
See related answer: Derivation of Euler's equations for rigid body rotation
Best Answer
Here is another approach that is essentially the same as the one suggested by John, but I find it to be less error-prone when doing calculations by hand.
You are given
$$ \mathbf{A} = 5\mathbf{i} - 6.5\mathbf{j}\\ \mathbf{B} = -3.5\mathbf{i} + 7\mathbf{j} $$
You know that $ \mathbf{C} \perp \mathbf{A}$. Therefore,
$$ \mathbf{C} = s \cdot (6.5 \mathbf{i} + 5\mathbf{j}) $$
with $s$ a scale factor. This is because in 2D, for any vector $\alpha = a\mathbf{i} + b\mathbf{j}$, the vector $\beta = -b\mathbf{i} + a\mathbf{j}$ satisfies $\alpha \cdot \beta = 0$ and thus $\alpha \perp \beta$.
Given that $\mathbf{B} \cdot \mathbf{C} = 15$, it should be fairly easy to solve for the scale factor $s$, thus solving for $\mathbf{C}$.