This question was originally titled:

What is the relation between the scalar product and dot product of two vectors?

But that didn’t match what the body of the question asks, so let’s start by clarifying some terms:

The scalar product and the dot product are two names for the same thing, written \$\vec a \cdot \vec b\$.

What’s written on the left side of the equation there is the magnitude of the vector product, also called the cross product.

With that terminology cleared up, we can say something about the equation in question:

$$\begin{align}
\left \| \vec a \times \vec b \right\| &= \vec a \cdot \vec b\\
ab \left| \sin \theta \right| &= ab \left| \cos \theta \right|
\end{align}$$

Where \$a \ge 0\$ and \$b \ge 0\$ are the lengths of vectors \$\vec a\$ and \$\vec b\$, respectively, and \$\theta\$ is the angle between them.

We can see that this equation will hold true if either \$a = 0\$ or \$b = 0\$ (either \$\vec a\$ or \$\vec b\$ or both are the zero vector), or if the angle obeys \$ \left| \sin \theta \right| = \left| \cos \theta \right|\$, which happens if there’s an angle of \$\frac {\pi} 4\$, \$\frac {- \pi} 4\$, \$\frac {4 \pi} 4\$, or \$\frac {-3 \pi} 4\$ between them.

(The latter two are the same separation between the vectors as the first two, just “in the opposite order”. This is the difference between \$\vec a\$ being clockwise or counterclockwise of \$\vec b\$, if you fix one view/orientation of the plane containing them. This matters for the sign of the cross product, but not its magnitude: \$ \vec a \times \vec b = – \vec b \times \vec a\$)