\[cos(\theta) = \frac{\mathbf{A}\cdot\mathbf{B}}{||\mathbf{A}||\cdot||\mathbf{B}||}=\frac{\sum_{i=1}^n A_{i} \cdot B_{i}}{\sqrt{\sum_{i=1}^{n}A_{i}^{2}}\cdot\sqrt{\sum_{i=1}^{n}B_{i}^{2}}}\]

To determine if two vectors are similar using the formula you provided, we need to calculate the cosine of the angle between them. The formula for the cosine of the angle $\theta$ between two vectors $\mathbf{A}$ and $\mathbf{B}$ is given by:

\[cos(\theta) = \frac{\mathbf{A}\cdot\mathbf{B}}{||\mathbf{A}||\cdot||\mathbf{B}||}\]

where $\mathbf{A}\cdot\mathbf{B}$ represents the dot product of the two vectors and $||\mathbf{A}||$ and $||\mathbf{B}||$ represent the magnitudes (or norms) of the two vectors.

In this case, we are given the vectors $\mathbf{A}=[50,10]$ and $\mathbf{B}=[14,12]$. To calculate their similarity, we first need to find their dot product:

\[\mathbf{A}\cdot\mathbf{B} = 50 \times 14 + 10 \times 12 = 700 + 120 = 820\]

Next, we calculate the magnitudes of $\mathbf{A}$ and $\mathbf{B}$:

\[\|\|\mathbf{A}\|\| = \sqrt{50^2 + 10^2} = \sqrt{2500 + 100} = \sqrt{2600} \approx 50.99\] \[\|\|\mathbf{B}\|\| = \sqrt{14^2 + 12^2} = \sqrt{196 + 144} = \sqrt{340} \approx 18.73\]

Now, we can use the formula to find the cosine of the angle between $\mathbf{A}$ and $\mathbf{B}$:

\[cos(\theta) = \frac{\mathbf{A}\cdot\mathbf{B}}{||\mathbf{A}||\cdot||\mathbf{B}||} = \frac{820}{\sqrt{2600 \times 340}} \approx 0.915\]

Since $cos(\theta) \approx 0.915$, the vectors $\mathbf{A}$ and $\mathbf{B}$ are similar, meaning they have a small angle between them (approximately 18 degrees).


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