91Ó°ÊÓ

The **dot product** is a crucial tool in understanding the relationship between two vectors. It allows us to find the angle between them. The dot product of two vectors, \(\textbf{u} = \langle u_1, u_2 \rangle\) and \(\textbf{v} = \langle v_1, v_2 \rangle\), is calculated using the formula:

\[ \textbf{u} \cdot \textbf{v} = u_1 v_1 + u_2 v_2 \]

This means you multiply the corresponding components of the vectors and then sum the results.

In the given exercise, with \(\textbf{u} = \langle-6, 5\rangle\) and \(\textbf{v} = \langle5, 6\rangle\), the dot product is:

\[(-6 \cdot 5) + (5 \cdot 6) = -30 + 30 = 0\]

The dot product being 0 indicates that the vectors are perpendicular to each other.

The **magnitude** of a vector indicates its length. It's calculated using the Pythagorean theorem. For a vector \(\textbf{u} = \langle u_1, u_2 \rangle\), the magnitude is:

\[||\textbf{u}|| = \sqrt{u_1^2 + u_2^2} \]

This formula finds the 'straight-line' distance from the origin to the point defined by the vector.

In the problem, for \(\textbf{u} = \langle-6, 5\rangle\), we get:

\[||\textbf{u}|| = \sqrt{(-6)^2 + 5^2} = \sqrt{36 + 25} = \sqrt{61} \]

Similarly, for \(\textbf{v} = \langle5, 6\rangle\), the procedure is the same:

\[ ||\textbf{v}|| = \sqrt{5^2 + 6^2} = \sqrt{25 + 36} = \sqrt{61} \]

Finding the magnitude is an essential step in using the dot product formula for the angle between the vectors.

To find the angle **between vectors**, we use the inverse cosine function (often written as \(\text{cos}^{-1}\)). The formula to get the angle \(\theta\) between two vectors \(\textbf{u}\) and \(\textbf{v}\) using the dot product and magnitudes is:

\[ \cos(\theta) = \frac{\textbf{u} \cdot \textbf{v}}{||\textbf{u}|| \cdot ||\textbf{v}||} \]

Then, we solve for \(\theta\) by taking the inverse cosine of the result:

\[\theta = \cos^{-1}\left(\frac{\textbf{u} \cdot \textbf{v}}{||\textbf{u}|| \cdot ||\textbf{v}||}\right) \]

In the given problem, since the dot product is 0:

\[ \cos(\theta) = \frac{0}{\sqrt{61} \cdot \sqrt{61}} = \frac{0}{61} = 0 \]

Thus, \(\theta = \cos^{-1}(0)\), which equals 90 degrees. This confirms that the vectors are perpendicular.