91Ó°ÊÓ

Vector normalization is the process of converting a vector into a unit vector. A unit vector has a length (or magnitude) of 1, but still points in the same direction as the original vector. To normalize a vector \(\textbf{V} \), you first find its magnitude using the formula:
\(\text{Magnitude of } \textbf{V} = \sqrt{V_x^2 + V_y^2 + V_z^2}\).
Next, you divide each component of the vector by its magnitude. For example, for the vector \(\textbf{V} = 3\textbf{i} + 3\textbf{j} + 3\textbf{k}\), the magnitude is \(\text{3} \sqrt{3}\).
The normalized vector then becomes:
\(\frac{\textbf{V}}{\text{3} \sqrt{3}} = \frac{\textbf{i}}{\text{\sqrt{3}}} + \frac{\textbf{j}}{\text{\sqrt{3}}} + \frac{\textbf{k}}{\text{\sqrt{3}}}\).
Notice how each component of the vector is now divided by the original vector's magnitude.

The dot product is an operation that takes two equal-length sequences of numbers (usually coordinate vectors), and returns a single number. The dot product of two vectors \(\textbf{A}\) and \(\textbf{B}\) is calculated as:
\(\textbf{A} \cdot \textbf{B} = A_xB_x + A_yB_y + A_zB_z\).
In the problem, we calculated the dot product of \(\textbf{V} = 3\textbf{i} + 3\textbf{j} + 3\textbf{k}\) and \(\textbf{i} + \textbf{j} + \textbf{k}\).
This resulted in \((3 \cdot 1) + (3 \cdot 1) + (3 \cdot 1) = 9\).
It's important to note here that the result was 9 and not 0, which means that vector \(\textbf{V}\) is not orthogonal to \(\textbf{i} + \textbf{j} + \textbf{k}\).

Two vectors are perpendicular (or orthogonal) if their dot product is zero. For example, if you have two vectors \(\textbf{A}\) and \(\textbf{B}\), they are perpendicular if:
\(\textbf{A} \cdot \textbf{B} = 0\).
In our exercise, to verify this with vectors \(\textbf{V}\) and \((2\textbf{i} + \textbf{j} - 3\textbf{k})\), we computed the dot product:
\(\textbf{V} \cdot (2\textbf{i} + \textbf{j} - 3\textbf{k}) = 3 \cdot 2 + 3 \cdot 1 + 3 \cdot (-3) = 6 + 3 - 9 = 0\).
Since the dot product is zero, vectors \(\textbf{V}\) and \((2\textbf{i} + \textbf{j} - 3\textbf{k})\) are indeed perpendicular.

Understanding the angles that a vector makes with the coordinate axes can provide insights into the vector's direction. For a vector \(\textbf{V} = V_x\textbf{i} + V_y\textbf{j} + V_z\textbf{k}\), the cosines of the angles \(\theta_x\), \(\theta_y\), and \(\theta_z\) with the x, y, and z-axes can be calculated as:

• \(\text{cos} \theta_x = \frac{V_x}{||\textbf{V}||}\)
• \(\text{cos} \theta_y = \frac{V_y}{||\textbf{V}||}\)
• \(\text{cos} \theta_z = \frac{V_z}{||\textbf{V}||}\)

Given our vector \(\textbf{V} = 3\textbf{i} + 3\textbf{j} + 3\textbf{k}\) with a magnitude of \(\text{3}\sqrt{3}\), the components along the axes are the same:
\(\text{cos} \theta_x = \text{cos} \theta_y = \text{cos} \theta_z = \frac{3}{3 \sqrt{3}} = \frac{1}{\text{\sqrt{3}}}\).
This symmetry confirms the vector makes equal angles with all three axes.