Problem 54 - Algebra and Trigonometry

Question

For \(\mathbf{v}=2 \mathbf{i}+3 \mathbf{j}\) and \(\mathbf{w}=3 \mathbf{i}-2 \mathbf{j}\) (a) Find the dot product \(\mathbf{v} \cdot \mathbf{w}\). (b) Find the angle between \(\mathbf{v}\) and \(\mathbf{w}\). (c) Are the vectors parallel, orthogonal, or neither?

Step-by-Step Solution

Answer

(a) 0 (b) \frac{\pi}{2} (c) Orthogonal

Step 1: Identify the Vectors

Given vectors are \(\textbf{v} = 2\textbf{i} + 3\textbf{j}\) and \(\textbf{w} = 3\textbf{i} - 2\textbf{j}\). Identify components: \(\textbf{v} = (2, 3)\) and \(\textbf{w} = (3, -2)\).

Step 2: Find the Dot Product

The formula for the dot product is \(\textbf{v} \cdot \textbf{w} = v_1w_1 + v_2w_2\). Here, \(\textbf{v} = (2, 3)\) and \(\textbf{w} = (3, -2)\). So, \(\textbf{v} \cdot \textbf{w} = (2\cdot3) + (3\cdot-2) = 6 - 6 = 0\).

Step 3: Calculate Magnitudes of Vectors

The magnitude of \(\textbf{v} \) is \[| \textbf{v} | = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}\.\] The magnitude of \(\textbf{w} \) is \[| \textbf{w} | = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}\.\]

Step 4: Find the Angle Between Vectors

Using the dot product formula, \[ \textbf{v} \cdot \textbf{w} = | \textbf{v} | | \textbf{w} | \cos \theta \] Since \(\textbf{v} \cdot \textbf{w} = 0 \), we have \[0 = \sqrt{13}\sqrt{13} \cos \theta \] which simplifies to \[ \theta = \cos^{-1}(0) = \frac{\pi}{2}\]

Step 5: Determine Relationship Between Vectors

Since the dot product of \(\textbf{v} \) and \(\textbf{w} \) is zero, \(\textbf{v} \) and \(\textbf{w} \) are orthogonal, or perpendicular, to each other.

Key Concepts

Dot Product

The dot product is an important operation in vector mathematics. To find the dot product of two vectors, \(\textbf{v} \cdot \textbf{w}\), we multiply their corresponding components and add the results. For vectors \(\textbf{v} = 2\textbf{i} + 3\textbf{j}\) and \(\textbf{w} = 3\textbf{i} - 2\textbf{j}\), we have components \( (2, 3) \) and \( (3, -2) \). Applying the formula: \(\textbf{v} \cdot \textbf{w} = v_1 w_1 + v_2 w_2\), we get:
\((2 \cdot 3) + (3 \cdot (-2)) = 6 - 6 = 0\).
Hence, the dot product is zero.

Angle Between Vectors

Finding the angle between vectors involves the dot product and magnitudes of the vectors. From the dot product formula, \(\textbf{v} \cdot \textbf{w} = | \textbf{v} | | \textbf{w} | \cos \theta\), we solve for \(\theta\). Given that \( \textbf{v} \cdot \textbf{w} = 0\) and magnitudes \(| \textbf{v} | = \sqrt{13}\) and \(| \textbf{w} | = \sqrt{13}\), we get:
\( 0 = \sqrt{13} \cdot \sqrt{13} \cdot \cos \theta\) which simplifies to: \( \theta = \cos^{-1}(0) = \frac{\pi}{2}\).
Therefore, the angle between \( \textbf{v} \) and \( \textbf{w} \) is \( \frac{\pi}{2}\) or 90 degrees.

Orthogonal Vectors

Vectors that are orthogonal are perpendicular or at a 90-degree angle to one another. This is identified when their dot product equals zero. In our example, since \( \textbf{v} \cdot \textbf{w} = 0\), vectors \( \textbf{v} \) and \( \textbf{w} \) are orthogonal. Orthogonality is a fundamental concept in geometry and linear algebra. If you ever get a dot product of zero between two vectors, it is a clear indication that the vectors are orthogonal.

Vector Magnitude

The magnitude of a vector represents its length or size, usually denoted by \(|\textbf{v}| \). To calculate the magnitude, we use the formula: \( |\textbf{v}| = \sqrt{v_1^2 + v_2^2}\). For \( \textbf{v} = 2\textbf{i} + 3\textbf{j} \), the magnitude is:
\( |\textbf{v}| = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \).
Similarly, for \( \textbf{w} = 3\textbf{i} - 2\textbf{j} \), the magnitude is: \( |\textbf{w}| = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \).
Understanding the magnitude of vectors is essential as it provides insight into the scale and dimensions represented by the vector.