91Ó°ÊÓ

The dot product is a fundamental concept in vector mathematics. When two vectors are perpendicular, their dot product is always zero. This stems from the formula for the dot product of two vectors:

• If we have vectors \( \mathbf{v} = (v_1, v_2) \) and \( \mathbf{w} = (w_1, w_2) \), their dot product is \( \mathbf{v} \cdot \mathbf{w} = v_1w_1 + v_2w_2 \).
• For perpendicular vectors, we set \( \mathbf{v} \cdot \mathbf{w} = 0 \).

In the exercise, vector \( \mathbf{v} \) is \((2, 3)\), and vector \( \mathbf{w} = (a, b) \) must be perpendicular to \( \mathbf{v} \). Thus, the equation becomes \( 2a + 3b = 0 \). This is our starting point for finding perpendicular vector \( \mathbf{w} \). This relationship confirms that when two vectors are perpendicular, their direction indicators, \(a\) and \(b\) in our case, must balance each other out to zero.

Understanding the magnitude or "length" of a vector is crucial, especially when given specific conditions, like in this exercise where \( \|\mathbf{w}\| = 5 \). The magnitude of a vector \( \mathbf{w} = (a, b) \) can be calculated using the formula:

• \( \|\mathbf{w}\| = \sqrt{a^2 + b^2} \)

This represents the distance from the origin to the endpoint of the vector in a 2D plane. Here, we are instructed that this distance is 5, so we set up the equation \( \sqrt{a^2 + b^2} = 5 \). By substituting \(a = -\frac{3}{2}b\) (derived from the dot product condition) into the magnitude formula, we establish a new equation: \( \sqrt{\left(-\frac{3}{2}b\right)^2 + b^2} = 5 \).This condition checks that \( \mathbf{w} \)'s length complies with the exercise requirements.

When equations involve multiple variables, solving them systematically is essential. This process typically involves simplifying one equation under a given condition before moving on to use any given secondary conditions, like magnitude.

• Starting with the condition \( 2a + 3b = 0 \), we solved for \(a\): thus \(a = -\frac{3}{2}b\).
• Next, plug this expression for \(a\) into the magnitude equation: \( \sqrt{(-\frac{3}{2}b)^2 + b^2} = 5 \).
• Simplifying this, we derived: \( \frac{9}{4}b^2 + b^2 = 25 \).
• Further simplification led to: \( \frac{13}{4}b^2 = 25 \), thus solving \( b^2 = \frac{100}{13} \).

Finding \(b\) involves taking the square root of \( \frac{100}{13} \), resulting in \( b = \pm \frac{10}{\sqrt{13}} \). Subsequently, \(a\) was obtained as \(a = -\frac{15}{\sqrt{13}} \) for one such vector \( \mathbf{w} = (a,b) \). This methodical approach ensures we find vector solutions that fulfil given conditions.