Question

2,3,4,5,6,7,8,9, and 10 find a · b. for exercises 9 - 10, refer to the notation introduced in exercises 8.2.30, 8.2.31, and 8.2.32.

1. |a| = 3, |b| = \sqrt{6}, the angle between a and b is 45°

Explanation:

Step1: Recall the formula for the dot product

The formula for the dot product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is \(\mathbf{a}\cdot\mathbf{b}=\vert\mathbf{a}\vert\vert\mathbf{b}\vert\cos\theta\), where \(\vert\mathbf{a}\vert\) and \(\vert\mathbf{b}\vert\) are the magnitudes of vectors \(\mathbf{a}\) and \(\mathbf{b}\) respectively, and \(\theta\) is the angle between them.

Step2: Substitute the given values into the formula

Given \(\vert\mathbf{a}\vert = 3\), \(\vert\mathbf{b}\vert=\sqrt{6}\), and \(\theta = 45^{\circ}\) (so \(\cos\theta=\cos45^{\circ}=\frac{\sqrt{2}}{2}\)).
Substitute into the formula: \(\mathbf{a}\cdot\mathbf{b}=3\times\sqrt{6}\times\frac{\sqrt{2}}{2}\).

Step3: Simplify the expression

First, simplify \(3\times\sqrt{6}\times\frac{\sqrt{2}}{2}\). Since \(\sqrt{6}\times\sqrt{2}=\sqrt{12} = 2\sqrt{3}\), then \(3\times\sqrt{6}\times\frac{\sqrt{2}}{2}=3\times\frac{2\sqrt{3}}{2}\).
Cancel out the common factor of 2: \(3\times\sqrt{3}=3\sqrt{3}\).

Answer:

\(3\sqrt{3}\)