Question

math 131 unit 3 section 6.5 written homewor
show your work as in class. exact value/form means; fractions (and \\( \pi \\) where appropr

1. \\( \boldsymbol { v } = \boldsymbol { i } + \boldsymbol { j } , \boldsymbol { w } = - \boldsymbol { i } + \boldsymbol { j } \\)

a) find the dot product \\( \mathbf { v } \cdot \mathbf { w } \\).
b) find the angle between \\( \mathbf { v } \\) and \\( \mathbf { w } \\).

Explanation:

Step1: Calculate the dot product

For vectors \(\mathbf{v}=a_1\mathbf{i}+b_1\mathbf{j}\) and \(\mathbf{w}=a_2\mathbf{i}+b_2\mathbf{j}\), the dot product \(\mathbf{v}\cdot\mathbf{w}=a_1a_2 + b_1b_2\). Here, \(\mathbf{v}=\mathbf{i}+\mathbf{j}\) (\(a_1 = 1\), \(b_1=1\)) and \(\mathbf{w}=-\mathbf{i}+\mathbf{j}\) (\(a_2=-1\), \(b_2 = 1\)).

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Step2: Calculate the magnitude of vectors

The magnitude of a vector \(\mathbf{v}=a\mathbf{i}+b\mathbf{j}\) is \(|\mathbf{v}|=\sqrt{a^{2}+b^{2}}\).
For \(\mathbf{v}=\mathbf{i}+\mathbf{j}\), \(|\mathbf{v}|=\sqrt{1^{2}+1^{2}}=\sqrt{2}\).
For \(\mathbf{w}=-\mathbf{i}+\mathbf{j}\), \(|\mathbf{w}|=\sqrt{(-1)^{2}+1^{2}}=\sqrt{2}\).

Step3: Use the formula for the angle between two vectors

The formula for the angle \(\theta\) between two vectors \(\mathbf{v}\) and \(\mathbf{w}\) is \(\cos\theta=\frac{\mathbf{v}\cdot\mathbf{w}}{|\mathbf{v}||\mathbf{w}|}\).
Since \(\mathbf{v}\cdot\mathbf{w} = 0\), \(|\mathbf{v}|=\sqrt{2}\), \(|\mathbf{w}|=\sqrt{2}\), then \(\cos\theta=\frac{0}{\sqrt{2}\times\sqrt{2}} = 0\).
We know that if \(\cos\theta=0\), then \(\theta=\frac{\pi}{2}\) (or \(90^{\circ}\))

Answer:

a) The dot product \(\mathbf{v}\cdot\mathbf{w}=0\)
b) The angle between \(\mathbf{v}\) and \(\mathbf{w}\) is \(\frac{\pi}{2}\) (or \(90^{\circ}\))