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The dot product is a way to multiply two vectors together which results in a scalar, not another vector. You can think of it as a measure of how much one vector goes in the direction of another. If you have two vectors \( \mathbf{u} = \langle u_1, u_2 \rangle \) and \( \mathbf{v} = \langle v_1, v_2 \rangle \), their dot product is calculated by multiplying the corresponding components and then summing the results. In this case, the vectors are \( \mathbf{u} = \langle 2, 7 \rangle \) and \( \mathbf{v} = \langle 3, 1 \rangle \). To find the dot product \( \mathbf{u} \cdot \mathbf{v} \), you need to do the following steps:

• Multiply the first components: \( 2 \times 3 = 6 \)
• Multiply the second components: \( 7 \times 1 = 7 \)
• Add these results: \( 6 + 7 = 13 \)

Thus, the dot product is 13.

The magnitude of a vector is like its size or length. For a vector \( \langle a, b \rangle \), you can find its magnitude using the Pythagorean theorem-like formula: \( \| \mathbf{a} \| = \sqrt{a^2 + b^2} \). This shows how far the vector moves from its start point to its endpoint. Let's find the magnitudes of the vectors \( \mathbf{u} = \langle 2, 7 \rangle \) and \( \mathbf{v} = \langle 3, 1 \rangle \):

• For \( \mathbf{u} \), calculate \( \sqrt{2^2 + 7^2} = \sqrt{4 + 49} = \sqrt{53} \)
• For \( \mathbf{v} \), calculate \( \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \)

These results mean \( \mathbf{u} \) has a magnitude of \( \sqrt{53} \) and \( \mathbf{v} \) has a magnitude of \( \sqrt{10} \). These values help when determining the angle between the vectors.

Finding the angle between two vectors involves both their dot product and magnitudes. To calculate it, you first find the cosine of the angle using the equation: \( \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{u} \| \| \mathbf{v} \|} \). This equation allows you to determine how aligned the vectors are with one another.Given our calculated values:

• \( \mathbf{u} \cdot \mathbf{v} = 13 \)
• \( \| \mathbf{u} \| = \sqrt{53} \) and \( \| \mathbf{v} \| = \sqrt{10} \)

Substitute these into the formula: \( \cos \theta = \frac{13}{\sqrt{53} \cdot \sqrt{10}} \).Calculate the product of the magnitudes: you get \( \cos \theta = \frac{13}{\sqrt{530}} \).Finally, find the angle \( \theta \) using the inverse cosine, or \( \cos^{-1} \), function: \( \theta = \cos^{-1} \left( \frac{13}{\sqrt{530}} \right) \).This calculation gives an approximate angle of 40 degrees between the vectors, showing how they relate directionally.