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Chapter 6: Problem 18

Find the angle between \(\mathbf{v}\) and \(\mathbf{w} .\) Round to the nearest tenth of a degree. $$\mathbf{v}=-2 \mathbf{i}+5 \mathbf{j}, \quad \mathbf{w}=3 \mathbf{i}+6 \mathbf{j}$$

Short Answer

Expert verified
The angle between the vectors \( \mathbf{v} \) and \( \mathbf{w} \) is approximately 45.6 degrees.

Step by step solution

01

Compute the dot product

Dot product of two vectors can be found by multiplying their corresponding components and adding the results together. Therefore, \( \mathbf{v} \cdot \mathbf{w} = -2*3 + 5*6 = 24 \).
02

Compute the magnitudes

The magnitude of a vector is found by squaring each component, summing the results and taking a square root. Hence, \( \| \mathbf{v} \| = \sqrt{(-2)^2 + 5^2} = \sqrt{29} \) and \( \| \mathbf{w} \| = \sqrt{3^2 + 6^2} = \sqrt{45} \).
03

Compute the cosine

Substitute the dot product and the magnitudes into the formula to find the cosine of the angle: \( cos(螛) = \frac{24}{\sqrt{29} * \sqrt{45}} \approx 0.707 \).
04

Determine the angle

To find the angle use inverse cosine function: \( 螛 = arccos(0.707) \). Convert this to degrees and round to the nearest tenth. The result is approximately \( 螛 = 45.6 \) degrees.

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