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Chapter 6: Problem 20

Find the angle between \(\mathbf{v}\) and \(\mathbf{w} .\) Round to the nearest tenth of a degree. $$\mathbf{v}=\mathbf{i}+2 \mathbf{j}, \quad \mathbf{w}=4 \mathbf{i}-3 \mathbf{j}$$

Short Answer

Expert verified
The angle between vectors \(\mathbf{v}\) and \(\mathbf{w}\) is approximately 101.5 degrees.

Step by step solution

01

Write down the given vectors

The given vectors are \(\mathbf{v} = \mathbf{i} + 2\mathbf{j}\) and \(\mathbf{w} = 4\mathbf{i} - 3\mathbf{j}\)
02

Calculate the dot product

The dot product is calculated as \(\mathbf{v}.\mathbf{w} = \mathbf{i}.\mathbf{i} + 2\mathbf{j}. -3\mathbf{j} = 4 - 6 = -2 \)
03

Calculate the magnitudes of the vectors

The magnitude of vector \(\mathbf{v}\) is \(\|\mathbf{v}\| = \sqrt{1^2 + 2^2} = \sqrt{5}\) and the magnitude of vector \(\mathbf{w}\) is \(\|\mathbf{w}\| = \sqrt{4^2 + (-3)^2} = \sqrt{25} = 5\)
04

Find the angle using dot product formula

The formula for a dot product is \(\mathbf{v}.\mathbf{w} = \|\mathbf{v}\|\|\mathbf{w}\|\cos(\theta)\), where \(\theta\) is the angle between the vectors. Solving for \(\theta\), we plug in the values from steps 2 and 3 and obtain \(\theta = \cos^{-1}(-2/(\sqrt{5}*5))\)
05

Calculate and round off the answer

Calculating the above expression in degrees and rounding off to the nearest tenth gives the final answer

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