Problem 32 - Algebra and Trigonometry Real Mathematics, Real People

Question

Graph the vectors and find the degree measure of the angle between the vectors. $$\begin{aligned} &\mathbf{u}=6 \mathbf{i}-2 \mathbf{j}\\\ &\mathbf{v}=8 \mathbf{i}-5 \mathbf{j} \end{aligned}$$

Step-by-Step Solution

Answer

The degree measure of the angle between the vectors \(\mathbf{u}\) and \(\mathbf{v}\) is\(\arccos(\frac{58}{\sqrt{40} * \sqrt{89}})\)

Step 1: Plot the vectors

First, plot the vectors \(\mathbf{u}=6\mathbf{i}-2\mathbf{j}\) and \(\mathbf{v}=8\mathbf{i}-5\mathbf{j}\) on a coordinate plane. Both vectors start from the origin (0, 0). Vector \(\mathbf{u}\) ends at (6,-2) and vector \(\mathbf{v}\) ends at (8,-5)

Step 2: Calculate the dot product of the vectors

Next, calculate the dot product of the vectors. The dot product of two vectors \(\mathbf{u}\) and \(\mathbf{v}\) is calculated by multiplying corresponding components of both vectors and adding them together. \( \mathbf{u} \cdot \mathbf{v} = (6*8) + (-2*-5) = 48 + 10 = 58\)

Step 3: Determine the magnitudes of the vectors

The magnitude of a vector is calculated as the square root of the sum of the squares of its components. So for vector \(\mathbf{u}\), it is \(\sqrt{(6^2)+(-2^2)} = \sqrt{36+4} = \sqrt{40}\), and for vector \(\mathbf{v}\), it is \(\sqrt{(8^2)+(-5^2)} = \sqrt{64+25} = \sqrt{89}\)

Step 4: Calculate the cosine of the angle between the vectors

The cosine of the angle \(\theta\) between vectors \(\mathbf{u}\) and \(\mathbf{v}\) can be found using the formula \(\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \| \mathbf{v} \|}\). Substituting the calculated values, we get \(\cos \theta = \frac{58}{\sqrt{40} * \sqrt{89}}\)

Step 5: Calculate the angle between the vectors

Finally, the angle \(\theta\) between vectors \(\mathbf{u}\) and \(\mathbf{v}\) can be found taking the arccosine of the cosine value calculated in the previous step. Therefore, \(\theta = \arccos(\frac{58}{\sqrt{40} * \sqrt{89}})\)