Find the angle θ between the vectors. $\begin{array} { l }
\mathbf { u } = 6 \mathbf { i } - 1 \mathbf { j } \
\mathbf { v } = 7 \mathbf { i } + 2 \mathbf { j }
\end{array}$
(Round the answer to 2 decimal places. )

A) $25.41 ^ { \circ }$
B) $45.41 ^ { \circ }$
C) $30.41 ^ { \circ }$
D) $35.41 ^ { \circ }$
E) $40.41 ^ { \circ }$

Question

Find the angle θ between the vectors. ​​ $\begin{array} { l } 
\mathbf { u } = 6 \mathbf { i } - 1 \mathbf { j } \
\mathbf { v } = 7 \mathbf { i } + 2 \mathbf { j }
\end{array}$ ​
(Round the answer to 2 decimal places. )
​
A) $25.41 ^ { \circ }$
B) $45.41 ^ { \circ }$
C) $30.41 ^ { \circ }$
D) $35.41 ^ { \circ }$
E) $40.41 ^ { \circ }$

Quizplus · Accepted Answer

The angle θ between two vectors $\mathbf{u}$ and $\mathbf{v}$ can be found using the dot product formula: $\cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|}$. For $\mathbf{u} = 6\mathbf{i} - 1\mathbf{j}$ and $\mathbf{v} = 7\mathbf{i} + 2\mathbf{j}$, the dot product $\mathbf{u} \cdot \mathbf{v} = (6)(7) + (-1)(2) = 42 - 2 = 40$. The magnitudes are $\|\mathbf{u}\| = \sqrt{6^2 + (-1)^2} = \sqrt{37}$ and $\|\mathbf{v}\| = \sqrt{7^2 + 2^2} = \sqrt{53}$. Thus, $\cos(\theta) = \frac{40}{\sqrt{37}\sqrt{53}}$, and solving for θ gives $\theta \approx 25.41^\circ$.

Find the Angle θ Between the Vectors u=6i−1jv=7i+2j\begin{array} { l } \mathbf { u } = 6 \mathbf { i } - 1 \mathbf { j } \\\mathbf { v } = 7 \mathbf { i } + 2 \mathbf { j }\end{array}u=6i−1jv=7i+2j​

Find the Angle θ Between the Vectors $\begin{array} { l } \mathbf { u } = 6 \mathbf { i } - 1 \mathbf { j } \\\mathbf { v } = 7 \mathbf { i } + 2 \mathbf { j }\end{array}$