Find the angle $\theta$ between the vectors.Round your answer to two decimal places. $\begin{array} { l }
\mathbf { u } = 12 \mathbf { j } - 28 \mathbf { k } \
\mathbf { v } = 14 \mathbf { i } - 9 \mathbf { k }
\end{array}$
A) $\Theta \approx 58.20 ^ { \circ }$
B) $\Theta \approx 62.20 ^ { \circ }$
C) $\Theta \approx 59.20 ^ { \circ }$
D) $\Theta \approx 61.20 ^ { \circ }$
E) $\Theta \approx 60.20 ^ { \circ }$

Question

Find the angle $\theta$ between the vectors.Round your answer to two decimal places.  $\begin{array} { l } 
\mathbf { u } = 12 \mathbf { j } - 28 \mathbf { k } \
\mathbf { v } = 14 \mathbf { i } - 9 \mathbf { k }
\end{array}$ 
A) $\Theta \approx 58.20 ^ { \circ }$
B) $\Theta \approx 62.20 ^ { \circ }$
C) $\Theta \approx 59.20 ^ { \circ }$
D) $\Theta \approx 61.20 ^ { \circ }$
E) $\Theta \approx 60.20 ^ { \circ }$

Quizplus · Accepted Answer

The angle $\theta$ 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 $\mathbf{u} = 12\mathbf{j} - 28\mathbf{k}$ and $\mathbf{v} = 14\mathbf{i} - 9\mathbf{k}$, the dot product $\mathbf{u} \cdot \mathbf{v} = 0\cdot14 + 12\cdot0 + (-28)\cdot(-9) = 252$. The magnitudes are $\|\mathbf{u}\| = \sqrt{0^2 + 12^2 + (-28)^2} = \sqrt{144 + 784} = \sqrt{928}$ and $\|\mathbf{v}\| = \sqrt{14^2 + 0^2 + (-9)^2} = \sqrt{196 + 81} = \sqrt{277}$. Thus, $\cos(\theta) = \frac{252}{\sqrt{928}\sqrt{277}}$, and solving for $\theta$ gives $\theta \approx 60.20^\circ$.

Find the Angle θ\thetaθ Between the Vectors u=12j−28kv=14i−9k\begin{array} { l } \mathbf { u } = 12 \mathbf { j } - 28 \mathbf { k } \\\mathbf { v } = 14 \mathbf { i } - 9 \mathbf { k }\end{array}u=12j−28kv=14i−9k​

Find the Angle $\theta$ Between the Vectors $\begin{array} { l } \mathbf { u } = 12 \mathbf { j } - 28 \mathbf { k } \\\mathbf { v } = 14 \mathbf { i } - 9 \mathbf { k }\end{array}$