Find the angle $\theta$ between the vectors. $\begin{array} { l }
\mathbf { u } = \langle 6,2 \rangle \
\mathbf { v } = \langle 7,0 \rangle
\end{array}$
(Round the answer to 2 decimal places.)

A) $18.43 ^ { \circ }$
B) $33.43 ^ { \circ }$
C) $23.43 ^ { \circ }$
D) $28.43 ^ { \circ }$
E) $38.43 ^ { \circ }$

Question

Find the angle $\theta$  between the vectors.   $\begin{array} { l } 
\mathbf { u } = \langle 6,2 \rangle \
\mathbf { v } = \langle 7,0 \rangle
\end{array}$  
(Round the answer to 2 decimal places.)
 
A) $18.43 ^ { \circ }$
B) $33.43 ^ { \circ }$
C) $23.43 ^ { \circ }$
D) $28.43 ^ { \circ }$
E) $38.43 ^ { \circ }$

Quizplus · Accepted Answer

The dot product of the two vectors is $$\mathbf { u } \cdot \mathbf { v } = ( 6 ) ( 7 ) + ( 2 ) ( 0 ) = 42$$The magnitudes of the two vectors are$$\| \mathbf { u } \| = \sqrt { 6 ^ { 2 } + 2 ^ { 2 } } = \sqrt { 40 }$$$$\| \mathbf { v } \| = \sqrt { 7 ^ { 2 } + 0 ^ { 2 } } = 7$$So, the angle between the two vectors is$$	heta = \cos ^ { - 1 } \left( \frac { \mathbf { u } \cdot \mathbf { v } } { \| \mathbf { u } \| \| \mathbf { v } \| } ight) = \cos ^ { - 1 } \left( \frac { 42 } { \sqrt { 40 } \cdot 7 } ight) \approx 18.43 ^ { \circ }$$

Find the Angle θ\thetaθ Between the Vectors u=⟨6,2⟩v=⟨7,0⟩\begin{array} { l } \mathbf { u } = \langle 6,2 \rangle \\\mathbf { v } = \langle 7,0 \rangle\end{array}u=⟨6,2⟩v=⟨7,0⟩​

Find the Angle $\theta$ Between the Vectors $\begin{array} { l } \mathbf { u } = \langle 6,2 \rangle \\\mathbf { v } = \langle 7,0 \rangle\end{array}$