Find the component form of v given its magnitude and the angle it makes with the positive x-axis.
$\begin{array}{lr}\text { Magnitude}&\text { Angle }\\|\mathbf{v}\|=6&45^o
\end{array}$

A) $\langle - 3 \sqrt { 2 } , 3 \sqrt { 2 } \rangle$
B) $\langle - 3 \sqrt { 2 } , - 3 \sqrt { 2 } \rangle$
C) $\langle 3,3 \rangle$
D) $\langle 3 \sqrt { 2 } , 3 \sqrt { 2 } \rangle$
E) $\langle 3 \sqrt { 2 } , - 3 \sqrt { 2 } \rangle$

Question

Find the component form of v given its magnitude and the angle it makes with the positive x-axis.  
$\begin{array}{lr}\text { Magnitude}&\text { Angle }\\|\mathbf{v}\|=6&45^o
\end{array}$

A)  $\langle - 3 \sqrt { 2 } , 3 \sqrt { 2 } \rangle$
B)  $\langle - 3 \sqrt { 2 } , - 3 \sqrt { 2 } \rangle$
C)  $\langle 3,3 \rangle$
D)  $\langle 3 \sqrt { 2 } , 3 \sqrt { 2 } \rangle$
E)  $\langle 3 \sqrt { 2 } , - 3 \sqrt { 2 } \rangle$

Quizplus · Accepted Answer

The component form of a vector given its magnitude $r$ and angle $\theta$ with the positive x-axis can be found using the formulas $x = r\cos(\theta)$ and $y = r\sin(\theta)$. For a magnitude of 6 and an angle of $45^\circ$, both $x$ and $y$ components are $6\cos(45^\circ) = 6\sin(45^\circ) = 6 \times \frac{\sqrt{2}}{2} = 3\sqrt{2}$, resulting in the vector $\langle 3\sqrt{2}, 3\sqrt{2} \rangle$.

Find the Component Form of V Given Its Magnitude and the Angle