Find the component form of v if $\| \mathbf { v } \| = 8$ and the angle it makes with the x-axis is 60°.
A) $\langle 4 \sqrt { 3 } , 4 \rangle$
B) $\langle 8,8 \sqrt { 3 } \rangle$
C) $\langle 4,4 \sqrt { 3 } \rangle$
D) $\langle 8 \sqrt { 3 } , 8 \rangle$
E) $\langle 4 \sqrt { 2 } , 4 \sqrt { 2 } \rangle$

Question

Find the component form of v if $\| \mathbf { v } \| = 8$ and the angle it makes with the x-axis is 60°.
A)  $\langle 4 \sqrt { 3 } , 4 \rangle$
B)  $\langle 8,8 \sqrt { 3 } \rangle$
C) $\langle 4,4 \sqrt { 3 } \rangle$
D) $\langle 8 \sqrt { 3 } , 8 \rangle$
E)  $\langle 4 \sqrt { 2 } , 4 \sqrt { 2 } \rangle$

Quizplus · Accepted Answer

The component form of a vector is given by $\langle x,y \rangle$, where $x$ is the horizontal component and $y$ is the vertical component. The magnitude of the vector is given by $\| \mathbf { v } \| = \sqrt { x^2 + y^2 }$, and the angle it makes with the x-axis is given by $\theta = \tan^{-1} \left( \frac { y } { x } \right)$.In this case, we are given that $\| \mathbf { v } \| = 8$ and $\theta = 60^\circ$. Therefore, we have:$$8 = \sqrt { x^2 + y^2 }$$$$60^\circ = \tan^{-1} \left( \frac { y } { x } \right)$$Solving the second equation for $y$, we get:$$y = x \tan 60^\circ = x \sqrt { 3 }$$Substituting this into the first equation, we get:$$8 = \sqrt { x^2 + ( x \sqrt { 3 } )^2 }$$$$8 = \sqrt { x^2 + 3x^2 }$$$$8 = \sqrt { 4x^2 }$$$$8 = 2x$$$$x = 4$$Therefore, $y = x \sqrt { 3 } = 4 \sqrt { 3 }$. Thus, the component form of $\mathbf { v }$ is $\langle 4,4 \sqrt { 3 } \rangle$.

Find the Component Form of V If ∥v∥=8\| \mathbf { v } \| = 8∥v∥=8

Find the Component Form of V If $\| \mathbf { v } \| = 8$