Question

Vector v is shown in the graph.Which are the magnitude and direction of v? Round the answers to the thousandths place.

Answer 1

Step 1:

First, draw and label the vertical and the horizontal units of the vector.

Step 2:

Write the column vector

$v\text{ = }\begin{bmatrix}{4} & \\ {8} & {}\end{bmatrix}$

Step 3:

$\begin{gathered} \text{Magnitude of v = }\sqrt[]{v^2_x+v^2_y} \\ v_x\text{ = 8 } \\ v_y\text{ = }4 \\ \text{Magnitude of v = }\sqrt[]{8^2+4^2} \\ =\text{ }\sqrt[]{64\text{ + 16}} \\ =\text{ }\sqrt[]{80} \\ MagnitudeofV=\text{ 8.944} \end{gathered}$

Step 4:

Find the direction using the formula below

$\begin{gathered} tan\theta\text{ = }\frac{v_y}{v_x} \\ \tan \theta\text{ = }\frac{4}{8} \\ \tan \theta\text{ = 0.5} \\ \theta=tan^{-1}(0.5) \\ \theta\text{ = }26.565 \end{gathered}$

Final answer

$\begin{gathered} \text{Magnitude = 8.944} \\ \text{Direction = 26.565} \\ \mleft\Vert v\mleft\Vert\text{ = 8.944 , }\theta\text{ = 26.565}\mright?\mright? \end{gathered}$