Question

Three displacement vectors $\overrightarrow{A}$and$\overrightarrow{B}$shown on figure both have magnitudes of 3.00m. The direction of vector $\overrightarrow{A}$is $\theta =30.0^{\circ }$. Find graphically

a) $A⃗+B⃗$b) $A⃗-B⃗$c) $B⃗-\left(2A\right)⃗$d) $A⃗-\left(2B\right)⃗$(Report all angles counterclockwise from the positive x axis.)

Short answer

$$\begin{gathered} a)A⃗+B⃗=5.2\hspace{0.33em}m,\hspace{0.33em}{60}^0 \\ b)A⃗-B⃗=3.00\hspace{0.33em}m,\hspace{0.33em}{330}^0 \\ c)B⃗-A⃗=3.00\hspace{0.33em}m,\hspace{0.33em}{150}^0 \\ d)A⃗-\left(2B\right)⃗=5.2\hspace{0.33em}m,\hspace{0.33em}{300}^0 \end{gathered}$$

Step-by-step solution

Define vector

A vector is a quantity that has both a direction and a magnitude, and is used to determine the relative location of two points in space.

State given data

Magnitudes of $\overrightarrow{A}$and $\overrightarrow{B}$are 3 m each.

$\theta =30.0^0$

Step 3: (a) Calculate

Magnitude: ${(2.6)}^2+{(3+1.5)}^2=5.2m$

$\begin{array}{l}\theta ={\tan }^1\left[\frac{3+1.5}{2.6}\right]\\ ={60}^0\end{array}$

Hence,$A⃗+B⃗=5.2\hspace{0.33em}m,\hspace{0.33em}{60}^0$