Question 3.10: Adding Vectors Three vectors A = 2i - 3j , B = 4i - 2j, and ......

Adding Vectors

Three vectors \overrightarrow{ A }=2 \hat{ \imath }-3 \hat{ \jmath }, \overrightarrow{ B }=4 \hat{ \imath }-2 \hat{\jmath} \text {, and } \overrightarrow{ C }=-5 \hat{ \imath }-7 \hat{ \jmath } are dimensionless. Find vector \overrightarrow{ R }=\overrightarrow{ A }+\overrightarrow{ B }+\overrightarrow{ C } in component form. Also give the magnitude and direction of  \overrightarrow{ R }.

Step-by-Step
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INTERPRET and ANTICIPATE
The best way to anticipate the result of adding three vectors is to sketch the vectors and add them geometrically (Fig.3.40). The three vectors are drawn head to tail. The resultant vector \vec{R} is drawn from the tail of \vec{A} to the head of \vec{C}.

From our sketch, we see that the x component of \vec{R} is positive and small and that the y component is negative and large. We see that angle \theta is between 270° and 360°.

SOLVE
According to Equation 3.18,

\overrightarrow{ R }=\left(A_x+B_x\right) \hat{\imath}+\left(A_y+B_y\right) \hat{\jmath}\quad \quad (3.18)

to find the vector sum, we add the x components together and the y components together.

\begin{aligned}& R_x=A_x+B_x+C_x=2+4-5=1 \\& R_y=A_y+B_y+C_y=-3-2-7=-12\end{aligned}

Writing \vec{R} in component form means adding the vector components of \vec{R}.

\begin{aligned}& \overrightarrow{ R }=R_x \hat{ \imath }+R_y \hat{ \jmath } \\& \overrightarrow{ R }=1 \hat{ \imath }-12 \hat{ \jmath }\end{aligned}

CHECK and THINK
As expected, the x component is positive and small, and the y component is negative and large.

SOLVE
The magnitude of the resultant comes from modifying Equation 3.12.

\begin{aligned}& R=\sqrt{R_x^2+R_y^2}=\sqrt{1^2+(-12)^2} \quad \quad (3.12)\\& R=12 \text { (two significant figures) }\end{aligned}

The direction comes from Equation 3.14.

\alpha=\tan ^{-1} \frac{R_y}{R_x}=\tan ^{-1} \frac{-12}{1}=-85^{\circ}\quad \quad (3.14)

Our calculator is giving us angle \alpha instead of angle \theta, so we must add 360° to find \theta. That is one reason our sketch is so important.

\begin{aligned}& \theta-\alpha=360^{\circ} \\& \theta=\alpha+360^{\circ}=-85^{\circ}+360^{\circ} \\& \theta=275^{\circ}\end{aligned}

CHECK and THINK
Because the y component is 12 times greater than the x component, the x component has little effect on the magnitude. From our sketch, we would estimate that the magnitude is near 12. Also, we can see that \theta is close to 270°, which matches our algebraic results.

3.40

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