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$$
\begin{array}{l}{\text { Given two vectors } A=4 i+7 j \text { and } B=5 i-2 j} \\ {\text { (a) find the magnitude of each vector; (b) write an expression for the vector difference } A-B \text { using unit vectors; }} \\ {\text { (c) find the magnitude and direction of the vector difference } A-B} \\ {\text { (d) In a vector diagram show } A, B \text { and } A-B \text { and also show that your }} \\ {\text { diagram agrees qualitatively with your answer in part (c). }}\end{array}
$$

$$
\vec{A}= 4 i+7 j
$$

$$
A_ x=+4
$$

$$
A _y=+7
$$

$$
\rightarrow A=\sqrt{A_ x^{2}+A _y^{2}}
$$

$$
=\sqrt{4^{2}+7^{2}}
$$

$$
=8.06
$$

$$\vec{B}=5 i-2 j$$

$$
B_ x=+5
$$

$$
B_ y=-2
$$

$$
\rightarrow B=\sqrt{B_ x^{2}+B_ y^{2}}
$$

$$
=\sqrt{5^{2}+(-2)^{2}}
$$

$$
=5.39
$$

$$
A-B=4\hat{i}+7 \hat{j}-\left(5{i}-2\hat j\right)=4\hat i+7\hat j-5 i+2\hat j
$$

$$
\rightarrow A-B=-1 \hat{i}+9 \hat{j}
$$

$$
A-B=R [\begin{aligned} R_{x} &=(A-B)_{x} \\  R_{y} &=(A-B) y \end{aligned}
$$

$$
R_{x}=-1, \quad R_{y}=+9
$$

$$
\rightarrow R=\sqrt{R_ x^{2}+R_ y^{2}}=\sqrt{(-1)^{2}+(9)^{2}}=9.06
$$

$$
\rightarrow \tan \theta=\frac{R_ y}{R_{x}}=\frac{9}{-1}=-9
$$

$$
\rightarrow \theta=-83.6^{\circ}+180^{\circ}=96.3^{\circ}
$$

$$
\begin{array}{l}{\text { (a) Write each vector in Fig. } E 1.43 \text { in terms of the unit vectors iand j }} \\ {\text { (b) Use unit vectors to express the vector } C \text { Where } C=3 A-4 B} \\ {\text { (c) Find the magnitude and direction of } C}\end{array}
$$

$$
A_{x}=A \cos \theta\quad, \quad A _y=A \sin \theta
$$

$$
\vec{A}=(A \cos \theta) \hat{\imath}+(A \sin \theta)\hat j
$$

$$
=3.6 \cos 70^{\circ} \hat{\imath}+3.6 \sin 70^{\circ} \hat{\jmath}=(1.23 \mathrm{m}) \hat{\imath}+(3.38) \hat{\mathrm{j}}
$$

$$
B_ x=B \cos \theta, \quad B_ y=B \sin \theta
$$

$$
\vec{B}=(B \cos \theta) \hat{\imath}+(B \sin \theta) \hat{\jmath}
$$

$$
=(-2.4) \cos 30\hat i-2.4 \sin 30\hat j=(-2.08 \mathrm{m}) \hat{i}+(-1.2 \mathrm{m}) \hat{\mathrm{j}}
$$

$$
\vec{C}=3 \vec{A}-4 \vec{B}
$$

$$
\vec{A}=(1.23) \hat{i}+(3.38) \hat{j}
$$

$$
\rightarrow 3 \vec{A}=3(1.23) \hat{\imath}+3(3.38) \hat{\jmath}
$$

$$
=3.69 \hat{\imath}+10.14 \hat j
$$

$$
\vec{B}=(-2.08) \hat{\imath}+(-1.2) \hat{\jmath}
$$

$$
{\rightarrow 4 \vec{B}=4(-2.08) \hat{i}+4(-1.2) \hat{j}}
$$

$$
=-8.32 \hat{\imath}-4.8 \hat{\jmath}
$$

$$
\rightarrow \quad C=3 A-4 B=(3.69 \hat{\imath})+(10.14) \hat{\jmath}+(8.32) \hat{\imath}
$$

$$
=12.01 \hat{i}+14.94 \hat j
$$

$$
C=\sqrt{c_{x}^{2}+c_{y}^{2}}
$$

$$
=\sqrt{(12.01)^{2}+(14.94)^{2}}=19.17 m
$$

$$
\theta=\tan ^{-1}\left(\frac{14.94}{12.01}\right)=51.2^{\circ}
$$

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