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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}

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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