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• Notes

$\begin{array}{l}{\text { Osama stands at the center of a football field while Ali stands at } 20 \mathrm{m} \text { east of Osama. }} \\ {\text { Bader stands at } 30 \mathrm{m} 36.9^{\circ} \text { west of north of Osama. What is the distance between Ali }} \\ {\text { and Bader? }}\end{array}$

$\text { Ali } \rightarrow A$

$\text { Bader } \rightarrow B$

$\vec{A}+\vec{R}=\vec{B}$

$(20) i+\vec{R}=(-30 \sin 36.9) \hat{\imath}+(30 \cos 36.9) \hat{\jmath}$

$\longrightarrow \vec{R}=(-38 \hat{\imath}+24 \hat{\jmath}) m$

$R=\sqrt{(-38)^{2}+(24)^{2}}=45 \mathrm{m}$

OR

$A B=\sqrt{A^{2}+B^{2}-2(A)(B) \cos \theta}$

$=\sqrt{(20)^{2}+(30)^{2}-2(20)(30)\cos 126.9}=45 \mathrm{m}$

$\begin{array}{l}{\text { If } A=2 i+j+2 K, B=-2 i+2 j-K \text { and } C=A+B \text { . }} \\ {\text { Find the angle between } C \text { and the } y \text { -axis. }}\end{array}$

$∵ \vec{C}=\vec{A}+\vec{B}$

$\vec{A}+\vec{B}=2 i+j+2 k+(-2 i+2 j-k)=2 i+j+2 k-2 i+2 j-k$

$=3 \hat j+\hat{k}$

$∴ \vec{C}=3 \hat{j}+\hat{k}$

$∵ Cy=C \cos \theta y \rightarrow \theta y=\cos ^{-1}\left(\frac{C y}{C}\right)$

$\theta y=\cos ^{-1}\left(\frac{3}{\sqrt{3^{2}+1^{2}}}\right) \rightarrow \cos ^{-1}\left(\frac{3}{\sqrt{10}}\right)=18.4$

$\begin{array}{l}{\text { Two vectors } A \text { and } B \text { have magnitudes } A=5.0 \text { and } B=4.0 .} \\ {\text { Their vector product is } A \times B=6 i+8 j . \text { What is the angle }} \\ {\text { between the } A \text { and } B \text { ? }}\end{array}$

$|\vec{A} \times \vec{B}|=A B \sin \theta$

$|\vec{A} \times \vec{B}|=\sqrt{6^{2}+8^{2}}=10$

$\theta=\sin ^{-1}\left(\frac{|\vec{A} \times \vec{B}|}{A B}\right)=\sin ^{-1}\left(\frac{10}{(5)(4)}\right)=30^{\circ}$