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• Notes

$\begin{array}{l}{\text { For the vectors } A \text { and } D \text { in Fig. } E 1.28 \text { , (a) find the magnitude }} \\ {\text { and direction of the vector product } A \times D(b) \text { find the magnitude }} \\ {\text { and direction of } D \times A ?}\end{array}$

$\theta=180^{\circ}-53^{\circ}=127^{\circ}$

$\vec{A} \times \vec{D}=A D \sin \theta \longrightarrow|\vec{A} \times \vec{D}|=(8)(10) \sin 127=63.9 \mathrm{m}^{2}$

$(-Z-\text { direction })(\text { into the page })$

$\vec{D} \times \vec{A}=\vec{A} \times \vec {D} \quad \text { ( } \text { in the opposite Direction) }$

$\begin{array}{l}{\text { Two vectors } A \text { and } B \text { have magnitudes } A=3 \text { and } B=3} \\ {\text { Their vector product is } A X B=-5 k+2 i} \\ {\text { What is the angle between } A \text { and } B ?}\end{array}$

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

$|\vec{A} \times \vec{B}|=\sqrt{(-5)^{2}+(2)^{2}}=\sqrt{25+4}=\sqrt{29}$

$\rightarrow \sin \theta=\frac{|\vec{A} \times \vec{B}|}{A B}=\frac{\sqrt{29}}{(3)(3)}=0.5984$

$\rightarrow \theta=\sin ^{-1}(0.5984)=36.8^{\circ}$