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Question Number 126414 by sdfg last updated on 20/Dec/20

Answered by physicstutes last updated on 20/Dec/20

$${\mathrm{Q}}_2.\mathrm{normally}\mathrm{we}\mathrm{should}\mathrm{be}\mathrm{computing}\mid \mathbf{A}\times \mathbf{B}\mid$$$$\mathrm{the}\mathrm{unit}\mathrm{vector}\mathbf{n}\mathrm{could}\mathrm{be}\mathrm{any}\mathrm{unit}\mathrm{vector}\mathrm{they}\mathrm{are}\mathrm{millions}\mathrm{of}\mathrm{them}\mathrm{so}$$$$\mathrm{i}\mathrm{guess}\mathrm{we}\mathrm{can}\mathrm{not}\mathrm{get}\mathrm{a}\mathrm{specific}\mathbf{A}\times \mathbf{B}\mathrm{vector}\mathrm{but}\mathrm{we}\mathrm{can}\mathrm{get}\mid \mathbf{A}\times \mathbf{B}\mid \mathrm{thus}:$$$$\mid \mathbf{A}\times \mathbf{B}\mid =\mid \mathbf{A}\mid \mid \mathbf{B}\mid \sin \theta \mid \stackrel{\wedge }{\mathbf{n}}\mid$$$$=\left(\frac{1}{\sqrt{3}}\right)\left(4\right)\sin 60,\mathrm{since}\mid \mathbf{n}\mid =1$$$$=\left(\frac{1}{\cancel{\sqrt{3}}}\right)\left(4\right)\left(\frac{\cancel{\sqrt{3}}}{2}\right)=2$$$${\mathrm{Q}}_3.0\mathrm{since}\mathbf{A},\mathbf{B},\mathbf{C}\mathrm{are}\mathrm{perpendicular}\mathrm{irrespectively}.$$

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