Physics Two Dimensional Motion Question 147
Question: If $ |\vec{A}\times \vec{B}|,=\sqrt{3},\vec{A},.,\vec{B},, $ then the value of $ |\vec{A}+\vec{B}| $ is:
Options:
A) $ {{( {{\text{A}}^{\text{2}}}\text{+}{{\text{B}}^{\text{2}}}\text{+}\frac{\text{AB}}{\sqrt{\text{3}}} )}^{1/2}} $ B) $ \text{A+B} $ C) $ {{( {{\text{A}}^{\text{2}}}\text{+}{{\text{B}}^{\text{2}}}\text{+}\sqrt{\text{3}}\text{AB} )}^{1/2}} $ D) $ {{( {{\text{A}}^{\text{2}}}\text{+}{{\text{B}}^{\text{2}}}\text{+AB} )}^{1/2}} $
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Answer:
Correct Answer: D
Solution:
[d] $ ! \vec{A}!!\times!!,\vec{B}!!|!!\text{=},\sqrt{\text{3}}\vec{A} \vec{B} $
$ \text{orAB sin}\theta ,,\text{=},\sqrt{3}\text{ AB cos}\theta $
$ \
Therefore ,,,,\tan \theta ,,\text{=},\sqrt{3}\text{,or}\theta ,\text{=},,\text{60}{}^\circ $ Thus $ !!|!!\vec{A}+\vec{B}!!|!!\text{ =}\sqrt{{{\text{A}}^{\text{2}}}\text{+}{{\text{B}}^{\text{2}}}\text{+2},\text{AB},\text{cos},\text{60 }!!{}^\circ!!} $
$ =\sqrt{{{\text{A}}^{\text{2}}}\text{+}{{\text{B}}^{\text{2}}}\text{+2AB}}. $
BETA
