Kinematics Question 599
Question: If $ |\vec{A}\times \vec{B}|=\sqrt{3}\vec{A}.\vec{B}, $ then the value of $ |\vec{A}+\vec{B}| $ it’s:
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}} $
Show Answer
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}}. $
