Kinematics Question 304
Question: A force $ \vec{F}=3\hat{i}+c\hat{j}+2\hat{k} $ acting on a particle causes a displacement $ \vec{S}=-4\hat{i}+2\hat{j}-3\hat{k} $ in it’s own direction. If the work done it’s 6J, then the value of c will be [DPMT 1997]
Options:
A) $ {{( A^{2}+B^{2}+\frac{AB}{\sqrt{3}} )}^{1/2}} $
B) $ A+B $
C) $ {{(A^{2}+B^{2}+\sqrt{3}AB)}^{1/2}} $
D) $ {{(A^{2}+B^{2}+AB)}^{1/2}} $
Correct Answer: D $ |\overrightarrow{A}\times \overrightarrow{B}|=\sqrt{3}(\overrightarrow{A}.\overrightarrow{B}) $ $ AB\sin \theta =\sqrt{3}AB\cos \theta $ Now $ |\overrightarrow{R}|=|\overrightarrow{A}+\overrightarrow{B}|=\sqrt{A^{2}+B^{2}+2AB\cos \theta } $ $ =\sqrt{A^{2}+B^{2}+2AB( \frac{1}{2} )} $ $ ={{(A^{2}+B^{2}+AB)}^{1/2}} $
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Answer:
Solution:
$ \Rightarrow $ $ \tan \theta =\sqrt{3} $ $ \theta =60{}^\circ $
