Kinematics Question 428

Question: Two forces $ {{\vec{F}}_1}=10\hat{i}-\hat{j}-15\hat{k} $ and $ {{\vec{F}}_2}=10\hat{i}-\hat{j}-15\hat{k} $ act on a single point. The angle between $ {{\vec{F}}_1} $ and $ {{\vec{F}}_2} $ it’s nearly

Options:

A) $ 30^{o} $

B) $ 45^{o} $

C) $ 60^{o} $

D) $ 90^{o} $

Show Answer

Answer:

Correct Answer: B

Solution:

[b] $ \cos \theta =\frac{{{{\vec{F}}}_1}.{{{\vec{F}}}_2}}{|{{{\vec{F}}}_1}||{{{\vec{F}}}_2}|} $ $ =\frac{(5\hat{i}+10\hat{j}-20\hat{k})\cdot (10\hat{i}-5\hat{j}-15\hat{k})}{\sqrt{25+100+400}\cdot \sqrt{100+25+225}} $ $ =\frac{50-50+300}{\sqrt{525}\cdot \sqrt{350}} $

$ \Rightarrow $ $ \cos \theta =\frac{1}{\sqrt{2}} $

$ \therefore $ $ \theta =45^{o} $



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