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Angle between the vectors \vec{2i}+\vec{6j}+\vec{3k} \:and \:\vec{12i}-\vec{4j}+\vec{3k} \:is

  • Option 1)

    cos^-(\frac{1}{10})

  • Option 2)

    cos^-(\frac{9}{11})

  • Option 3)

    cos^-(\frac{9}{91})

  • Option 4)

    cos^-(\frac{1}{9})

 

Answers (1)

best_answer

As we learnt

Angle between vector a and vector b -

\cos \Theta =\frac{\vec{a}.\vec{b}}{\left | \vec{a} \right |\left | \vec{b} \right |}

- wherein

Here 0\leq \Theta \leq \pi??????

 

 \cos \Theta =\frac{\vec{a.\vec{b}}}{\left | \vec{a} \right |\left | \vec{b} \right |}

= \frac{\left ( 2i+6j+3\vec{k} \right )}{\sqrt{2^{2}+6^{2}+3^{2}}}.\frac{\left ( 12i-4j+3k \right )}{\sqrt{12^{2}+4^{2}+3^{2}}}

=\frac{24-24+9}{\sqrt{49}\times \sqrt{169}}

=\frac{9}{7\times 13}=\frac{9}{7\times 13}= \frac{9}{91}

\Theta =\cos ^{-1}\left ( \frac{9}{91} \right )


Option 1)

cos^-(\frac{1}{10})

Incorrect Option

Option 2)

cos^-(\frac{9}{11})

Incorrect Option

Option 3)

cos^-(\frac{9}{91})

Correct Option

Option 4)

cos^-(\frac{1}{9})

Incorrect Option

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Plabita

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