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If \vec{a} and \vec{b} are two vectors such that \left | \vec{a} \right | = 3 and \left | \vec{b} \right | = 2 then \left | \vec{a}*\vec{b} \right |^{2} +(\vec{a}\cdot \vec{b})^{2}equals

Option: 1

25


Option: 2

13


Option: 3

36


Option: 4

48


Answers (1)

best_answer

As we learn

lagrange's identity -

(\vec{a}\times \vec{b})=\left | \vec{a} \right |^2\left | \vec{b} \right |^2-(\vec{a}.\vec{b})^2

- wherein

Here \vec{a} and \vec{b} are two vectors

 

 \left | \vec{a}*\vec{b} \right |^{2} +(\vec{a}\cdot \vec{b})^{2}=\left | \vec{a} \right |^{2}\left | \vec{b} \right |^{2}sin^{2\Theta }+\left | \vec{a} \right |^{2}\left | \vec{b} \right |^{2} cos^{2\Theta }

=\left | \vec{a} \right |^{2}\left | \vec{b} \right |^{2}=9*4=36

 

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