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UPSC NDA Math Model Paper 4

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Question : 54 of 120

Marks: +1, -0

If

|\vec{a}|=3,|\vec{b}|=4

|\vec{a} \times \vec{b}|=6

, then find the value of

\vec{a} \cdot \vec{b}

$4\sqrt{3}$
6
$6\sqrt{3}$
12

Solution:

Given:

|\vec{a}|=3,|\vec{b}|=4

and

|\vec{a} \times \vec{b}|=6

As we know,

\vec{a} \times \vec{b}=|\vec{a}| \times |\vec{b}| \times \sin \theta \times \hat{n}

|\vec{a} \times \vec{b}|=|\vec{a}| \times |\vec{b}| \times |\sin \theta| \times |\hat{n}|

=|\vec{a}| \times |\vec{+b}| \times \sin \theta

(∵ Magnitude of a unit vector is one)

\Rightarrow \sin \theta=\frac{\vec{|a \times \vec{b}|}}{\vec{|a| \times |\vec{b}|}}=\frac{6}{12}=\frac{1}{2}

\therefore \theta=30^{\circ}

Now,

\cos \theta=\cos 30=\frac{\sqrt{3}}{2}

As we know that, If

\vec{a}

and

\vec{b}

are two vectors, then the scalar product between the given by:

\vec{a} \cdot \vec{b}=|\vec{a}| \times |\vec{b}| \times \cos \theta

\therefore \vec{a} \cdot \vec{b}=3 \times 4 \times \frac{\sqrt{3}}{2}=6\sqrt{3}

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