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OA and $\mathbf{B O}$ are two vectors of magnitudes 5 and 6 respectively. If $\angle B O A=60^{\circ}$, then $\mathbf{O A} \cdot \mathbf{O B}$ is equal to
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The correct answer is:
15
We have magnitudes of $\mathbf{O A}$ and $\mathbf{O B}$ are 5 and 6 respectively
$$
\begin{array}{ll}
\text { And } & \angle B O A=60^{\circ} \\
\text { Now, } & \mathbf{O A} \cdot \mathbf{O B}=|\mathbf{O A}| \cdot \mathbf{O B} \mid \cos \theta \\
\Rightarrow & \mathbf{O A} \cdot \mathbf{O B}=5 \cdot 6 \cdot \cos 60^{\circ} \\
\Rightarrow & \mathbf{O A} \cdot \mathbf{O B}=30 \times \frac{1}{2} \\
\Rightarrow & \mathbf{O A} \cdot \mathbf{O B}=15
\end{array}
$$
$$
\begin{array}{ll}
\text { And } & \angle B O A=60^{\circ} \\
\text { Now, } & \mathbf{O A} \cdot \mathbf{O B}=|\mathbf{O A}| \cdot \mathbf{O B} \mid \cos \theta \\
\Rightarrow & \mathbf{O A} \cdot \mathbf{O B}=5 \cdot 6 \cdot \cos 60^{\circ} \\
\Rightarrow & \mathbf{O A} \cdot \mathbf{O B}=30 \times \frac{1}{2} \\
\Rightarrow & \mathbf{O A} \cdot \mathbf{O B}=15
\end{array}
$$
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