KCET · Maths · Vector Algebra
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
- A 0
- B 15
- C \(-15\)
- D \(15 \sqrt{3}\)
Answer & Solution
Correct Answer
(B) 15
Step-by-step Solution
Detailed explanation
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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