KCET · Maths · Vector Algebra
If \( 2 \vec{a} \cdot \vec{b}=|\vec{a}| \cdot|\vec{b}| \) then the angle between \( \vec{a} \& \vec{b} \) is
- A \( 30^{\circ} \)
- B \( 0^{\circ} \)
- C \( 90^{\circ} \)
- D \( 60^{\circ} \)
Answer & Solution
Correct Answer
(D) \( 60^{\circ} \)
Step-by-step Solution
Detailed explanation
Given that, \( 2 \vec{a} \cdot \vec{b}=|\vec{a}| \cdot|\vec{b}| \)
As we know, \( \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \)
So, \( 2|\vec{a}||\vec{b}| \cos \theta=|\vec{a}||\vec{b}| \)
\( \Rightarrow 2 \cos \theta=1 \)
\( \Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=60^{\circ} \)
As we know, \( \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \)
So, \( 2|\vec{a}||\vec{b}| \cos \theta=|\vec{a}||\vec{b}| \)
\( \Rightarrow 2 \cos \theta=1 \)
\( \Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=60^{\circ} \)
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