KCET · Maths · Vector Algebra
If \( \vec{a} \) and \( \vec{b} \) are unit vectors then what is the angle between \( \vec{a} \) and \( \vec{b} \) for \( \sqrt{3} \vec{a}-\vec{b} \) to be unit
vector?
- A \( 30^{\circ} \)
- B \( 45^{\circ} \)
- C \( 60^{\circ} \)
- D \( 90^{\circ} \)
Answer & Solution
Correct Answer
(A) \( 30^{\circ} \)
Step-by-step Solution
Detailed explanation
Given that, \(|\vec{a}|=|\vec{b}|=1\)
and and \((\sqrt{3} \vec{a}-\vec{b}) \cdot(\sqrt{3} \vec{a}-\vec{b})=1\)
\(\Rightarrow 3|\vec{a}|^{2}+|\vec{b}|^{2}-2 \sqrt{3} \vec{a} \cdot \vec{b}=1\)
\(\Rightarrow 3+1-2 \sqrt{3} \cos \theta=1\)
\(\Rightarrow \cos \theta=\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \theta=30^{\circ}\)
and and \((\sqrt{3} \vec{a}-\vec{b}) \cdot(\sqrt{3} \vec{a}-\vec{b})=1\)
\(\Rightarrow 3|\vec{a}|^{2}+|\vec{b}|^{2}-2 \sqrt{3} \vec{a} \cdot \vec{b}=1\)
\(\Rightarrow 3+1-2 \sqrt{3} \cos \theta=1\)
\(\Rightarrow \cos \theta=\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \theta=30^{\circ}\)
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