KCET · Maths · Three Dimensional Geometry
If \( \vec{a} \& \vec{b} \) are unit vectors, then the angle between \( \vec{a} \) and \( \vec{b} \) for \( \sqrt{3} \vec{a}-\vec{b} \) to be unit vector is
- 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
Since \(\vec{a}\) and \(\vec{b}\) are unit vectors. Then,
\(|\vec{a}|=|\vec{b}|=1\)
also, \(|\sqrt{3} \vec{a}-\vec{b}|=1 \rightarrow(1)\)
which is also unit vector.
Squaring both the sides of Eq. (1), we get
\(3|\vec{a}|^{2}+|\vec{b}| \vec{r}-2 \sqrt{3}|\vec{a}| \cdot|\vec{b}|=(1)^{2}\)
\(3 \cdot 1+1-2 \sqrt{3} \cdot 1 \cdot 1 \cdot \cos \theta=1\)
\(\Rightarrow 2 \sqrt{3} \cos \theta=3\)
\(\Rightarrow \cos \theta=\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \theta=30^{\circ}\)
\(|\vec{a}|=|\vec{b}|=1\)
also, \(|\sqrt{3} \vec{a}-\vec{b}|=1 \rightarrow(1)\)
which is also unit vector.
Squaring both the sides of Eq. (1), we get
\(3|\vec{a}|^{2}+|\vec{b}| \vec{r}-2 \sqrt{3}|\vec{a}| \cdot|\vec{b}|=(1)^{2}\)
\(3 \cdot 1+1-2 \sqrt{3} \cdot 1 \cdot 1 \cdot \cos \theta=1\)
\(\Rightarrow 2 \sqrt{3} \cos \theta=3\)
\(\Rightarrow \cos \theta=\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \theta=30^{\circ}\)
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