JEE Mains · Maths · STD 11 - 9. straight line
For \(p\,>\,0\), a vector \(\vec{v}_{2}=2 \hat{i}+(p+1) \hat{j}\) is obtained by rotating the vector \(\vec{v}_{1}=\sqrt{3} p \hat{i}+\hat{j}\) by an angle \(\theta\) about origin in counter clockwise direction. If \(\tan \theta=\frac{(\alpha \sqrt{3}-2)}{4 \sqrt{3}+3}\), then the value of \(\alpha\) is equal to \(....\)
- A \(6\)
- B \(5\)
- C \(4\)
- D \(3\)
Answer & Solution
Correct Answer
(A) \(6\)
Step-by-step Solution
Detailed explanation
\(\left|\vec{V}_{1}\right|=\left|\overrightarrow{\mathrm{V}}_{2}\right|\) \(3 \mathrm{P}^{2}+1=4+(\mathrm{P}+1)^{2}\) \(2 \mathrm{P}^{2}-2 P-4=0\) \(\Rightarrow \mathrm{P}^{2}-\mathrm{P}-2=0\) \(\mathrm{P}=2,-1\) (rejected)…
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