JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(a, b , c \in R\) be such that \(a ^{2}+ b ^{2}+ c ^{2}=1\) If \(a \cos \theta=b \cos \left(\theta+\frac{2 \pi}{3}\right)=\operatorname{ccos}\left(\theta+\frac{4 \pi}{3}\right)\) where \(\theta=\frac{\pi}{9},\) then the angle between the vectors \(a \hat{i}+b \hat{j}+c \hat{k}\) and \(b \hat{i}+c \hat{j}+a \hat{k}\) is
- A \(\frac{\pi}{2}\)
- B \(0\)
- C \(\frac{\pi}{9}\)
- D \(\frac{2 \pi}{3}\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi}{2}\)
Step-by-step Solution
Detailed explanation
\(\cos \phi=\frac{\bar{p} \cdot \bar{q}}{|\bar{p}||\bar{q}|}=\frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}}=\frac{\sum a b}{1}\) \(=\operatorname{abc}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)…
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