JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\hat{a}\) and \(\hat{b}\) be two unit vectors such that the angle between them is \(\frac{\pi}{4}\). If \(\theta\) is the angle between the vectors \((\hat{a}+\hat{b})\) and \((\hat{a}+2 \hat{b}+2(\hat{a} \times \hat{b}))\) then the value of \(164 \cos ^{2} \theta\) is equal to.
- A \(90+27 \sqrt{2}\)
- B \(45+18 \sqrt{2}\)
- C \(90+3 \sqrt{2}\)
- D \(54+90 \sqrt{2}\)
Answer & Solution
Correct Answer
(A) \(90+27 \sqrt{2}\)
Step-by-step Solution
Detailed explanation
\(\hat{a}^{\wedge} \hat{b}=\frac{\pi}{4}=\phi\) \(\hat{a} \cdot \hat{b}=|\hat{a}||\hat{b}| \cos \phi\) \(\hat{a} \cdot \hat{b}=\cos \phi=\frac{1}{\sqrt{2}}\)…
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