MHT CET · Maths · Vector Algebra
If \(\bar{a}\) and \(\bar{b}\) are two vectors such that \(|\vec{a}|=|\vec{b}|=\sqrt{2}\) with \(\vec{a} \cdot \vec{b}=-1\), then the angle between \(\bar{a}\) and \(\bar{b}\) is
- A \(\frac{3 \pi}{4}\)
- B \(\frac{5 \pi}{6}\)
- C \(\frac{5 \pi}{9}\)
- D \(\frac{2 \pi}{3}\)
Answer & Solution
Correct Answer
(D) \(\frac{2 \pi}{3}\)
Step-by-step Solution
Detailed explanation
\(\theta=\cos ^{-1}\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\right)=\cos ^{-1}\left(\frac{-1}{\sqrt{2} \sqrt{2}}\right)=\cos ^{-1}\left(\frac{-1}{2}\right)=\frac{2 \pi}{3}\)
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