JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\vec{a}\) and \(\vec{b}\) be the vectors along the diagonal of a parallelogram having area \(2 \sqrt{2}\). Let the angle between \(\vec{a}\) and \(\vec{b}\) be acute. \(|\vec{a}|=1\) and \(|\vec{a} . \vec{b}|=|\vec{a} \times \vec{b}| .\) If \(\vec{c}=2 \sqrt{2}(\vec{a} \times \vec{b})-2 \vec{b}\), then an angle between \(\vec{b}\) and \(\vec{c}\) is
- A \(\frac{\pi}{4}\)
- B \(-\frac{\pi}{4}\)
- C \(\frac{5 \pi}{6}\)
- D \(\frac{3 \pi}{4}\)
Answer & Solution
Correct Answer
(D) \(\frac{3 \pi}{4}\)
Step-by-step Solution
Detailed explanation
Area \(=\frac{1}{2}|\overrightarrow{ a } \times \overrightarrow{ b }|=2 \sqrt{2} \Rightarrow|\overrightarrow{ a } \times \overrightarrow{ b }|=4 \sqrt{2}\) \(|\overrightarrow{ a }|=1\) and…
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