JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\vec{a}, \vec{b}, \vec{c}\) be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle \(\theta\), with the vector \(\vec{a}+\vec{b}+\vec{c}\). Then \(36 \cos ^{2} 2 \theta\) is equal to \(.....\)
- A \(1\)
- B \(2\)
- C \(3\)
- D \(4\)
Answer & Solution
Correct Answer
(D) \(4\)
Step-by-step Solution
Detailed explanation
\(|\vec{a}+\vec{b}+\vec{c}|^{2}=|\vec{a}|^{2}+|\bar{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\overrightarrow{a \cdot c}+\vec{b} \cdot \vec{c})=3\) \(\Rightarrow|\vec{a}+\vec{b}+\vec{c}|=\sqrt{3}\)…
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