JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\vec{a}\) and \(\vec{b}\) be two vectors such that \(|\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+2|\vec{b}|^{2}, \vec{a} \cdot \vec{b}=3 \quad\) and \(\quad|\vec{a} \times \vec{b}|^{2}=75\).Then \(|\vec{a}|^{2}\) is equal to \(.......\)
- A \(14\)
- B \(13\)
- C \(12\)
- D \(11\)
Answer & Solution
Correct Answer
(A) \(14\)
Step-by-step Solution
Detailed explanation
\(|\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+2|\vec{b}|^{2} ; \vec{a} \cdot \vec{b}=3\) As \(|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=|\vec{a}|^{2}+2|\vec{b}|^{2}\) \(|\vec{b}|^{2}=2 \vec{a} \cdot \vec{b}=6\) \(|\vec{a} \times \vec{b}|^{2}=75\)…
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