JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\vec{a}\) and \(\vec{b}\) be two vector such that \(|\vec{a}|=\sqrt{14}\), \(|\vec{b}|=\sqrt{6}\) and \(|\vec{a} \times \vec{b}|=\sqrt{48}\). Then \((\vec{a} \cdot \vec{b})^2\) is equal to \(...........\).
- A \(36\)
- B \(35\)
- C \(37\)
- D \(39\)
Answer & Solution
Correct Answer
(A) \(36\)
Step-by-step Solution
Detailed explanation
\(|\vec{a}|=\sqrt{14},|\vec{b}|=\sqrt{6} \quad|\vec{a} \times \vec{b}|=\sqrt{48}\) \(|\vec{a} \times \vec{b}|^2+|\vec{a} \cdot \vec{b}|^2=|\vec{a}|^2 \times|\vec{b}|^2\) \(\Rightarrow(\vec{a} \cdot \vec{b})^2=84-48=36\)
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