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Physics · SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
हिन्दीIf \( |\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = 144 \) and \( |\vec{a}| = 3 \), then the magnitude of vector \( \vec{b} \) is:
Correct answer · Option 4
\( 4 \)
Explanation
Lagrange's identity gives \( |\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 \). Thus \( 144 = 3^2 |\vec{b}|^2 = 9 |\vec{b}|^2 \implies |\vec{b}|^2 = 16 \implies |\vec{b}| = 4 \).
