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Introduction to vector algebra - class-XI

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If $(\vec{a}\times\vec{b})^{2}+(\vec{a}.\vec{b})^{2}=144$ and $|\vec{a}|=4,\ |\vec{b}|=$

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$3$

💡 Explanation:

Using the identity |a x b|^2 + (a . b)^2 = |a|^2 * |b|^2, we have 144 = 4^2 * |b|^2. Thus, 144 = 16 * |b|^2, which means |b|^2 = 9, so |b| = 3.