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Vector Algebra

KCET 2019

MCQ+1 / -02019

If \(|\mathbf{a}|=16,|\mathbf{b}|=4\), then \(\sqrt{|\mathbf{a} \times \mathbf{b}|^2+|\mathbf{a} \cdot \mathbf{b}|^2}=\)

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