Let a and b be two vectors such that | a | = 14 , | b | = 6 and | a × b | = 48 . Then ( a · b ) 2 is equal to _________ . [2023]

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Solution

Q.
Let $\vec{a}$ and $\vec{b}$ be two vectors 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 _________ . [2023]

Ans.
(36)

Here, $| \vec{a} |$ $= \sqrt{14},$ $| \vec{b} |$ $= \sqrt{6}$ and $| \vec{a} \times \vec{b} |$ $= \sqrt{48}$

${(\vec{a} \cdot \vec{b})}^{2}$ $+$ ${(\vec{a} \times \vec{b})}^{2} =$ $| \vec{a} |^{2}$ $| \vec{b} |^{2}$ $(Lagrange's identity)$

${(\vec{a} \cdot \vec{b})}^{2} + {(\sqrt{48})}^{2} = {(\sqrt{14})}^{2} {(\sqrt{6})}^{2}$ $\Rightarrow {(a \cdot b)}^{2} = 14 \times 6 - 48 = 36$

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