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]

Question

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]

Smart Achievers · Accepted Answer

(36)

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

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

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

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]

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Solution

Q. 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]

Ans. (36) Here, |a→|=14,|b→|=6 and |a→×b→|=48 (a→·b→)2+(a→×b→)2=|a→|2|b→|2 (Lagrange's identity) (a→·b→)2+(48)2=(14)2(6)2⇒(a·b)2=14×6-48=36

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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]