Let a→, b→, c→ be three vectors of magnitude 3, 4, 5 respectively, satisfying a→b→c→=60. If a→+2b→+3c→·a→×c→×b→+b→=λ, then λ is equal to

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MARKS Content Team · Accepted Answer

Correct answer: (B) 32. Given, a→b→c→=3·4·5=60 Also, a→b→c→=60⇒a→b→c→=a→b→c→ ⇒a→,b→,c→ are mutually perpendicular vectors Also, a→×c→×b→+b→=a→·b→c→-c→·b→a→+b→ =b→ Hence, a→+2b→+3c→·b→=2b→^2=32

Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors of magnitude $3, 4, 5$ respectively, satisfying $|[\vec{a} \vec{b} \vec{c}]| = 60$ . If $(\vec{a} + 2 \vec{b} + 3 \vec{c}) \cdot ((\vec{a} \times \vec{c}) \times \vec{b} + \vec{b}) = λ$ , then $λ$ is equal to

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