If 2a→+3b→+4c→=0→⇒then a→×b→+b→×c→+c→×a→=

2a→+3b→+4c→=0¯⇒(2a→+3b→+4c→)×a→=0→⇒4c→×a→=3a→×b→→(1)(2a→+3b→+4c→)×b→=0→⇒2a→×b=4b→×c→a→×b+b→×c→+c→+a→=43(c→×a→)+12⋅43c→×a→+c→×a→=3c→×a→

If 2a→+3b→+4c→=0→⇒then a→×b→+b→×c→+c→×a→=

2a→+3b→+4c→=0¯⇒(2a→+3b→+4c→)×a→=0→⇒4c→×a→=3a→×b→→(1)(2a→+3b→+4c→)×b→=0→⇒2a→×b=4b→×c→a→×b+b→×c→+c→+a→=43(c→×a→)+12⋅43c→×a→+c→×a→=3c→×a→

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$If 2 \vec{a} + 3 \vec{b} + 4 \vec{c} = \vec{0} \Rightarrow$ $t h e n \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} =$

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$3 \vec{c} \times \vec{a}$

$3 \vec{b} \times \vec{c}$

$3 \vec{a} \times \vec{b}$

$\vec{a}$

answer is D.

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Detailed Solution

$2 \vec{a} + 3 \vec{b} + 4 \vec{c} = \bar{0} \Rightarrow (2 \vec{a} + 3 \vec{b} + 4 \vec{c}) \times \vec{a} = \vec{0} \Rightarrow 4 \vec{c} \times \vec{a} = 3 \vec{a} \times \vec{b} \to (1)$

$(2 \vec{a} + 3 \vec{b} + 4 \vec{c}) \times \vec{b} = \vec{0} \Rightarrow 2 \vec{a} \times b = 4 \vec{b} \times \vec{c} \vec{a} \times b + \vec{b} \times \vec{c} + \vec{c} + \vec{a} = \frac{4}{3} (\vec{c} \times \vec{a}) + \frac{1}{2} \cdot \frac{4}{3} \vec{c} \times \vec{a} + \vec{c} \times \vec{a} = 3$