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Question

Mathematics Question on Vector Algebra

Ifa+2b+3c=0\vec{a}+2\vec{b}+3\vec{c}=\vec{0} then a×b+b×c+c×a\vec{a} \times \vec{b}+\vec{b} \times \,\vec{c}+\vec{c} \times \vec{a} =

A

6(b×c)6 (\vec{b} \times \vec{c})

B

2(b×c)2 (\vec{b} \times \vec{c})

C

3(c×a)3 (\vec{c} \times \vec{a})

D

O\overrightarrow{ O }

Answer

6(b×c)6 (\vec{b} \times \vec{c})

Explanation

Solution

Given, a+2b+3c=0...(i)\vec{ a }+2 \vec{ b }+3 \vec{ c }=\vec{ 0 } \,\,\,\,\,\,\,\,\,...(i) Taking cross product with b\vec{ b }, we get a×b+2b×b+3c×b=0×b\vec{ a } \times \vec{ b }+ 2 \vec{ b } \times \vec{ b }+3 \vec{ c } \times \vec{ b }=\vec{ 0 } \times \vec{ b } a×b=3b×c\Rightarrow \vec{ a } \times \vec{ b }=3 \vec{ b } \times \vec{ c } Again taking cross product with c\vec{ c } of E (i), we get a×c+2b×c+3c×c=0×c\vec{ a } \times \vec{ c }+2 \vec{ b } \times \vec{ c }+3 \vec{ c } \times \vec{ c }=\vec{ 0 } \times \vec{ c } c×a=2b×c\Rightarrow \vec{ c } \times \vec{ a }=2 \vec{ b } \times \vec{ c } a×b+b×c+c×a\therefore \vec{ a } \times \vec{ b }+\vec{ b } \times \vec{ c }+\vec{ c } \times \vec{ a } =3b×c+b×c+2b×c= 3 \vec{ b } \times \vec{ c }+\vec{ b } \times \vec{ c }+2 \vec{ b } \times \vec{ c } =6(b×c) = 6(\vec{ b } \times \vec{ c })