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Question

If a,b,c are three non coplanar, non zero vectors, then (a.a)b×c+(a.b)c×a+(a.c)a×b is equal to

A
[bca]a
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B
[cab]b
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C
[abc]c
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D
none of these
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Solution

The correct option is A [bca]a
Let a =l(b×c)+m(c×a)+n(a×b) ...(1)
We have to find the values of l,m and n.
Multiply both sides of (1) scalarly by a.
a.a=la.(b×c)+ma.(c×a)+na.(a×b)
a.a=l[abc]+m[aca]+n[aab]=l[abc].
Scalar triple product is zero when two vectors are equal
l=a.a[abc]
Similarly, multiplying both sides of (1) salarly by b and c, we get m=a.b(abc),n=a.c[abc]
a=a.a[abc](b×c)+a.b[abc](c×a)+a.c[abc](a×b)
Similarly we can write the values b and c as b=b.a[abc](b×c)+b.b[abc](c×a)+b.c[abc](a×b)
and c=c.a[abc](b×c)+c.b[abc](c×a)+c.c[abc](a×b).


a=(a.a)b×c[bca]+(a.b)c×a[cab]+(a.c)a×b[abc](a.a)b×c+(a.b)c×a+(a.c)a×b=[bca]a

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