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Question

lf ¯¯¯a,¯¯b, ¯¯c are three non-coplanar, non-zero vectors, then (¯¯¯a.¯¯¯a)(¯¯bׯ¯c)+(¯¯¯a.¯¯b)(¯¯cׯ¯¯a)+(¯¯¯a.¯¯c)(¯¯¯aׯ¯b) is equal to

A
[¯¯¯a¯¯b¯¯c]¯¯c
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B
[¯¯b¯¯c ¯¯¯a]¯¯¯a
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C
[¯¯c ¯¯¯a¯¯b]¯¯b
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D
¯¯¯0
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Solution

The correct option is B ¯¯¯0

We have,

(a.a)(b×c)+(a.b)(c×a)+(a.c)(a×b)

Now,

(a.a)(b×c)=[a.a.c]b[a.a.b]c

(a.b)(c×a)=[a.b.c]c[a.b.c]a

(a.c)(a×b)=[a.c.b]a[a.c.a]b

On adding and we get.

(a.a)(b×c)+(a.b)(c×a)+(a.c)(a×b)=[a.a.c]b[a.a.b]c+[a.b.c]c[a.b.c]a+[a.c.b]a[a.c.a]b

(a.a)(b×c)+(a.b)(c×a)+(a.c)(a×b)=0

Hence, this is the answer.

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