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
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.