Let →r=(→a×→b)×→c
since, cross product of two vectors is a vector perpendicular to both the vectors
∴→r=→a×(→b×→c) -----(1)
⇒→r⊥→aand→r⊥(→b×→c)
But (→b×→c) is a vector perpendicular to the plane of →b and →c therefore
→r⊥(→b×→c)
⇒→r lies in the plane →b and →c
$ \Rightarrow \overrightarrow r $ is expressible as linear combination of →b and →c
⇒Thereexistscalerx,y
such that
→r=x→b+y→c−−−−−(2)
Now
→r⊥→a
⇒→r.→a=0
⇒(x→b+y→c).→a=0
⇒x(→b.→a)+y(→c.→a)=0
⇒x(→a.→b)+y(→a.→c)=0
⇒x(→a.→b)=−y(→a.→c)
⇒x(→a.→c)=y−(→a.→b)=λ(say)
⇒x=λ(→a.→c)
y=−λ(→a.→b)
Substituting the vector of x and y in (2) we get
→r=λ(→a.→c)→b−λ(→a.→b)→c
⇒→r=λ[(→a.→c)→b−λ(→a.→b)→c]
⇒→a×(→b×→c)=λ[(→a.→c)→b−λ(→a.→b)→c]
This is vector identity and so it is true ∀→a,→b,→c
∴→a×(→b×→c)=(→a.→c)→b−(→a.→b)→c
Hence, the given question is proved