1- proff that a cross (across(acrossb)= (a.a)(b cross a ), 2- if a cross b= c cross d and a cross c = b cross d prove that a-d is parallel to b-c prove in depth also see under what i m asked
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1) Here in this question you have not mentioned that a is perpendicular to b. Without this condition you will not be able to prove.
So lets assume that a is perpendicular to b
LHS = a x (a x (a x b)
= a x {(a.b)a - (a.a)b} [ as a x (b x c)] = (a.c).b - (a.b)c]
= a x {-(a.a) b} [ as a.b = 0 , they are perpendicular to each other]
= a x {-|a|2 b} [ as a.a = |a|2 ]
= - |a|2 (a x b)
= |a|2 ( b x a) = RHS [ as -(a x b) = (b x a)]
Hope you have got it.
2) For a -d parallel to b-c , there cross product should be equals to zero.
So ( a-d ) x ( b-c )
= a x b - a x c - d x b + d x c [expanding it]
= a x b - a x c + b x d - c x d [ using the property A X B = - ( B X A) ]
= a x b - c x d + b x d - a x c [ using the property A X B = - ( B X A) ]
= 0 + 0 [as a x b = c x d and b x d = a x c]
= 0
Hence the cross product is zero, so they are parallel to each other.
So lets assume that a is perpendicular to b
LHS = a x (a x (a x b)
= a x {(a.b)a - (a.a)b} [ as a x (b x c)] = (a.c).b - (a.b)c]
= a x {-(a.a) b} [ as a.b = 0 , they are perpendicular to each other]
= a x {-|a|2 b} [ as a.a = |a|2 ]
= - |a|2 (a x b)
= |a|2 ( b x a) = RHS [ as -(a x b) = (b x a)]
Hope you have got it.
2) For a -d parallel to b-c , there cross product should be equals to zero.
So ( a-d ) x ( b-c )
= a x b - a x c - d x b + d x c [expanding it]
= a x b - a x c + b x d - c x d [ using the property A X B = - ( B X A) ]
= a x b - c x d + b x d - a x c [ using the property A X B = - ( B X A) ]
= 0 + 0 [as a x b = c x d and b x d = a x c]
= 0
Hence the cross product is zero, so they are parallel to each other.