If ab and c are three non-coplanar vectors, then (a+b+c).[(a+b)×(a+c)] equals

Question

If ab and c are three non-coplanar vectors, then (a+b+c).[(a+b)×(a+c)] equals (A) 0 (B) [abc] (C) 2.[cbc] (D) -[abc]

Tardigrade Admin · Accepted Answer

( a + b + c ) ⋅[( a + b ) ×( a + c )] =( a + b + c ) ⋅[ a × a + a × c + b × a + b × c ] = a ⋅( a × c )+ a ⋅ ( b × a )+ a ⋅( b × c ) + b ⋅( mid a × c ) + b ⋅( b × a )+ b ⋅( b × c ) + c ⋅( a × c )+ c ⋅( b × a )+ c ⋅( b × c ) ⇒ [ a b c ]+[ b a c ]+[ c b a ] ⇒[ a b c ]

Q. If $a b$ and $c$ are three non-coplanar vectors, then $(a + b + c) . [(a + b) \times (a + c)]$ equals

0

$[a b c]$

$2. [c b c]$

$- [a b c]$

Q. If ab and c are three non-coplanar vectors, then (a+b+c).[(a+b)×(a+c)] equals

0

[abc]

2.[cbc]

−[abc]

(a+b+c)⋅[(a+b)×(a+c)] =(a+b+c)⋅[a×a+a×c+b×a+b×c] ={a⋅(a×c)+a⋅(b×a)+a⋅(b×c)}+{b⋅(∣a×c) +b⋅(b×a)+b⋅(b×c)}+{c⋅(a×c)+c⋅(b×a)+c⋅(b×c)} ⇒[abc]+[bac]+[cba]⇒[abc]

Q. If $a b$ and $c$ are three non-coplanar vectors, then $(a + b + c) . [(a + b) \times (a + c)]$ equals

$[a b c]$

$2. [c b c]$

$- [a b c]$