If a , b , c , d are four distinct vectors satisfying the conditions a × b = c × d and a × c = b × d , then prove that a ⋅ b + c ⋅ d ≠ a ⋅ c + b ⋅ d

Question

If a , b , c , d are four distinct vectors satisfying the conditions a × b = c × d and a × c = b × d , then prove that a ⋅ b + c ⋅ d ≠ a ⋅ c + b ⋅ d  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Given, a × b = c × d and a × c = b × d ⇒ a × b - a × c = c × d - b × d ⇒ a ×( b - c )=( c - b ) × d ⇒ a ×( b - c )-( c - b ) × d =0 ⇒ a ×( b - c )- d ×( b - c )=0 ⇒ ( a - d ) ×( b - c )=0 ⇒( a - d ) |( b - c ) ∴ ( a - d ) ⋅( b - c ) ≠ 0 ⇒ a ⋅ b + d ⋅ c ≠ d ⋅ b + a ⋅ c

Q. If a,b,c,d are four distinct vectors satisfying the conditions a×b=c×d and a×c=b×d, then prove that a⋅b+c⋅d=a⋅c+b⋅d

Given, a×b=c×d and a×c=b×d ⇒a×b−a×c=c×d−b×d ⇒a×(b−c)=(c−b)×d ⇒a×(b−c)−(c−b)×d=0 ⇒a×(b−c)−d×(b−c)=0 ⇒(a−d)×(b−c)=0⇒(a−d)∥(b−c) ∴(a−d)⋅(b−c)=0 ⇒a⋅b+d⋅c=d⋅b+a⋅c

Q. If $a, b, c, d$ are four distinct vectors satisfying the conditions $a \times b = c \times d$ and $a \times c = b \times d$ , then prove that $a \cdot b + c \cdot d \neq = a \cdot c + b \cdot d$