Let B = a i + b j + c k. Then the cross product of A = i + 2j - k with B is
A × B = ( i + 2j - k ) × ( a i + b j + c k )
A × B = a ( i × i ) + 2a ( j × i ) - a ( k × i )
… … … + b ( i × j ) + 2b ( j × j ) - b ( k × j )
… … … + c ( i × k ) + 2c ( j × k ) - c ( k × k )
A × B = 0 - 2a k - a j
… … … + b k + 0 + b i
… … … - c j + 2c i - 0
A × B = (b + 2c) i + (-a - c) j + (b - 2a) k
So we have
3 i - j + 5 k = (b + 2c) i + (c - a) j + (b - 2a) k
which gives us the system of equations,
{ b + 2c = 3
{ -a - c = -1
{ -2a + b = 5
Solve for a, b, and c.
• Eliminate c from the first two equations:
(b + 2c) + 2 (-a - c) = 3 + 2 (-1)
-2a + b = 1
But -2a + b = 5, and 5 ≠ 1, so there is no such vector B that satisfies the cross product!