Let B = b₁ i + b₂ j + b₃ k.
Since A and B are anti-parallel, the angle between them is 180°. Then the dot product of A and B is
A • B = ||A|| ||B|| cos(180°)
⇒ A • B = - ||A|| ||B||
The magnitude of A is
||A|| = √(2² + 3² + (-1)²) = √14
so that
A • B = 2b₁ + 3b₂ - b₃ = -6√14
Because A and B are anti-parallel, their cross product is the zero vector.
A × B = (3b₃ + b₂) i - (2b₃ + b₁) j + (2b₂ - 3b₁) k = 0 i + 0 j + 0 k
⇒ 3b₃ + b₂ = 0, 2b₃ + b₁ = 0, and 2b₂ - 3b₁ = 0
Now,
2b₂ - 3b₁ = 0 ⇒ b₂ = 3/2 b₁
2b₃ + b₁ = 0 ⇒ b₃ = -1/2 b₁
so that
2b₁ + 3/2 b₁ - (-1/2 b₁) = -6√14
7b₁ = -6√14
b₁ = -6/7 √14 = -6 √(2/7)
Similarly,
2b₂ - 3b₁ = 0 ⇒ b₁ = 2/3 b₂
3b₃ + b₂ = 0 ⇒ b₃ = -1/3 b₂
so that
2 (2/3 b₂) + 3b₂ - (-1/3 b₂) = -6√14
14/3 b₂ = -6√14
b₂ = -18/14 √14 = -9 √(2/7)
Finally,
2 (-6 √(2/7)) + 3 (-9 √(2/7)) - b₃ = -6√14
-39 √(2/7) - b₃ = -6√14
b₃ = 6√14 - 39 √(2/7) = 3 √(2/7)
and so the vector B is
B = -6 √(2/7) i - 9 √(2/7) j + 3 √(2/7) k
or equilvalently,
B = -3 √(2/7) (2 i + 3 j - k)
B = -3 √(2/7) A
