Voorbeelden

1. Als ${\bf A} = 2{\bf i} -3{\bf j} + {\bf k}$, dan is $A = \sqrt{ 2^2 +(-3)^2 + 1^2 } = \sqrt{14}$.
2. Gegeven: ${\bf A} = 2{\bf i} -3{\bf j} + {\bf k}$ en ${\bf B} = 5{\bf i} +{\bf j} -7{\bf k}$. Te bewijzen: ${\bf A} \perp {\bf B}$.

   
   Bewijs:
   

   $$\begin{array}{rl} {\bf A} \cdot {\bf B} & = A_1B_1 + A_2B_... ...cdot 1 +1 \cdot (-7) \\ & = 10 -3 -7 = 0, \\ \end{array}$$ (15)

   
   
   dus ${\bf A} \perp {\bf B}$.
3. Gegeven: ${\bf A} = 3{\bf i} -4{\bf j} + 5{\bf k}$ en ${\bf B} = {\bf i} +2{\bf j} -{\bf k}$. Te berekenen: $\cos{\angle{ ({\bf A};{\bf B}) }}$.

   
   Oplossing:
   

   $$\begin{array}{rl} {\bf A} \cdot {\bf B} & = AB \cdot \cos{... ...6} \cdot \cos{\angle{ ({\bf A};{\bf B}) }}. \\ \end{array}$$ (16)

   
   
   Verder geldt
   

   $${\bf A} \cdot {\bf B} = A_1B_1 + A_2B_2 + A_3B_3 = 3 -8 -5 = -10.$$ (17)

   
   
   Dus $10 \sqrt{3} \cdot \cos{\angle{ ({\bf A};{\bf B}) }} = -10$, ofwel $\cos{\angle{ ({\bf A};{\bf B}) }} = -{1 \over 3} \sqrt{3}$.
4. Als ${\bf A} = 2{\bf i} -3{\bf j} + {\bf k}$ en ${\bf B} = 5{\bf i} +{\bf j} -7{\bf k}$, dan is
   

   $$\begin{array}{rl} {\bf A} \times {\bf B} & = \left\vert \... ... \\ & = 20{\bf i} + 19{\bf j} + 17{\bf k}. \\ \end{array}$$ (18)