Dolžina vektorja

$\overset{\rightharpoonup}{a}=(-2,1)$ $$|\overset{\rightharpoonup}{a}|=|(-2,1)|=\sqrt{(-2)^2+1^2}=\sqrt{5}$$ $\overset{\rightharpoonup}{b}=(2,2)-(1,-1)=(1,3)$ $$|\overset{\rightharpoonup}{b}|=|(1,3)|=\sqrt{1^2+3^2}=\sqrt{10}$$

Določimo najprej komponente vektorja $2\overset{\rightharpoonup}{a}-3\overset{\rightharpoonup}{b}$: $$2\overset{\rightharpoonup}{a}-3\overset{\rightharpoonup}{b}=2(4,-1,0)-3(2,1,1)=(2,-5,-3)$$ Od tod je: $$|2\overset{\rightharpoonup}{a}-3\overset{\rightharpoonup}{b}|=\sqrt{2^2+(-5)^2+(-3)^2}=\sqrt{38}$$

Iz $|\overset{\rightharpoonup}{a}|=|\overset{\rightharpoonup}{b}|$ izhaja: $$\sqrt{x^2+5}=\sqrt{y^2+2},$$ iz $\overset{\rightharpoonup}{a}\cdot \overset{\rightharpoonup}{b}=0$ pa: $$-x+2-y=0.$$ Dobljeni sitem enačb ima rešitev $$x=\frac{1}{4},y=\frac{7}{4}$$

$$d(A,B)=|\overset{\Large\rightharpoonup}{AB}|=|(x_2-x_1,y_2-y_1,z_2-z_1)|=$$ $$=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$