Parallel vectors

Definition

Two non-zero vectors are parallel if one of the vectors can be presented as scalar multiplication of the other.

If \(\overline{u}\parallel\overline{v}\), then \(\overline{u}=r\overline{v}\), where \(r \ne 0\).

If such a scalar doesn’t exist, vectors are nonparallel, \(\overline{u}\nparallel\overline{v}\).

EXAMPLE 1. Parallel vectors

Let’s examine if vectors \(\overline{u}=(6, -4)\) and \(\overline{v}=(-\frac{1}{2},\frac{1}{3})\) are parallel.

We have to find if such a scalar \(r\) exists so that \(\overline{u}=r\overline{v}\).

\[\begin{split}\begin{align}(6, -4) & =r(-\frac{1}{2},\frac{1}{3}) \\ \\ (6, -4) & =(-\frac{1}{2}r,\frac{1}{3}r) \end{align}\end{split}\]

Now we can write two equations corresponding the two components.

\[\begin{split}\begin{align}6 & =-\frac{1}{2}r & -4 & =\frac{1}{3}r \\ \\ r & =-12 & r & =-12 \\ \\ \end{align}\end{split}\]

We have found such a scalar \(r\) that satifies \(\overline{u}=r\overline{v}\) so vectors are parallel. In fact, vectors are in the opposite directions since \(r < 0\).

We can also examine the parallelism of two vectors by forming their unit vectors and comparing the unit vectors components.

Unit vector (kesken)

Sometimes it is necessary to form a unit vector \(\overline{a}^0\) of a given vector \(\overline{a}\). The unit vector has the same direction as the original vector but it has the norm of 1. The unit vector of formed by dividing a vector by its magnitude. In practise, every component of a vector will be divided separately. In general for a three dimensional vector, the unit vector is given by

\(\begin{align}\overline{a}^0 & =\frac{\overline{a}}{|\overline{a}|} \\ \\ & =(\frac{a_x}{|\overline{a}|},\frac{a_y}{|\overline{a}|},\frac{a_z}{|\overline{a}|}) \\ \\ \end{align}\)

EXAMPLE 2. Where is the head of a displacement vector from coordinate point C, when the magnitude of the vector is 30 and it is parallel to line AB?

Kuva 1. 23 Esimerkki yksikkövektorin hyödyntämisestä

Huom! Pisteen C koordinaatteihin ei saa suoraan lisätä arvoa 30. Tämä veisi pisteen D väärään suuntaan suoraan koilliseen.

Muodostetaan ensin vektori \(\overline{AB}=\overline{OB}-\overline{OA}=(400,140)-(40,100)=(360,40)\).

Jotta pisteestä C päästään 30 m vektorin \(\overline{AB}\) suuntaisesti, muodostetaan ensin vektorin \(\overline{AB}\) yksikkövektori.

\(\begin{align}\{\overline{AB}}^0&=\frac{\overline{AB}}{|\overline{AB}|}=\frac{(360,40)}{\sqrt{{360}^2+{20}^2}}=\frac{(360,40)}{40\sqrt{82}} \\ & =(\frac{360}{40\sqrt{82}},\frac{40}{40\sqrt{82}})=(\frac{9}{\sqrt{82}},\frac{1}{\sqrt{82}}) \\ & =(0.993\ldots,0.110\ldots)\end{align}\)

Seuraavaksi kerrotaan yksikkövektori pituudella 30, jotta saadaan siirtymävektori pisteiden C ja D välille. \( \overline{CD}=30\bullet{\overline{AB}}^0=30\bullet(\frac{9}{\sqrt{82}}\)

\(\frac{1}{\sqrt{82}})=(\frac{270}{\sqrt{82}}\)

\(\frac{30}{\sqrt{82}})\)

Täten paikkavektori \(\overline{OD}=\overline{OC}+\overline{CD}=(40,130)+(\frac{270}{\sqrt{82}}\)

\(\frac{30}{\sqrt{82}})\)

\(=(69.816\ldots,133.312\ldots)\approx(70,133)\) eli piste D = (70, 133).

Esimerkki.