The Properties of Vector Operations

This property can be proven with the following formula. || $$\begin{align} \overrightarrow{u} + \overrightarrow{v} &= \overrightarrow{AB} + \overrightarrow{BC} \\ &= \overrightarrow{AD} +  \overrightarrow{DC}\\ &=\overrightarrow{v} + \overrightarrow{u}\end{align}$$ ||

||\color{Red}{\overrightarrow{u}+(\overrightarrow{v}+\overrightarrow{w}) = (\overrightarrow{u}+\overrightarrow{v})+\overrightarrow{w}}||This property can be proven by using vectors |\overrightarrow{AB}, \overrightarrow{BC}| and |\overrightarrow{CD}| and Chasles' relation. || $$\begin{align}(\overrightarrow{AB} + \overrightarrow{BC})+\overrightarrow{CD} &= \overrightarrow{AC}+\overrightarrow{CD} \\ &= \overrightarrow{AD}\end{align}$$ ||A similar calculation can be performed by moving the parentheses.|| $$\begin{align} \overrightarrow{AB} + (\overrightarrow{BC}+\overrightarrow{CD}) &= \overrightarrow{AB}+\overrightarrow{BD} \\ &= \overrightarrow{AD}\end{align}$$ ||Thus, we obtain the associative property.

Additive identity:||\color{Red}{\overrightarrow{u}+\overrightarrow{0}=\overrightarrow{u}}||The components of the vector |\overrightarrow{u}| are |(a,b)| and the components of the |\overrightarrow{0}| vector are |(0,0)|.

Thus: || $$\begin{align} \overrightarrow{u} + \overrightarrow{0} &= (a,b)+(0,0)\\ &= (a+0,b+0)\\ &=(a,b) \\ &= \overrightarrow{u}\end{align}$$ ||

Multiplicative identity: ||\color{Red}{1\overrightarrow{u}=\overrightarrow{u}}||The components of the vector |\overrightarrow{u}| are |(a,b)|.

Therefore: || $$\begin{align}1 \times \overrightarrow{u}&=1 \times (a,b) \\ &= (1a,1b) \\ &=(a,b) \\ &= \overrightarrow{u}\end{align}$$ ||

||\color{Red}{\overrightarrow{u}+(-\overrightarrow{u})=\overrightarrow{0}}||
The components of the vector |\overrightarrow{u}| are |(a,b)| and the components of the vector |-\overrightarrow{u}| are |(-a,-b)|.

Then: || $$\begin{align}\overrightarrow{u} + (-\overrightarrow{u})&=(a,b)+(-a,-b) \\ &= (a-a,b-b) \\ &= (0,0) \\ &= \overrightarrow{0}\end{align}$$ ||

||\color{Red}{k(c\overrightarrow{u})=(kc)\overrightarrow{u}}||
The components of the vector |\overrightarrow{u}| are |(a,b)|.

So: || $$\begin{align}k(c \overrightarrow{u}) &= k(c(a,b))\\ &=k(ca,cb)\\ &=(kca,kcb)\\ &=(kc)(a,b) \\ &= (kc)\overrightarrow{u}​\end{align}$$ ||

||\color{Red}{k(\overrightarrow{u}+\overrightarrow{v})=k\overrightarrow{u}+k\overrightarrow{v}}||
The components of the vector |\overrightarrow{u}| are |(a,b)| and the components of the vector |\overrightarrow{v}| are |(c,d).|

Then:|| $$\begin{align}k(\overrightarrow{u}+\overrightarrow{v}) &= k((a,b)+(c,d)) \\ &= k(a+c,b+d) \\ &= (k(a+c),k(b+d))\\ &=(ka+kc,kb+kd) \\ &= (ka,kb)+(kc,kd)\\ &=k\overrightarrow{u}+k\overrightarrow{v}\end{align}$$ ||

||\color{Red}{(k+c)\overrightarrow{u}=k\overrightarrow{u}+c\overrightarrow{u}}||
The components of the vector |\overrightarrow{u}| are |(a,b).|

Thus:|| $$\begin{align} (k+c)\overrightarrow{u} &=(k+c)(a,b) \\ &=((k+c)a,(k+c)b) \\ &= (ka+ca,kb+cb) \\ &= (ka,kb)+(ca+cb)\\ &=k(a,b)+c(a,b) \\ &= k\overrightarrow{u}+c\overrightarrow{u}\end{align}$$ ||

In this relation, |A|, |B|, and |C| represent points in the Cartesian plane.||\color{Red}{\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}}|| ​

||\color{Red}{ \overrightarrow{u} \cdot (\overrightarrow{v} + \overrightarrow{w}) = \overrightarrow{u} \cdot \overrightarrow{v} + \overrightarrow{u} \cdot \overrightarrow{w}}||

Let |\overrightarrow{u}=(a,b)|, |\overrightarrow{v}=(c,d)| et |\overrightarrow{w}=(e,f)|.

Then:
|| $$\begin{align}\overrightarrow{u} \cdot (\overrightarrow{v} + \overrightarrow{w}) &= (a,b) \cdot ((c,d)+(e,f))\\ &= (a,b) \cdot (c+e,d+f) \\ &=(a(c+e)+b(d+f))\\ &=(ac+ae+bd+bf)\\ &=(ac+bd)+(ae+bf) \\ &= (a,b) \cdot (c,d) + (a,b) \cdot (e,f)\\ &= \overrightarrow{u} \cdot \overrightarrow{v} + \overrightarrow{u} \cdot \overrightarrow{w}\end{align}$$ ||

Let |\overrightarrow{u}=(a,b)|, |\overrightarrow{v}=(c,d)| be vectors, and |k_1| and |k_2|, scalars.||\color{Red}{k_1 \overrightarrow{u} \cdot k_2 \overrightarrow{v}=(k_1k_2)\overrightarrow{u} \cdot \overrightarrow{v}}||

|| $$\begin{align}k_1 \overrightarrow{u} \cdot k_2 \overrightarrow{v} &= k_1(a,b) \cdot k_2(c,d)\\ &= (k_1a,k_1b) \cdot(k_2c,k_2d)\\ &= k_1k_2ac+k_1k_2bd\\ &=(k_1k_2)(ac+bd)\\ &= (k_1k_2)(a,b) \cdot (c,d)\\ &=(k_1k_2)\overrightarrow{u} \cdot \overrightarrow{v}\end{align}$$ ||

Pour valider ta compréhension à propos des vecteurs de façon interactive, consulte la MiniRécup suivante :