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Next: Cross product. Up: Analytic geometry in the Previous: Planes and lines.

Dot product.

Definition 1.6   Let $\overrightarrow{u} $ and $\overrightarrow{v} $ be two vectors in $\mathcal{E}$. The dot product, or scalar product of $\overrightarrow{u} $ and $\overrightarrow{v} $ is the real number

\begin{displaymath}\overrightarrow{u}\cdot \overrightarrow{v} = \vert \overright...
...row{v} \vert
\cos ( \overrightarrow{u} , \overrightarrow{v} )
\end{displaymath}

Proposition 1.7       

1.
$\forall \overrightarrow{u} $, $\overrightarrow{u}\cdot \overrightarrow{u}\geq 0$.
2.
$\overrightarrow{u}\cdot \overrightarrow{u} = 0 \Longleftrightarrow \overrightarrow{u} = \overrightarrow{0} $.
3.
$\forall \overrightarrow{u} ,\overrightarrow{v} $, $\overrightarrow{u}\cdot \overrightarrow{v} = \overrightarrow{v}\cdot \overrightarrow{u} .$
4.
$\forall \overrightarrow{u} ,\overrightarrow{v} ,\overrightarrow{w} $, $\overrightarrow{u}\cdot (\overrightarrow{v} + \overrightarrow{w} ) =
\overrightarrow{u}\cdot \overrightarrow{v} + \overrightarrow{u}\cdot \overrightarrow{w} $.
5.
$\forall \overrightarrow{u} ,\overrightarrow{v} ,\overrightarrow{w} $, $ ( \overrightarrow{u} + \overrightarrow{v} ) \cdot \overrightarrow{w} ) =
\over...
...tarrow{u}\cdot \overrightarrow{w} + \overrightarrow{v}\cdot \overrightarrow{w} $.
6.
$\forall \overrightarrow{u} ,\overrightarrow{v} $, $\forall \lambda \in \mathbb{R} $, $\overrightarrow{u}\cdot \lambda \overrightarrow{v} = \lambda \overrightarrow{u}\cdot \overrightarrow{v} = \lambda (\overrightarrow{u}\cdot \overrightarrow{v} ) .$

Proposition 1.8   Two vectors $\overrightarrow{u} $ and $\overrightarrow{v} $ are orthogonal if, and only if, $\overrightarrow{u}\cdot \overrightarrow{v} = 0$.

Example 1.9   If $\overrightarrow{u} = \begin{pmatrix}1\\ 2\\ 3 \end{pmatrix}$ and $\overrightarrow{v} =
\begin{pmatrix}1 \\ 1\\ -1 \end{pmatrix}$, then $ \overrightarrow{u}\cdot \overrightarrow{v} = 1 \cdot 1 + 1 \cdot 2 + (-1) \cdot 3 = 0$. Thus $\overrightarrow{u}\perp \overrightarrow{v} $.

Definition 1.10   Let $\mathcal{P}$ be a plane whose cartesian equation is ax+by+cz+d=0. The vector $\overrightarrow{N} =
\begin{pmatrix}a \\ b \\ c \end{pmatrix}$ is called a normal vector for $\mathcal{P}$.


  
Figure 5: A vector normal to a plane.
\begin{figure}
\mbox{\epsfig{file=NormalVector.eps,height=4cm} }
\end{figure}

Example 1.11   The vector $\overrightarrow{N} =\begin{pmatrix}1 \\ 2 \\ -3 \end{pmatrix}$ is normal to the plane $\mathcal{P}$ whose equation is x + 2y -3z +5 =0.

Proposition 1.12   Let $\mathcal{P_1}$ and $\mathcal{P_2}$ be two planes with respective normal vectors $\overrightarrow{N} _1$ and $\overrightarrow{N} _2$. The two planes $\mathcal{P_1}$ and $\mathcal{P_2}$ are orthogonal if, and only if, the vectors $\overrightarrow{N} _1$ and $\overrightarrow{N} _2$ are orthogonal.

\fbox{
$\mathcal{P}_1 \perp \mathcal{P}_2 \Longleftrightarrow \overrightarrow{N}_1 \perp \overrightarrow{N}_2$ }

Example 1.13   Let $\mathcal{P}_1: \; 2x-y+3z-1=0$ and $\mathcal{P}_2: \; 4x+2y-2z+5=0$. These planes have respective normal vectors $\overrightarrow{N} _1= \begin{pmatrix}2 \\ -1 \\ 3 \end{pmatrix}$ and $\overrightarrow{N} _2= \begin{pmatrix}4 \\ 2 \\ -2 \end{pmatrix}$. We have: $\overrightarrow{N} _1 \cdot \overrightarrow{N} _2 = 2 \cdot 4 + (-1) \cdot 2 + 3 \cdot (-2) = 0$, thus $\mathcal{P}_1 \perp \mathcal{P}_2$.

Proposition 1.14   A line $\mathcal{L}$ and a plane $\mathcal{P}$ are perpendicular if, and only if, a direction vector for $\mathcal{L}$ is a normal vector for $\mathcal{P}$.

Example 1.15   Let $\mathcal{L}$ be the line given by the parametric equations

\begin{displaymath}\begin{cases}
x=1+2t \\ y=-5+4t \\ z=3-3t
\end{cases}\; , \; t \in \mathbb{R} .
\end{displaymath}

and let $\mathcal{P}$ be the plane whose cartesian equation is 2x+4y-3z-1=0. Then $\mathcal{L}$ and $\mathcal{P}$ are perpendicular.

Example 1.16   If $\mathcal{P}$ is defined by the equation 3x-y+5z-6=0, then the line $\mathcal{L}$ through the point A(5,-2,7) and perpendicular to $\mathcal{P}$ has the following parametric equations:

\begin{displaymath}\begin{cases}
x=5+3t \\ y= -2-t \\ z= 7+5t
\end{cases}\; , \; t \in \mathbb{R} .
\end{displaymath}


next up previous contents
Next: Cross product. Up: Analytic geometry in the Previous: Planes and lines.
Noah Dana-Picard
2001-05-30