Area vector

If one side of the plane $\pi$ is specified as its front and the other side of the front is back, the front and back can be specified for this plane $\pi$. A plane whose front and back are specified in this way is called a directed plane. To display the orientation of the directed plane, consider a shape on the directed plane $\pi$, and when you turn the edge of the shape so that you see this shape on your left shoulder, it is perpendicular to this plane and the right-hand thread the direction of travel represents the front and back of the directed plane. A vector of magnitude 1 perpendicular to this plane and in the direction of the right-handed screw is called a unit normal vector, andit is represented by a $\boldsymbol{n}$.

When the area of?the figure on this directed plane $\pi$ is $S$, the vector ${\bf S} = S \boldsymbol{n}$ is called the area vector of this figure. From $\vert{\bf S}\vert = \vert S \boldsymbol{n}\vert = S \vert\boldsymbol{n}\vert = S$, the size of the area vector represents the area of this figure, The direction of ${\bf S}$ represents the slope of this figure in space.

As shown in the figure above, let the plane figure with the area vector ${\bf S}$ be the bottom, and the volume of the pillar with the generatrix parallel to the vector $\boldsymbol{A}$ be $V$. At this time, assume that the angle between $\boldsymbol{A}$ and ${\bf S}$ is an acute angle. At this time, it is assumed that the angle between $\boldsymbol{A}$ and ${\bf S}$ is an acute angle. Then, the height $h$ of this pillar is expressed as

$$h = \vert\boldsymbol{A}\vert\cos{\theta}$$

and

$$V = h\vert{\bf S}\vert = \vert{bf A}\vert\vert{\bf S}\vert\cos{\theta} = \boldsymbol{A} \cdot{\bf S}$$

Then we say $\boldsymbol{A} \cdot{\bf S}$ directed volume．

Consider a rectangle ABCD with an area vector ${\bf S}$ and a directed plane $\pi$, as shown in the figure above. At this time, the image obtained by projecting the rectangle ABCD onto the directed plane $\pi$ is A'B'C'D'. Find the area of ??the rectangle A'B'C'D'when the angle between the rectangle ABCD and the directed plane $\pi$ is $\theta$.． First, notice that the angle between the area vector ${\bf S}$ and the normal vector $\boldsymbol{n}$ of the directed plane $\pi$ is $\theta$. Then, the area $S'$ of the rectangle A'B'C'D'is $\ overhead {\rm A'B'} \ \overline {\rm B'C'}$．Also， $\overline{\rm A'B'} = \overline{\rm AB}\cos{\theta}$, $\overline{\rm B'C'} = \overline{\rm BC}$．Therefore，

$$S' = \overline{\rm AB}\ \overline{\rm BC}\cos{\theta} = \vert{\bf... ...rt{\bf S}\vert\vert\boldsymbol{n}\vert\cos{\theta} = {\bf S}\cdot\boldsymbol{n}$$

Generalizing this argument, when there is a plane figure with an area vector of ${\bf S}$ and a directed plane $\pi$, this plane figure is obtained by projecting it onto the plane $\pi$. The area of ?the figure $S$ is given by

$$S'= {\bf S} \cdot \boldsymbol{n}$$

At this time, ${\bf S} \cdot \ boldsymbol{n}$ is called the bf directed area of the orthodox projection on the directed plane $\pi$ of this plane.