[size=150][b]Equation of a line in 3D Space[/b][/size][br][br]As we know, we can uniquely determine a line in 3D space by specifying a point on the line and the direction of the line. Suppose a line [math]L[/math] passing through the point [math]P_0=(x_0,y_0,z_0)[/math] and is in the same direction as the vector [math]\vec{v}=\langle a,b,c\rangle[/math] (this non-zero vector is called the [b]direction vector[/b] of [math]L[/math]). Let [math]P=(x,y,z)[/math] be any point on line [math]L[/math] and [math]O=(0,0,0)[/math]. We have[br][br][math]\overrightarrow{OP}=\overrightarrow{OP_0}+\overrightarrow{P_0P}[/math][br][br][math]\overrightarrow{OP_0}=\langle x_0,y_0,z_0\rangle[/math] i.e. the position vector of [math]P_0[/math].[br][br]As [math]\overrightarrow{P_0P}[/math] is parallel to [math]\vec{v}[/math], [math]\overrightarrow{P_0P}=t\vec{v}=t\langle a,b,c\rangle[/math] for some real number [math]t[/math]. Therefore, we have[br][br][math]\langle x,y,z\rangle=\langle x_0,y_0,z_0\rangle+t\langle a,b,c\rangle=\langle x_0+at,y_0+bt,z_0+ct\rangle[/math][br][br]The parametric equation of line [math]L[/math] is as follows:[br][br][math]L \ : \ \begin{cases} x & = x_0+at \\ y & = y_0+bt \\ z & = z_0+ct \end{cases}[/math][br][br]where [math]t[/math] is any real number.[br][br]The line is illustrated in the applet below.[br][br][br]

[u]Remarks[/u][list=1][*]The parametric equation of a line [math]L[/math] is not unique i.e. different parametric equations can represent the same line. For example, you can choose another point [math]P_0[/math] on the line or another direction vector that is a scalar multiple of [math]\vec{v}[/math].[/*][*]If [math]a,b,c[/math] are all non-zero, we can eliminate the parameter [math]t[/math] and write the equation of line in the following [b]symmetric form[/b]: [math]\frac{x-x_0}a=\frac{y-y_0}b=\frac{z-z_0}c[/math].[/*][/list][br][br]Two lines in 3D space are non-parallel if their direction vectors are non-parallel i.e. one is not a scalar multiple of another. In 2D space, any two non-parallel lines must intersect each other. However, this is generally not true in 3D space, as shown in the following example:[br][br][u]Example[/u]: [br][br]Let [math]L_1[/math] be the line passing through [math](1,2,3)[/math] in the direction [math]\vec{v_1}=\langle -1,1,4\rangle[/math].[br]Let [math]L_2[/math] be the line passing through [math](1,0,1)[/math] in the direction [math]\vec{v_2}=\langle 1,5,0\rangle[/math].[br][br]Show that they are non-parallel lines that do not intersect each other.[br][br][u]Solution[/u]: Obviously, [math]\vec{v_1}[/math] and [math]\vec{v_2}[/math] are not parallel vectors. Therefore line [math]L_1[/math] and [math]L_2[/math] are non-parallel.[br][br]The following are their parametric equations:[br][br][math]L_1 \ : \ \begin{cases} x & = 1-t \\ y & = 2 + t \\ z & = 3 + 4t \end{cases}[/math][br][br]for any real number [math]t[/math][br][br][math]L_2 \ : \ \begin{cases} x & = 1+s \\ y & = 5s \\ z & = 1 \end{cases}[/math][br][br]for any real number [math]s[/math][br][br]Assume there exists an intersection point [math](x,y,z)[/math] between the two lines i.e. there exists [math]t[/math] and [math]s[/math] such that they satisfy the above two parametric equation simultaneously. We should be able to solve the following system of equations:[br][br][math]\begin{cases} 1-t&=1+s \\ 2+t&=5s \\ 3+4t&=1 \end{cases}[/math][br][br]However, it can easily be shown that the above system of equations has no solution! Hence, the two lines never intersect each other.[br][br]

[b][size=150]Equation of a plane in 3D space[/size][/b][br][br]For any plane in 3D space, we can identify its "direction" by its [b]normal vector[/b] - a non-zero vector that is perpendicular to any vector contained in the plane. We can uniquely determine a plane if we are given a point [math]P_0=(x_0,y_0,z_0)[/math] and a normal vector [math]\vec{n}=\langle a,b,c\rangle[/math]. Let [math]P=(x,y,z)[/math] be any point on the plane [math]S[/math]. Then we have[br][br][math]\overrightarrow{P_0P}\cdot \vec{n}=0[/math] (because [math]\overrightarrow{P_0P}[/math] is contained in [math]S[/math]), which implies that[br][br][math]\langle x-x_0,y-y_0,z-z_0\rangle \cdot \langle a,b,c \rangle =0[/math][br][br]Therefore, the following is the equation of the plane [math]S[/math]:[br][br][math]a(x-x_0)+b(y-y_0)+c(z-z_0)=0[/math][br][br]This is called the [b]point-normal form[/b] of the equation of the plane.[br][br]Expanding the left hand side of the above equation, we get[br][br][math]ax+by+cz+d=0[/math][br][br]where [math]d=-ax_0-by_0-cz_0[/math]. This is called the [b]general form[/b] of the equation of the plane.[br][br][br]The plane is illustrated in the applet below.[br][br]

Two distinct planes are [b]parallel[/b] if their normal vectors are parallel. Moreover, they do not intersect each other. If two distinct planes are non-parallel, they intersect at a line, called the [b]line of intersection[/b].[br][br]Two distinct planes are [b]orthogonal[/b] if their normal vectors are orthogonal vectors. More generally, the angle between two planes is the angle between the normal vectors. If we specify that such angle, called [math]\theta[/math], is smaller or equal to [math]\frac{\pi}2[/math], we have the following formula:[br][br][math]\cos\theta = \left|\frac{\vec{n_1}\cdot\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}\right|[/math], where [math]\vec{n_1}, \vec{n_2}[/math] are the normal vectors of the two given planes.[br][br][br][br]

You can use the applet below to visualize the planes and lines in the exercises.[br]

[u]Distance Problem[/u][br][br]Given a plane [math]ax+by+cz+d=0[/math] and a point [math]P=(x',y',z')[/math] not lying on the plane. How can we find the (shortest) distance [math]D[/math] from the point to the plane?[br][br]Suppose point [math]Q[/math] is the foot of the perpendicular from point [math]P[/math] to the plane. By definition, the distance [math]D[/math] between [math]P[/math] and [math]Q[/math] is the distance from [math]P[/math] to the plane. Let [math]P_0=(x_0,y_0,z_0)[/math] be a point on the plane and [math]\vec{n}=\langle a,b,c \rangle[/math] be the normal vector of the plane. Then we have[br][br][math]\overrightarrow{P_0P}=\langle x'-x_0,y'-y_0,z'-z_0\rangle[/math] and the equation of the plane is [math]ax+by+cz+d=0[/math], where [math]d=-ax_0-by_0-cz_0[/math].[br][br][math]D=\left|\text{Proj}_{\vec{n}}\overrightarrow{P_0P}\right|=\frac{|\vec{n}\cdot\overrightarrow{P_0P}|}{|\vec{n}|}=\frac{|a(x'-x_0)+b(y'-y_0)+c(z'-z_0)|}{\sqrt{a^2+b^2+c^2}}=\frac{|ax'+by'+cz'+d|}{\sqrt{a^2+b^2+c^2}}[/math][br][br]Hence, the distance formula from point [math]P=(x',y',z')[/math] to the plane [math]ax+by+cz+d=0[/math] is as follows:[br][br][math]D=\frac{|ax'+by'+cz'+d|}{\sqrt{a^2+b^2+c^2}}[/math][br][br]The distance from a point to a plane is illustrated in the applet below.[br][br]

This formula can also be used to compute the distance between two parallel planes and the distance between two non-parallel lines.[br]