Line in space, all possible equations.
Literature: Collection of problems in mathematics. Part 1. Edited by A. V. Efimov, B. P. Demidovich.
There are such forms of writing the equation of a line in space:
1) $\left\{\begin{array}{lcl}A_1x+B_1y+C_1z+D_1=0\quad (P_1)\\ A_2x+B_2y+C_2z+D_2=0\quad (P_2)\end{array}\right. - $ the general equation of the line $L$ in space, as the line of intersection of two planes $P_1$ and $P_2.$
2) $\frac{x-x_0}{m}=\frac{y-y_0}{n}=\frac{z-z_0}{p} -$ the canonical equation of the line $L,$ which passes through the point $M(x_0, y_0, z_0)$ parallel to the vector $\overline{S}=(m, n, p).$ The vector $\overline S$ is the direction vector of the line $L.$
3) $\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1} -$ the equation of the line that passes through two points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2).$
4) By equating each of the parts of the canonical equation 2 to the parameter $t,$ we obtain the parametric equation of the line:
$$\left\{\begin{array}{lcl}x=x_0+mt\\ y=y_0+nt\\z=z_0+pt\end{array}\right. $$
The arrangement of two lines in space.
Let $L_1:$ $\frac{x-x_1}{m_1}=\frac{y-y_1}{n_1}=\frac{z-z_1}{p_1}$ and $\overline{S}_1=(m_1, n_1, p_1);$
$L_2:$ $\frac{x-x_2}{m_2}=\frac{y-y_2}{n_2}=\frac{z-z_2}{p_2},$ and $\overline{S}_2=(m_2, n_2, p_2).$
Condition for parallelism of two lines: Lines $L_1$ and $L_2$ are parallel if and only if $\overline{S}_1\parallel\overline{S}_2\Leftrightarrow$ $\frac{m_1}{m_2}=\frac{n_1}{n_2}=\frac{p_1}{p_2}.$
Condition for perpendicularity of two lines: $L_1\perp L_2\Leftrightarrow$ $\overline{S}_1\perp\overline{S}_2\Leftrightarrow$ ${m_1}\cdot{m_2}+{n_1}\cdot{n_2}+p_1\cdot p_2=0.$
Angle between lines:
$\cos\widehat{(L_1, L_2)}=$ $\frac{\overline{S}_1\cdot\overline{S}_2}{|\overline S_1|\cdot|\overline S_2|}=\frac{{m_1}\cdot{m_2}+{n_1}\cdot{n_2}+p_1\cdot p_2}{\sqrt{m_1^2+n_1^2+p_1^2}\cdot\sqrt{m_2^2+n_2^2+p_2^2}}.$
The distance from a point to a line is equal to the length of the perpendicular dropped from the point to the given line.
Let the line $L$ be defined by the equation $\frac{x-x_0}{m}=\frac{y-y_0}{n}=\frac{z-z_0}{p},$ hence $\overline S=(m, n, p).$ Let also $M_2=(x_2, y_2, z_2)$ be an arbitrary point belonging to the line $L.$ Then the distance from the point $M_1=(x_1, y_1, z_1)$ to the line $L$ can be found using the formula:
$$d(M_1, L)=\frac{|[\overline{M_1M_2}, \overline S]|}{|\overline S|}.$$
Examples.
2.198. Write the canonical equation of the line passing through the point $M_0(2, 0, -3)$ parallel to:
a) the vector $q(2, -3, 5);$
The canonical form of the line is given by $\frac{x-2}{2} = \frac{y-0}{-3} = \frac{z+3}{5}.$
b) the line $\frac{x-1}{5}=\frac{y+2}{2}=\frac{z+1}{-1};$
Since the line is parallel, it has the same direction ratios. The canonical equation is $\frac{x-2}{5} = \frac{y-0}{2} = \frac{z+3}{-1}.$
c) the $OX$ axis;
The direction vector of $OX$ is $(1, 0, 0)$. The canonical equation is $\frac{x-2}{1} = \frac{y-0}{0} = \frac{z+3}{0}$ (noting that division by zero indicates the component is absent).
d) the line $\left\{\begin{array}{lcl}3x-y+2z-7=0,\\ x+3y-2z-3=0; \end{array}\right.$
First, find the direction vector by crossing the normal vectors of the planes. The canonical equation can then be formed using this direction vector and the point $M_0$.
e) the line $x=-2+t, y=2t, z=1-\frac{1}{2}t.$
The direction vector can be taken from the coefficients of $t$, which are $(1, 2, -\frac{1}{2})$. The canonical equation is $\frac{x-2}{1} = \frac{y-0}{2} = \frac{z+3}{-\frac{1}{2}}.$
Solution.
a) Let's use formula (2) for the equation of a line in space:
$\frac{x-x_0}{m}=\frac{y-y_0}{n}=\frac{z-z_0}{p} -$ the canonical equation of the line $L,$ which passes through the point $M(x_0, y_0, z_0)$ parallel to the vector $\overline{S}=(m, n, p).$
Given $M_0(2, 0, -3)$ and $\overline{S}=q(2,-3,5)$, the equation becomes $\frac{x-2}{2}=\frac{y-0}{-3}=\frac{z+3}{5}\Rightarrow\frac{x-2}{2}=\frac{y}{-3}=\frac{z+3}{5}.$
Answer: $\frac{x-2}{2}=\frac{y}{-3}=\frac{z+3}{5}.$
b) A line parallel to another line should have the same direction vector. The direction vector for the line $\frac{x-1}{5}=\frac{y+2}{2}=\frac{z+1}{-1}$ is $\overline S(5, 2, -1).$ Using this for the line through $M_0(2, 0, -3)$ gives $\frac{x-2}{5}=\frac{y}{2}=\frac{z+3}{-1}.$
Answer: $\frac{x-2}{5}=\frac{y}{2}=\frac{z+3}{-1}.$
c) The OX axis has a direction vector $i=(1, 0, 0).$ Thus, the equation of the line through $M_0(2, 0, -3)$ parallel to $i(1, 0, 0)$ is $\frac{x-2}{1}=\frac{y}{0}=\frac{z+3}{0}.$
Answer: $\frac{x-2}{1}=\frac{y}{0}=\frac{z+3}{0}.$
d) The line defined by the intersection of two planes is perpendicular to the normals of both planes. Hence, the direction vector of the line $\lef\t{\begin{array}{lcl}3x-y+2z-7=0,\\ x+3y-2z-3=0; \end{array}\right.$ can be found as the cross product of the normal vectors of the given planes.
For $P_1:$ $3x-y+2z-7=0,$ the normal vector is $N_1(3, -1, 2);$
For $P_2:$ $x+3y-2z-3=0,$ the normal vector is $N_2(1, 3, -2).$
Calculate the cross product:
$[N_1, N_2]=\begin{vmatrix}i&j&k\\3&-1&2\\1&3&-2\end{vmatrix}=i(2-6)-j(-6-2)+k(9+1)=-4i+8j+10k.$
Hence, the direction vector $\overline S (-4, 8, 10)$ gives the line equation through $M_0(2, 0, -3)$ as $\frac{x-2}{-4}=\frac{y}{8}=\frac{z+3}{10}.$
Answer: $\frac{x-2}{-4}=\frac{y}{8}=\frac{z+3}{10}.$
e) Let's find the direction vector of the line $x=-2+t, y=2t, z=1-\frac{1}{2}t.$ To do this, we'll express the line's equation in canonical form:
$\left\{\begin{array}{lcl}x=-2+t,\\ y=2t,\\z=1-\frac{1}{2}t \end{array}\right.\Rightarrow$ $\left\{\begin{array}{lcl}t=x+2,\\ t=\frac{y}{2},\\t=\frac{z-1}{-\frac{1}{2}} \end{array}\right.$ $\Rightarrow\frac{x+2}{1}=\frac{y}{2}=\frac{z-1}{-\frac{1}{2}}.$
From this, we find the direction vector $\overline S\left(1, 2, -\frac{1}{2}\right).$ To eliminate the fraction, we can multiply the components of the direction vector by 2: $\overline S_1(2, 4, -1).$
Next, we need to find the equation of the line passing through the point $M_0(2, 0, -3)$ parallel to the vector $\overline S(2, 4, -1):$
$\frac{x-2}{2}=\frac{y-0}{4}=\frac{z+3}{-1}\Rightarrow\frac{x-2}{2}=\frac{y}{4}=\frac{z+3}{-1}.$
Answer: $\frac{x-2}{2}=\frac{y}{4}=\frac{z+3}{-1}.$
2.199(a). Write the equation of the line passing through two given points $M_1 (1, -2, 1)$ and $M_2(3, 1, -1).$
Solution.
We will use formula (3) for the line equation in space:
$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1} - $ the equation of the line passing through two points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2).$
Substitute the given points:
$\frac{x-1}{3-1}=\frac{y+2}{1+2}=\frac{z-1}{-1-1} \Rightarrow$ $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-1}{-2}.$
Answer: $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-1}{-2}.$
2.204. Find the distance between parallel lines
$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z}{2}$ and $\frac{x-7}{3}=\frac{y-1}{4}=\frac{z-3}{2}.$
Solution.
The distance between parallel lines $L_1$ and $L_2$ is equal to the distance from any point on line $L_1$ to line $L_2$. Therefore, it can be found using the formula $$d(L_1, L_2)=d(M_1, L_2)=\frac{|[\overline{M_1M_2}, \overline S]|}{|\overline S|},$$ where $M_1$ is an arbitrary point on line $L_1$, $M_2$ is an arbitrary point on line $L_2$, and $\overline S$ is the direction vector of line $L_2$.
From the canonical equations of the lines, we take points $M_1=(2, -1, 0) \in L_1$, $M_2=(7, 1, 3) \in L_2$, and $\overline S=(3, 4, 2)$.
From here, we find $\overline{M_1M_2}=(7-2, 1+1, 3-0)=(5, 2, 3);$
$[\overline{M_1M_2}, \overline S]=\begin{vmatrix}i&j&k\\5&2&3\\3&4&2\end{vmatrix}=i(4-12)-j(10-9)+k(20-6)=$ $=-8i-j+14k.$
$|[\overline{M_1M_2},\overline S]|=\sqrt{8^2+1^2+14^2}=\sqrt{64+1+196}=\sqrt{261}=\sqrt{9*29}=3\sqrt{29}.$
$|\overline S|=\sqrt{3^2+4^2+2^2}=\sqrt{9+16+4}=\sqrt{29}$
$$d(L_1, L_2)=\frac{|[\overline{M_1M_2}, \overline S]|}{|\overline S|}=\frac{3\sqrt{29}}{\sqrt{29}}=3.$$
Answer: 3.
2.205 (a). Find the distance from the point $A(2, 3, -1)$ to the given line $L:$
$\left\{\begin{array}{lcl}2x-2y+z+3=0,\\ 3x-2y+2z+17=0 \end{array}\right.$
Solution.
To find the distance from point $A$ to line $L,$ we need to choose an arbitrary point $M$ on line $L$ and find the direction vector of this line.
Let's choose point $M.$ Assume $z=0.$ Substitute this value into the system:
$\left\{\begin{array}{lcl}2x-2y+0+3=0,\\ 3x-2y+0+17=0 \end{array}\right.\Rightarrow$ $\left\{\begin{array}{lcl}2x-2y+3=0,\\ 3x-2y+17=0 \end{array}\right.-\Rightarrow$ $\left\{\begin{array}{lcl}x+14=0,\\ 2x-2y+3=0 \end{array}\right.\Rightarrow$ $\left\{\begin{array}{lcl}x=-14,\\ -28-2y+3=0 \end{array}\right.\Rightarrow$ $\left\{\begin{array}{lcl}x=-14,\\ y=-\frac{25}{2}. \end{array}\right.$
Thus, $M=(-14, -\frac{25}{2}, 0)$
The direction vector is found as the cross product of the normals to the given planes:
For the plane $P_1: 2x-2y+z+3=0$ the normal vector has coordinates $N_1(2, -2, 1);$
for the plane $P_2: 3x-2y+2z+17=0,$ the normal vector has coordinates $N_2(3, -2, 2).$
We find the cross product:
$[N_1, N_2]=\begin{vmatrix}i & j & k \ 2 & -2 & 1 \ 3 & -2 & 2 \end{vmatrix} = i(-4+2) - j(4-3) + k(-4+6) = -2i - j + 2k.$
Thus, the direction vector for the line $\left\{\begin{array}{lcl} 2x-2y+z+3=0, \\ 3x-2y+2z+17=0 \end{array}\right.$ has coordinates $\overline S (-2, -1, 2).$
Now we can use the formula $$d(A, L)=\frac{|[\overline{AM}, \overline S]|}{|\overline S|}.$$
$\overline{AM}=\left(2-(-14),3-\left(-\frac{25}{2}\right),-1-0\right)=\left(16, 15\frac{1}{2}, -1\right)$
$[\overline{AM}, \overline S]=\begin{vmatrix}i&j&k\\16&15,5&-1\\-2&-1&2\end{vmatrix}=i(31-1)-j(32-2)+k(-16+31)=$ $=30i-30j+15k.$
$|[\overline{AM},\overline S]|=\sqrt{30^2+30^2+15^2}=\sqrt{900+900+225}=\sqrt{2025}=45.$
$|\overline S|=\sqrt{2^2+1^2+2^2}=\sqrt{4+1+4}=3$
$$d(A, L)=\frac{|[\overline{AM}, \overline S]|}{|\overline S|}=\frac{45}{3}=15.$$
Answer: The distance from point $A$ to line $L$ is $15$.
2.212. Write the canonical equation of the line passing through the point $M_0(3, -2, -4)$, parallel to the plane $P$: $3x-2y-3z-7=0$, and intersecting the line $L$: $\frac{x-2}{3}=\frac{y+4}{-2}=\frac{z-1}{2}$.
Solution:
Let's write the equation of the plane $P_1$ passing through the point $M_0(3, -2, -4)$ parallel to the plane $3x-2y-3z-7=0$:
$P: 3x-2y-3z-7=0\Rightarrow \overline N=(3; -2; -3).$ Искомая плоскость проходит через точку $M_0(3, -2, -4)$ perpendicular to the vector $\overline N(3, -2, -3)$.
$P_1: 3(x-3)-2(y+2)-3(z+4)=0\Rightarrow $
$P_1: 3x-9-2y-4-3z-12=0 \Rightarrow$
$P_1: 3x-2y-3z-25=0.$
Further, let's find the point of intersection of the plane $P_1$ and the line $L.$ To do this, we'll write the equation of the line $L$ in parametric form:
$L: \frac{x-2}{3}=\frac{y+4}{-2}=\frac{z-1}{2}=t\Rightarrow$
$\left\{\begin{array}{lcl}x=3t+2,\\ y=-2t-4,\\z=2t+1. \end{array}\right.$
Next, we'll substitute the values of $x, y,$ and $z$ expressed in terms of $t$ into the equation of the plane $P_1,$ and from the resulting equation, we'll solve for $t:$
$3x-2y-3z-25=0$
$3(3t+2)-2(-2t-4)-3(2t+1)-25=0$
$9t+6+4t+8-6t-3-25=0$
$7t-14=0$
$t=\frac{14}{7}=2$
Substituting the found value of $t$ into the equation of the line $L,$ we find the coordinates of the point of intersection:
$\left\{\begin{array}{lcl}x=3t+2,\\ y=-2t-4,\\z=2t+1. \end{array}\right.\Rightarrow $ $\left\{\begin{array}{lcl}x=6+2=8,\\ y=-4-4=-8,\\z=4+1=5. \end{array}\right.$
Thus, the line $L$ and the plane $P_1$ intersect at the point $M_1(8, -8, 5).$
Now, let's write the equation of the line passing through the points $M_0(3, -2, -4)$ and $M_1(8, -8, 5)$-- this will be the desired line. We'll use formula (3) $\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1} :$
$\frac{x-3}{8-3}=\frac{y+2}{-8+2}=\frac{z+4}{5+4}\Rightarrow$ $\frac{x-3}{5}=\frac{y+2}{-6}=\frac{z+4}{9}.$
Answer: $\frac{x-3}{5}=\frac{y+2}{-6}=\frac{z+4}{9}.$
Homework:
2.199.
b) Write the equation of the line passing through two given points $M_1 (3, -1, 0)$ and $M_2(1, 0, -3).$
Answer: $\frac{x-3}{-2}=\frac{y+1}{1}=\frac{z}{-3}.$
2.205.
b) Find the distance from the point $A(2, 3, -1)$ to the given line $L:$ $\left\{\begin{array}{lcl}x=3t+5,\\ y=2t,\\z=-2t-25. \end{array}\right.$
Answer: 21.
2.206. Prove that the lines $L_1: \left\{\begin{array}{lcl}2x+2y-z-10=0,\\ x-y-z-22=0, \end{array}\right.$ and $L_2: \frac{x+7}{3}=\frac{y-5}{-1}=\frac{z-9}{4}.$ are parallel and find the distance $\rho(L_1, L_2)$
Answer: 25.
2.207. Write the equations of the line passing through the intersection points of the plane $x-3y+2z+1=0$ with the lines $\frac{x-5}{5}=\frac{y+1}{-2}=\frac{z-3}{-1}$ and $\frac{x-3}{4}=\frac{y+4}{-6}=\frac{z-5}{2}.$
Answer: $\frac{x+1}{7}=\frac{y-2}{-1}=\frac{z-3}{-5}.$
2.211. Write the equation of the line passing through the point $M_0(7, 1, 0)$ parallel to the plane $2x+3y-z-15=0$ and intersecting the line $\frac{x}{1}=\frac{y-1}{4}=\frac{z-3}{2}.$
Answer: $\frac{x-7}{67}=\frac{y-1}{-28}=\frac{z}{70}.$
