Line in space, all possible equations.

There are such forms of writing the equation of a line in space:

1) ${\begin{array}{lcl} A_{1} x + B_{1} y + C_{1} z + D_{1} = 0 (P_{1}) \\ A_{2} x + B_{2} y + C_{2} z + D_{2} = 0 (P_{2}) \end{array} -$ 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 $\overset{―}{S} = (m, n, p) .$ The vector $\overset{―}{S}$ 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:

${\begin{array}{lcl} x = x_{0} + m t \\ y = y_{0} + n t \\ z = z_{0} + p t \end{array}$

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 ${\overset{―}{S}}_{1} = (m_{1}, n_{1}, p_{1});$

$L_{2} :$ $\frac{x - x_{2}}{m_{2}} = \frac{y - y_{2}}{n_{2}} = \frac{z - z_{2}}{p_{2}},$ and ${\overset{―}{S}}_{2} = (m_{2}, n_{2}, p_{2}) .$

Condition for parallelism of two lines: Lines $L_{1}$ and $L_{2}$ are parallel if and only if ${\overset{―}{S}}_{1} ∥ {\overset{―}{S}}_{2} \Leftrightarrow$ $\frac{m_{1}}{m_{2}} = \frac{n_{1}}{n_{2}} = \frac{p_{1}}{p_{2}} .$

Condition for perpendicularity of two lines: $L_{1} ⊥ L_{2} \Leftrightarrow$ ${\overset{―}{S}}_{1} ⊥ {\overset{―}{S}}_{2} \Leftrightarrow$ $m_{1} \cdot m_{2} + n_{1} \cdot n_{2} + p_{1} \cdot p_{2} = 0.$

Angle between lines:

$\cos \hat{(L_{1}, L_{2})} =$ $\frac{{\overset{―}{S}}_{1} \cdot {\overset{―}{S}}_{2}}{| {\overset{―}{S}}_{1} | \cdot | {\overset{―}{S}}_{2} |} = \frac{m_{1} \cdot m_{2} + n_{1} \cdot n_{2} + p_{1} \cdot p_{2}}{\sqrt{m_{1}^{2} + n_{1}^{2} + p_{1}^{2}} \cdot \sqrt{m_{2}^{2} + n_{2}^{2} + p_{2}^{2}}} .$

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 $\overset{―}{S} = (m, n, p) .$ 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:

$d (M_{1}, L) = \frac{| [\overset{―}{M_{1} M_{2}}, \overset{―}{S}] |}{| \overset{―}{S} |} .$

Examples.

1. 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 $O X$ axis;

The direction vector of $O X$ 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 ${\begin{array}{lcl} 3 x - y + 2 z - 7 = 0, \\ x + 3 y - 2 z - 3 = 0; \end{array}$

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 $x = - 2 + t, y = 2 t, z = 1 - \frac{1}{2} t .$

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 $\overset{―}{S} = (m, n, p) .$

Given $M_{0} (2, 0, - 3)$ and $\overset{―}{S} = q (2, - 3, 5)$ , 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 $\overset{―}{S} (5, 2, - 1) .$ 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 ${\begin{array}{lcl} 3 x - y + 2 z - 7 = 0, \\ x + 3 y - 2 z - 3 = 0; \end{array}$ can be found as the cross product of the normal vectors of the given planes.

For $P_{1} :$ $3 x - y + 2 z - 7 = 0,$ the normal vector is $N_{1} (3, - 1, 2);$

For $P_{2} :$ $x + 3 y - 2 z - 3 = 0,$ the normal vector is $N_{2} (1, 3, - 2) .$

Calculate the cross product:

$[N_{1}, N_{2}] = | \begin{matrix} i & j & k \\ 3 & - 1 & 2 \\ 1 & 3 & - 2 \end{matrix} | = i (2 - 6) - j (- 6 - 2) + k (9 + 1) = - 4 i + 8 j + 10 k .$

Hence, the direction vector $\overset{―}{S} (- 4, 8, 10)$ 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 $x = - 2 + t, y = 2 t, z = 1 - \frac{1}{2} t .$ To do this, we'll express the line's equation in canonical form:

${\begin{array}{lcl} x = - 2 + t, \\ y = 2 t, \\ z = 1 - \frac{1}{2} t \end{array} \Rightarrow$ ${\begin{array}{lcl} t = x + 2, \\ t = \frac{y}{2}, \\ t = \frac{z - 1}{- \frac{1}{2}} \end{array}$ $\Rightarrow \frac{x + 2}{1} = \frac{y}{2} = \frac{z - 1}{- \frac{1}{2}} .$

From this, we find the direction vector $\overset{―}{S} (1, 2, - \frac{1}{2}) .$ To eliminate the fraction, we can multiply the components of the direction vector by 2: ${\overset{―}{S}}_{1} (2, 4, - 1) .$

Next, we need to find the equation of the line passing through the point $M_{0} (2, 0, - 3)$ parallel to the vector $\overset{―}{S} (2, 4, - 1) :$

$\frac{x - 2}{2} = \frac{y - 0}{4} = \frac{z + 3}{- 1} \Rightarrow \frac{x - 2}{2} = \frac{y}{4} = \frac{z + 3}{- 1} .$

Answer: $\frac{x - 2}{2} = \frac{y}{4} = \frac{z + 3}{- 1} .$

2. 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: $\frac{x - 1}{2} = \frac{y + 2}{3} = \frac{z - 1}{- 2} .$

3. 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 $d (L_{1}, L_{2}) = d (M_{1}, L_{2}) = \frac{| [\overset{―}{M_{1} M_{2}}, \overset{―}{S}] |}{| \overset{―}{S} |},$ where $M_{1}$ is an arbitrary point on line $L_{1}$ , $M_{2}$ is an arbitrary point on line $L_{2}$ , and $\overset{―}{S}$ 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 $\overset{―}{S} = (3, 4, 2)$ .

From here, we find $\overset{―}{M_{1} M_{2}} = (7 - 2, 1 + 1, 3 - 0) = (5, 2, 3);$

$[\overset{―}{M_{1} M_{2}}, \overset{―}{S}] = | \begin{matrix} i & j & k \\ 5 & 2 & 3 \\ 3 & 4 & 2 \end{matrix} | = i (4 - 12) - j (10 - 9) + k (20 - 6) =$ $= - 8 i - j + 14 k .$

$| [\overset{―}{M_{1} M_{2}}, \overset{―}{S}] | = \sqrt{8^{2} + 1^{2} + 14^{2}} = \sqrt{64 + 1 + 196} = \sqrt{261} = \sqrt{9 * 29} = 3 \sqrt{29} .$

$| \overset{―}{S} | = \sqrt{3^{2} + 4^{2} + 2^{2}} = \sqrt{9 + 16 + 4} = \sqrt{29}$

$d (L_{1}, L_{2}) = \frac{| [\overset{―}{M_{1} M_{2}}, \overset{―}{S}] |}{| \overset{―}{S} |} = \frac{3 \sqrt{29}}{\sqrt{29}} = 3.$

Answer: 3.

4. Find the distance from the point $A (2, 3, - 1)$ to the given line $L :$

${\begin{array}{lcl} 2 x - 2 y + z + 3 = 0, \\ 3 x - 2 y + 2 z + 17 = 0 \end{array}$

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:

${\begin{array}{lcl} 2 x - 2 y + 0 + 3 = 0, \\ 3 x - 2 y + 0 + 17 = 0 \end{array} \Rightarrow$ ${\begin{array}{lcl} 2 x - 2 y + 3 = 0, \\ 3 x - 2 y + 17 = 0 \end{array} - \Rightarrow$ ${\begin{array}{lcl} x + 14 = 0, \\ 2 x - 2 y + 3 = 0 \end{array} \Rightarrow$ ${\begin{array}{lcl} x = - 14, \\ - 28 - 2 y + 3 = 0 \end{array} \Rightarrow$ ${\begin{array}{lcl} x = - 14, \\ y = - \frac{25}{2} . \end{array}$

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 $P_{1} : 2 x - 2 y + z + 3 = 0$ the normal vector has coordinates $N_{1} (2, - 2, 1);$

for the plane $P_{2} : 3 x - 2 y + 2 z + 17 = 0,$ the normal vector has coordinates $N_{2} (3, - 2, 2) .$

We find the cross product:

$[N_{1}, N_{2}] = | \begin{matrix} i & j & k 2 & - 2 & 1 3 & - 2 & 2 \end{matrix} | = i (- 4 + 2) - j (4 - 3) + k (- 4 + 6) = - 2 i - j + 2 k .$

Thus, the direction vector for the line ${\begin{array}{lcl} 2 x - 2 y + z + 3 = 0, \\ 3 x - 2 y + 2 z + 17 = 0 \end{array}$ has coordinates $\overset{―}{S} (- 2, - 1, 2) .$

Now we can use the formula $d (A, L) = \frac{| [\overset{―}{A M}, \overset{―}{S}] |}{| \overset{―}{S} |} .$

$\overset{―}{A M} = (2 - (- 14), 3 - (- \frac{25}{2}), - 1 - 0) = (16, 15 \frac{1}{2}, - 1)$

$[\overset{―}{A M}, \overset{―}{S}] = | \begin{matrix} i & j & k \\ 16 & 15, 5 & - 1 \\ - 2 & - 1 & 2 \end{matrix} | = i (31 - 1) - j (32 - 2) + k (- 16 + 31) =$ $= 30 i - 30 j + 15 k .$

$| [\overset{―}{A M}, \overset{―}{S}] | = \sqrt{30^{2} + 30^{2} + 15^{2}} = \sqrt{900 + 900 + 225} = \sqrt{2025} = 45.$

$| \overset{―}{S} | = \sqrt{2^{2} + 1^{2} + 2^{2}} = \sqrt{4 + 1 + 4} = 3$

$d (A, L) = \frac{| [\overset{―}{A M}, \overset{―}{S}] |}{| \overset{―}{S} |} = \frac{45}{3} = 15.$

Answer: The distance from point $A$ to line $L$ is $15$ .

5. Write the canonical equation of the line passing through the point $M_{0} (3, - 2, - 4)$ , parallel to the plane $P$ : $3 x - 2 y - 3 z - 7 = 0$ , 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 $3 x - 2 y - 3 z - 7 = 0$ :

$P : 3 x - 2 y - 3 z - 7 = 0 \Rightarrow \overset{―}{N} = (3; - 2; - 3) .$ The desired plane passes through the point $M_{0} (3, - 2, - 4)$ perpendicular to the vector $\overset{―}{N} (3, - 2, - 3)$ .

$P_{1} : 3 (x - 3) - 2 (y + 2) - 3 (z + 4) = 0 \Rightarrow$

$P_{1} : 3 x - 9 - 2 y - 4 - 3 z - 12 = 0 \Rightarrow$

$P_{1} : 3 x - 2 y - 3 z - 25 = 0.$

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:

$L : \frac{x - 2}{3} = \frac{y + 4}{- 2} = \frac{z - 1}{2} = t \Rightarrow$

${\begin{array}{lcl} x = 3 t + 2, \\ y = - 2 t - 4, \\ z = 2 t + 1. \end{array}$

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 $t :$

$3 x - 2 y - 3 z - 25 = 0$

$3 (3 t + 2) - 2 (- 2 t - 4) - 3 (2 t + 1) - 25 = 0$

$9 t + 6 + 4 t + 8 - 6 t - 3 - 25 = 0$

$7 t - 14 = 0$

$t = \frac{14}{7} = 2$

Substituting the found value of $t$ into the equation of the line $L,$ we find the coordinates of the point of intersection:

${\begin{array}{lcl} x = 3 t + 2, \\ y = - 2 t - 4, \\ z = 2 t + 1. \end{array} \Rightarrow$ ${\begin{array}{lcl} x = 6 + 2 = 8, \\ y = - 4 - 4 = - 8, \\ z = 4 + 1 = 5. \end{array}$

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:

b) Write the equation of the line passing through two given points $M_{1} (3, - 1, 0)$ and $M_{2} (1, 0, - 3) .$

Answer: $\frac{x - 3}{- 2} = \frac{y + 1}{1} = \frac{z}{- 3} .$

b) Find the distance from the point $A (2, 3, - 1)$ to the given line $L :$ ${\begin{array}{lcl} x = 3 t + 5, \\ y = 2 t, \\ z = - 2 t - 25. \end{array}$

Answer: 21.

3. Prove that the lines $L_{1} : {\begin{array}{lcl} 2 x + 2 y - z - 10 = 0, \\ x - y - z - 22 = 0, \end{array}$ and $L_{2} : \frac{x + 7}{3} = \frac{y - 5}{- 1} = \frac{z - 9}{4} .$ are parallel and find the distance $ρ (L_{1}, L_{2})$

Answer: 25.

4. Write the equations of the line passing through the intersection points of the plane $x - 3 y + 2 z + 1 = 0$ 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: $\frac{x + 1}{7} = \frac{y - 2}{- 1} = \frac{z - 3}{- 5} .$

5. Write the equation of the line passing through the point $M_{0} (7, 1, 0)$ parallel to the plane $2 x + 3 y - z - 15 = 0$ and intersecting the line $\frac{x}{1} = \frac{y - 1}{4} = \frac{z - 3}{2} .$

Answer: $\frac{x - 7}{67} = \frac{y - 1}{- 28} = \frac{z}{70} .$

Tags: Angle between lines, canonical equation of the line, distance, line in space

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