Equation of a Line in Space

Introduction to the Equation of a Line

In two-dimensional geometry, a line can be determined by a point and a slope, or by two distinct points.

In three-dimensional geometry, the same idea appears in a slightly richer form. A line is uniquely determined if:

1. it passes through a known point and has a known direction, or
2. it passes through two distinct points.

The direction of a line in space is described by a vector parallel to it. Once we know one point on the line and one direction vector, the entire line is fixed.

The equation of a line is commonly written in two equivalent forms:

• Vector form
• Cartesian (or symmetric) form

Both forms describe the same set of points in space and can be converted into one another.

Line through a Given Point and Parallel to a Given Vector

Vector Form

Let a line pass through a point $A$ whose position vector is $\vec{a}$, and let the line be parallel to a vector $\vec{b}$. If $P$ is any point on the line with position vector $\vec{r}$, then the vector $\overrightarrow{AP}$ must be parallel to $\vec{b}$. Hence, $$\overrightarrow{AP} = \lambda \vec{b}$$ for some real parameter $\lambda$. Since $$\vec{r} = \vec{a} + \overrightarrow{AP},$$ we get the vector equation of the line: $$\vec{r} = \vec{a} + \lambda \vec{b}.$$

Here, $\lambda$ is called the parameter. Different values of $\lambda$ give different points on the line.

Cartesian Form

Suppose the point $A$ has coordinates $(x_1, y_1, z_1)$, so that $$\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}.$$ Also suppose the direction vector is $$\vec{b} = a\hat{i} + b\hat{j} + c\hat{k},$$ where $a, b, c$ are direction ratios of the line. If $P(x,y,z)$ is any point on the line, then $$\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}.$$ Substituting in $$\vec{r} = \vec{a} + \lambda \vec{b},$$ and comparing coefficients of $\hat{i}, \hat{j}, \hat{k}$, we get $$x = x_1 + \lambda a, \qquad y = y_1 + \lambda b, \qquad z = z_1 + \lambda c.$$ So, $$x-x_1 = \lambda a, \qquad y-y_1 = \lambda b, \qquad z-z_1 = \lambda c.$$ Eliminating $\lambda$, we obtain the symmetric Cartesian equation: $$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}.$$

If $a, b, c$ are replaced by the direction cosines $l, m, n$, then the equation becomes $$\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}.$$ In that case the common value is the actual directed distance from the fixed point along the line, because $l, m, n$ are components of a unit direction vector.

Line Passing Through Two Given Points

Vector Form

Let the line pass through two points $A$ and $B$ whose position vectors are $\vec{a}$ and $\vec{b}$ respectively. Then the direction of the line is given by $$\overrightarrow{AB} = \vec{b} - \vec{a}.$$ Using the point-direction form of a line, the vector equation becomes $$\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a}).$$

This is the standard vector equation of the line through the two given points.

Cartesian Form

If $$A(x_1, y_1, z_1), \qquad B(x_2, y_2, z_2),$$ then the direction ratios of the line are $$x_2-x_1, \qquad y_2-y_1, \qquad z_2-z_1.$$ Therefore the Cartesian equation is $$\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}.$$

This is valid provided the denominators are interpreted carefully when one of them is zero.

Important Considerations and General Point on a Line

• When a denominator is zero: If one direction ratio is zero, say $a=0$, then the expression $$\frac{x-x_1}{0}$$ means $x-x_1=0$, so $x=x_1$. It does not mean division in the usual arithmetic sense; it is shorthand notation used in symmetric line equations.

• General point on a line: If $$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = \lambda,$$ then any point on the line can be written as $$(x,y,z) = (x_1 + a\lambda,\; y_1 + b\lambda,\; z_1 + c\lambda).$$

This parametric representation is extremely useful in finding intersections with planes, distances, feet of perpendiculars, and reflections.

Example 1: Vector and Cartesian Equation (Point & Direction)

Find the vector and the Cartesian equations of the line through the point $(5, 2, -4)$ and which is parallel to the vector $3\hat{i} + 2\hat{j} - 8\hat{k}$.

Solution: Step 1: The given point is $$A(5,2,-4).$$ Its position vector is $$\vec{a} = 5\hat{i} + 2\hat{j} - 4\hat{k}.$$

Step 2: The given direction vector is $$\vec{b} = 3\hat{i} + 2\hat{j} - 8\hat{k}.$$ So the direction ratios are $3,2,-8$.

Step 3: Use the vector form of the line: $$\vec{r} = \vec{a} + \lambda \vec{b}.$$ Hence, $$\vec{r} = (5\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 8\hat{k}).$$

Step 4: Use the Cartesian symmetric form: $$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}.$$ Substituting $x_1=5, y_1=2, z_1=-4$ and $a=3, b=2, c=-8$, we get $$\frac{x-5}{3} = \frac{y-2}{2} = \frac{z+4}{-8}.$$

Answer: Vector: $\vec{r} = (5\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 8\hat{k})$. Cartesian: $\frac{x-5}{3} = \frac{y-2}{2} = \frac{z+4}{-8}$.

Example 2: Vector and Cartesian Equation (Two Points)

Find the vector and the Cartesian equations of the line that passes through the origin and $(5, -2, 3)$.

Solution: Step 1: Let the two points be $$O(0,0,0), \qquad A(5,-2,3).$$ The position vectors are $$\vec{0} \quad \text{and} \quad 5\hat{i} - 2\hat{j} + 3\hat{k}.$$

Step 2: The direction vector of the line is $$\overrightarrow{OA} = 5\hat{i} - 2\hat{j} + 3\hat{k}.$$

Step 3: The vector equation is $$\vec{r} = \lambda(5\hat{i} - 2\hat{j} + 3\hat{k}).$$ This is because the line passes through the origin.

Step 4: The direction ratios are $5,-2,3$, and the point is $(0,0,0)$, so the Cartesian equation is $$\frac{x}{5} = \frac{y}{-2} = \frac{z}{3}.$$

Answer: Vector: $\vec{r} = \lambda(5\hat{i} - 2\hat{j} + 3\hat{k})$. Cartesian: $\frac{x}{5} = \frac{y}{-2} = \frac{z}{3}$.

Example 3: Cartesian to Vector Conversion

The Cartesian equation of a line is $\frac{x+3}{2} = \frac{y-5}{4} = \frac{z+6}{2}$. Find the vector equation for the line.

Solution: Step 1: Compare with the standard form $$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}.$$ From $$\frac{x+3}{2} = \frac{y-5}{4} = \frac{z+6}{2},$$ we get $$x_1 = -3, \qquad y_1 = 5, \qquad z_1 = -6.$$ So one point on the line is $(-3,5,-6)$.

Step 2: The direction ratios are the denominators: $$2,4,2.$$ So a direction vector is $$2\hat{i} + 4\hat{j} + 2\hat{k}.$$

Step 3: Therefore the vector equation is $$\vec{r} = (-3\hat{i} + 5\hat{j} - 6\hat{k}) + \lambda(2\hat{i} + 4\hat{j} + 2\hat{k}).$$

Answer: $\vec{r} = (-3\hat{i} + 5\hat{j} - 6\hat{k}) + \lambda(2\hat{i} + 4\hat{j} + 2\hat{k})$.

Example 4: Standardizing Cartesian Equations

Find the Cartesian equation of the line passing through the point $(-2, 4, -5)$ and parallel to the line given by $\frac{x+3}{3} = \frac{4-y}{5} = \frac{z+8}{6}$.

Solution: Step 1: Rewrite the given line in standard form. Since $$4-y = -(y-4),$$ we have $$\frac{4-y}{5} = \frac{y-4}{-5}.$$ So the line becomes $$\frac{x+3}{3} = \frac{y-4}{-5} = \frac{z+8}{6}.$$ Therefore, its direction ratios are $$3, -5, 6.$$

Step 2: A parallel line has the same direction ratios. The required line passes through $$(-2,4,-5).$$

Step 3: Using the symmetric form, $$\frac{x-(-2)}{3} = \frac{y-4}{-5} = \frac{z-(-5)}{6}.$$ Hence, $$\frac{x+2}{3} = \frac{y-4}{-5} = \frac{z+5}{6}.$$

Answer: $\frac{x+2}{3} = \frac{y-4}{-5} = \frac{z+5}{6}$.

Example 5: Intersection with a Plane (xy-plane)

Find the point where the line passing through the points $A(3, 4, 1)$ and $B(5, 1, 6)$ crosses the xy-plane.

Solution: Step 1: Find the direction ratios of the line through $A$ and $B$: $$5-3=2, \qquad 1-4=-3, \qquad 6-1=5.$$ So the line is $$\frac{x-3}{2} = \frac{y-4}{-3} = \frac{z-1}{5} = \lambda.$$

Step 2: Write a general point on the line: $$x = 3 + 2\lambda, \qquad y = 4 - 3\lambda, \qquad z = 1 + 5\lambda.$$ So any point is $$P(3+2\lambda,\;4-3\lambda,\;1+5\lambda).$$

Step 3: The xy-plane is given by $$z=0.$$ So, $$1 + 5\lambda = 0 \implies \lambda = -\frac{1}{5}.$$

Step 4: Substitute into $x$ and $y$: $$x = 3 + 2\left(-\frac15\right) = 3 - \frac25 = \frac{13}{5},$$ $$y = 4 - 3\left(-\frac15\right) = 4 + \frac35 = \frac{23}{5}.$$ Thus the required point is $$\left(\frac{13}{5}, \frac{23}{5}, 0\right).$$

Answer: $\left(\frac{13}{5}, \frac{23}{5}, 0\right)$.

Example 6: Equation of Coordinate Axes

Find the Cartesian equation of the y-axis.

Solution: Step 1: The y-axis passes through the origin $$(0,0,0).$$

Step 2: Its direction vector is along $\hat{j}$, so the direction ratios are $$0,1,0.$$

Step 3: Therefore the symmetric equation is $$\frac{x-0}{0} = \frac{y-0}{1} = \frac{z-0}{0}.$$ That is, $$\frac{x}{0} = \frac{y}{1} = \frac{z}{0}.$$

This is equivalent to the system $$x=0, \qquad z=0,$$ with $y$ free, which is exactly the y-axis.

Answer: $\frac{x}{0} = \frac{y}{1} = \frac{z}{0}$, equivalently $x=0, z=0$.

Example 7: Line Parallel to an Axis

Find the Cartesian equation of a line passing through $(2, -1, 3)$ and parallel to the z-axis.

Solution: Step 1: A line parallel to the z-axis has direction ratios $$0,0,1.$$

Step 2: The given point is $$(2,-1,3).$$

Step 3: Using the symmetric form, $$\frac{x-2}{0} = \frac{y-(-1)}{0} = \frac{z-3}{1}.$$ So, $$\frac{x-2}{0} = \frac{y+1}{0} = \frac{z-3}{1}.$$ This means $$x=2, \qquad y=-1,$$ and $z$ can vary freely.

Answer: $\frac{x-2}{0} = \frac{y+1}{0} = \frac{z-3}{1}$, equivalently $x=2, y=-1$.

Example 8: Checking Collinearity using Line Equations

Show that the points $A(1, 2, 3)$, $B(4, 0, 4)$, and $C(-2, 4, 2)$ are collinear using the equation of a line.

Solution: Step 1: First find the line through $A$ and $B$. Direction ratios of $AB$ are $$4-1=3, \qquad 0-2=-2, \qquad 4-3=1.$$ So its equation is $$\frac{x-1}{3} = \frac{y-2}{-2} = \frac{z-3}{1}.$$

Step 2: Check whether point $C(-2,4,2)$ satisfies this equation: $$\frac{-2-1}{3} = \frac{-3}{3} = -1,$$ $$\frac{4-2}{-2} = \frac{2}{-2} = -1,$$ $$\frac{2-3}{1} = -1.$$ All three ratios are equal.

Step 3: Since point $C$ satisfies the equation of the line through $A$ and $B$, it lies on the same line. Hence the three points are collinear.

Answer: Proved.

Example 9: Foot of Perpendicular from a Point to a Line

Find the coordinates of the foot of the perpendicular drawn from the point $P(0, 2, 3)$ to the line $\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}$.

Solution: Step 1: Let the foot of the perpendicular be $Q$ on the given line. Write the line in parametric form: $$\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3} = \lambda.$$ Then $$x = 5\lambda - 3, \qquad y = 2\lambda + 1, \qquad z = 3\lambda - 4.$$ So, $$Q = (5\lambda-3,\;2\lambda+1,\;3\lambda-4).$$

Step 2: Since $PQ$ is perpendicular to the line, the vector $$\overrightarrow{PQ} = (5\lambda-3,\;2\lambda-1,\;3\lambda-7)$$ must be perpendicular to the direction vector of the line $$\vec{d} = (5,2,3).$$

Step 3: Therefore, $$\overrightarrow{PQ} \cdot \vec{d} = 0.$$ So, $$5(5\lambda-3) + 2(2\lambda-1) + 3(3\lambda-7) = 0.$$ Expanding, $$25\lambda - 15 + 4\lambda - 2 + 9\lambda - 21 = 0,$$ $$38\lambda - 38 = 0,$$ $$\lambda = 1.$$

Step 4: Substitute $\lambda=1$ into the point $Q$: $$Q = (5-3,\;2+1,\;3-4) = (2,3,-1).$$

Answer: $(2, 3, -1)$.

Example 10: Image of a Point in a Line

Find the image of the point $P(0, 2, 3)$ in the line $\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}$.

Solution: Step 1: To find the image of a point in a line, first find the foot of the perpendicular from the point to the line. Let that foot be $Q$. From the previous example, the foot is $$Q(2,3,-1).$$

Step 2: If $P'$ is the image of $P$ in the line, then the line acts as the perpendicular bisector of the segment $PP'$. Hence $Q$ is the midpoint of $PP'$. Let $$P' = (x',y',z').$$ Then, using the midpoint formula, $$\frac{0+x'}{2} = 2 \implies x' = 4,$$ $$\frac{2+y'}{2} = 3 \implies y' = 4,$$ $$\frac{3+z'}{2} = -1 \implies z' = -5.$$

Therefore, $$P' = (4,4,-5).$$

Answer: $(4, 4, -5)$.