Mathematics · Ch 11 — Three-Dimensional Geometry

Equation of a Line Through a Given Point and Parallel to Given Vector

11.3.1

Equation of a Line Through a Given Point and Parallel to Given Vector

Equation of a Line Through a Given Point and Parallel to a Given Vector

The fundamental problem is to find the equation of a line when we know one point it passes through and its direction. In three-dimensional space, a line is completely determined by a point on it and a vector parallel to it. This vector gives the line its orientation, while the point fixes its location.

Vector Form of the Equation

Let $O$ be the origin of the rectangular coordinate system. Let $A$ be the given point on the line, with position vector $\vec{a}$ . Let $\vec{b}$ be a given vector parallel to the line $l$ . Now consider any arbitrary point $P$ on the line $l$ , with position vector $\vec{r}$ .

The vector $\overrightarrow{AP}$ is the vector from point $A$ to point $P$ . Since the line passes through $A$ and $P$ , and the line is parallel to $\vec{b}$ , the vector $\overrightarrow{AP}$ must also be parallel to $\vec{b}$ . Therefore, there exists some real number $\lambda$ such that:

\overrightarrow{AP} = \lambda \vec{b}

But we can express $\overrightarrow{AP}$ in terms of position vectors:

\overrightarrow{AP} = \overrightarrow{OP} - \overrightarrow{OA} = \vec{r} - \vec{a}

Equating these two expressions gives:

\vec{r} - \vec{a} = \lambda \vec{b}

Rearranging, we obtain the vector equation of the line:

\vec{r} = \vec{a} + \lambda \vec{b}

Here, $\lambda$ is a parameter that can take any real value. For each distinct value of $\lambda$ , this equation gives the position vector of a different point $P$ on the line. When $\lambda = 0$ , we get the given point $A$ itself.

\vec{a}

Note

If $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$ , then the numbers $a$ , $b$ , $c$ are called the direction ratios of the line. Conversely, if $a$ , $b$ , $c$ are the direction ratios of a line, then the vector $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$ is parallel to that line. Do not confuse $\vec{b}$ with its magnitude $|\vec{b}|$ .

Cartesian Form of the Equation

We now derive the Cartesian (or scalar) form from the vector equation. Let the coordinates of the given point $A$ be $(x_1, y_1, z_1)$ . Let the direction ratios of the line be $a$ , $b$ , $c$ . Let the coordinates of any point $P$ on the line be $(x, y, z)$ .

Then we can write the position vectors as:

\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}

\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}

\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}

Substituting these into the vector equation $\vec{r} = \vec{a} + \lambda \vec{b}$ :

x\hat{i} + y\hat{j} + z\hat{k} = (x_1\hat{i} + y_1\hat{j} + z_1\hat{k}) + \lambda(a\hat{i} + b\hat{j} + c\hat{k})

x\hat{i} + y\hat{j} + z\hat{k} = (x_1 + \lambda a)\hat{i} + (y_1 + \lambda b)\hat{j} + (z_1 + \lambda c)\hat{k}

Equating the coefficients of $\hat{i}$ , $\hat{j}$ , and $\hat{k}$ on both sides gives the parametric equations of the line:

x = x_1 + \lambda a

y = y_1 + \lambda b

z = z_1 + \lambda c

These are called parametric equations because the coordinates $x$ , $y$ , $z$ are expressed in terms of the parameter $\lambda$ .

›Proof

Derivation of the Cartesian equation by eliminating $\lambda$ :

From the parametric equations, we can solve for $\lambda$ in each case:

From $x = x_1 + \lambda a$ : $\lambda = \frac{x - x_1}{a}$ , provided $a \neq 0$

From $y = y_1 + \lambda b$ : $\lambda = \frac{y - y_1}{b}$ , provided $b \neq 0$

From $z = z_1 + \lambda c$ : $\lambda = \frac{z - z_1}{c}$ , provided $c \neq 0$

Since $\lambda$ is the same in all three equations, we equate these expressions:

$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$

This is the Cartesian equation of the line. …

Figure 11.3A line l in space passing through point A with position vector a, parallel to direction vector b, and an arbitrary point P with position vector r drawn from the origin of the X, Y, Z axes.

Fig. 11.3 — A line l in space passing through point A with position vector a, parallel to direction vector b, and an arbitrary point P with position vector r drawn from the origin of the X, Y, Z axes.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your NCERT textbook's own diagram.

Fig 11.3 is the foundational picture for the entire topic of lines in 3D. It shows a standard rectangular coordinate system with axes $X$ , $Y$ , $Z$ meeting at the origin $O$ . The key object is a straight line $l$ (drawn in indigo, with arrows on both ends to show it extends infinitely in both directions) that does not pass through $O$ . Instead, it sits somewhere in space, offset from the origin.

Two points are marked on this line. Point $A$ is a fixed, given point on $l$ . Its position vector from $O$ is labelled $\vec{a} = \overrightarrow{OA}$ . Point $P$ is an arbitrary (any) point on the same line, with position vector $\vec{r} = \overrightarrow{OP}$ . The figure also shows a separate arrow labelled $\vec{b}$ , drawn floating above the line (not attached to $O$ or $A$ ). This arrow represents the direction vector of the line — it is parallel to $l$ and gives the line its orientation in space.

The physical idea is simple: to describe every point on $l$ , you start at $A$ and move some distance along the direction $\vec{b}$ . The vector from $A$ to $P$ is $\overrightarrow{AP} = \vec{r} - \vec{a}$ . Since $\overrightarrow{AP}$ lies along the line, it must be parallel to $\vec{b}$ . That means $\overrightarrow{AP}$ is a scalar multiple of $\vec{b}$ : $\vec{r} - \vec{a} = \lambda \vec{b}$ , where $\lambda$ is a real number (the parameter). Rearranging gives the central result.

\vec{r} = \vec{a} + \lambda \vec{b}

This is the vector equation of a line through point $A$ (position $\vec{a}$ ) parallel to direction $\vec{b}$ . Every value of $\lambda$ gives a different point $P$ on $l$ : $\lambda = 0$ gives $A$ itself, $\lambda > 0$ gives points on one side of $A$ , and $\lambda < 0$ gives points on the other side.

From this vector form, the textbook derives the Cartesian form. If $A = (x_1, y_1, z_1)$ and $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$ (where $a, b, c$ are the direction ratios), then equating components of $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ with $\vec{a} + \lambda\vec{b}$ gives the parametric equations:

x = x_1 + \lambda a, \quad y = y_1 + \lambda b, \quad z = z_1 + \lambda c

Eliminating $\lambda$ yields the symmetric Cartesian form:

$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$ …

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