Mathematics · Ch 10 — Vector Algebra

Vector Joining Two Points

10.5.2

Vector Joining Two Points

Concept: The Vector from One Point to Another

In coordinate geometry, every point in space is represented by its position vector relative to the origin. For two points $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$ , the vector that starts at $P_1$ and ends at $P_2$ is the vector joining $P_1$ to $P_2$ , denoted $\overrightarrow{P_1P_2}$ .

Draw the position vectors $\overrightarrow{OP_1}$ and $\overrightarrow{OP_2}$ from the origin $O$ . In triangle $OP_1P_2$ , the vector $\overrightarrow{P_1P_2}$ is the third side, going from $P_1$ to $P_2$ .

Note

By the triangle law, $\overrightarrow{OP_1} + \overrightarrow{P_1P_2} = \overrightarrow{OP_2}$ .

Deriving the Formula for $\overrightarrow{P_1P_2}$

Rearranging the triangle law:

\overrightarrow{P_1P_2} = \overrightarrow{OP_2} - \overrightarrow{OP_1}

Substituting the position vectors in component form:

\overrightarrow{P_1P_2} = (x_2\hat{i} + y_2\hat{j} + z_2\hat{k}) - (x_1\hat{i} + y_1\hat{j} + z_1\hat{k})

\overrightarrow{P_1P_2} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}

\overrightarrow{P_1P_2} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}

Magnitude of $\overrightarrow{P_1P_2}$

The magnitude equals the distance between $P_1$ and $P_2$ :

|\overrightarrow{P_1P_2}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

Important

This is the distance formula in three-dimensional space, derived directly from vector algebra.

Direction Matters: Initial and Terminal Points …

Figure 10.153D coordinate axes with points P1(x1,y1,z1) and P2(x2,y2,z2), showing position vectors OP1 and OP2 and the vector P1P2 joining the two points as one side of triangle OP1P2.

Fig. 10.15 — 3D coordinate axes with points P1(x1,y1,z1) and P2(x2,y2,z2), showing position vectors OP1 and OP2 and the vector P1P2 joining the two points as one side of triangle OP1P2.

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.

The figure shows a three-dimensional right-handed coordinate system with origin $O$ . The axes are labelled: the $Z$ -axis points vertically upward, the $Y$ -axis points to the right, and the $X$ -axis points down and to the left. Along each axis, the unit vectors $\hat{i}$ , $\hat{j}$ , and $\hat{k}$ are drawn, pointing in the positive $X$ , $Y$ , and $Z$ directions respectively.

Two points are plotted in space. $P_1(x_1, y_1, z_1)$ is located in the middle-right region of the diagram, and $P_2(x_2, y_2, z_2)$ is higher up and to the right. From the origin $O$ , a dotted indigo arrow reaches $P_1$ — this is the position vector $\overrightarrow{OP_1}$ . A dashed indigo arrow reaches $P_2$ — this is $\overrightarrow{OP_2}$ . A solid indigo arrow, starting at $P_1$ and ending at $P_2$ , represents the vector joining the two points, $\overrightarrow{P_1P_2}$ . The three arrows together form triangle $OP_1P_2$ .

The physical idea is simple: the vector from one point to another is the difference of their position vectors. If you know where each point is relative to the origin, you can find the displacement between them by subtracting the starting point's position from the ending point's position.

\overrightarrow{P_1P_2} = \overrightarrow{OP_2} - \overrightarrow{OP_1}

Writing the position vectors in component form:

\overrightarrow{OP_1} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}, \qquad \overrightarrow{OP_2} = x_2\hat{i} + y_2\hat{j} + z_2\hat{k}

Applying the triangle law from the figure gives:

\overrightarrow{P_1P_2} = (x_2\hat{i} + y_2\hat{j} + z_2\hat{k}) - (x_1\hat{i} + y_1\hat{j} + z_1\hat{k})

\overrightarrow{P_1P_2} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}

The magnitude of this vector — the straight-line distance between $P_1$ and $P_2$ — follows from the distance formula in three dimensions:

|\overrightarrow{P_1P_2}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

Start your 14-day free trial to unlock the full solution →

← Previous · 10.5.1Example 9 — Write the direction ratios of the vector veca=hati+hatj-2hatk and henc…Next · 10.5.2 →Example 10 — Find the vector joining the points P(2, 3, 0) and Q(-1, -2, -4) direct…

Vector Joining Two Points