Tutorials › Calculus III › Finding Endpoints from Position Vectors and Displacements

Vector Components and Operations · Tutorial 44 of 999

Finding Endpoints from Position Vectors and Displacements

Use a displacement vector to find the endpoint of a directed segment, then verify the result by subtracting the starting point.

Beginner 9 min read

What You'll Learn

  • Distinguish a point’s position vector from a displacement vector.
  • Find an endpoint by adding its displacement to the starting point’s position vector.
  • Carry out the calculation separately in the three coordinate components.
  • Check a proposed endpoint by subtracting the starting coordinates.
  • Recognize how reversing the direction of a displacement changes the calculation.
  • Keep coordinate units consistent throughout the computation.

From a Starting Point to an Endpoint

A displacement describes a change in position. If you know where a point starts and how far it moves in each coordinate direction, you can determine where it ends. The calculation is direct, but two details matter: the displacement must point from the given starting point toward the endpoint, and its components must be combined with the starting coordinates using the correct signs.

In this tutorial, points are specified by their Cartesian coordinates, and vectors are written in bold lowercase. The main relationship is that the endpoint’s position vector equals the starting point’s position vector plus the displacement from the start to the endpoint.

Key idea: If a directed displacement begins at \(P\) and ends at \(Q\), then the position vector of \(Q\) is the position vector of \(P\) plus that displacement.

Position Vectors and Displacements

Let \(O\) be the origin of a three-dimensional Cartesian coordinate system. A point \(P=(p_1,p_2,p_3)\) has position vector \(\mathbf{p}=\overrightarrow{OP}=\langle p_1,p_2,p_3\rangle\). The position vector records the point’s location relative to the origin.

For two points \(P=(p_1,p_2,p_3)\) and \(Q=(q_1,q_2,q_3)\), the displacement from \(P\) to \(Q\) is the vector obtained by subtracting the starting coordinates from the ending coordinates. It points from \(P\) toward \(Q\), so its direction depends on the order of the points.

$$ \overrightarrow{PQ} = \mathbf{q}-\mathbf{p} = \langle q_1-p_1,\ q_2-p_2,\ q_3-p_3\rangle. $$

This subtraction rule is useful in both directions. If the two endpoints are known, subtract to find the displacement. If the starting point and displacement are known, add to recover the endpoint.

Definition: If \(P\) is the starting point and \(\mathbf{d}\) is the displacement from \(P\) to an endpoint \(Q\), then \(\mathbf{d}=\overrightarrow{PQ}\). The endpoint’s position vector is \(\mathbf{q}=\mathbf{p}+\mathbf{d}\), so its coordinates are the starting coordinates plus the corresponding displacement components.

The addition is component by component. For example, if the start is \((p_1,p_2,p_3)\) and the displacement is \(\langle d_1,d_2,d_3\rangle\), then the endpoint is \((p_1+d_1,p_2+d_2,p_3+d_3)\). A negative component means the coordinate decreases; it is not a request to subtract every component or to reverse the entire vector.

Why Adding the Displacement Gives the Endpoint

The endpoint formula follows directly from the definition of displacement. Suppose \(P=(p_1,p_2,p_3)\) is the start and \(Q=(q_1,q_2,q_3)\) is the endpoint. By the component formula for the displacement from \(P\) to \(Q\), its components are \(q_1-p_1\), \(q_2-p_2\), and \(q_3-p_3\). Adding these changes to the starting coordinates recovers the endpoint coordinates.

$$ \begin{aligned} \mathbf{p}+\overrightarrow{PQ} &=\langle p_1,p_2,p_3\rangle +\langle q_1-p_1,q_2-p_2,q_3-p_3\rangle\\ &=\langle p_1+q_1-p_1,\ p_2+q_2-p_2,\ p_3+q_3-p_3\rangle\\ &=\langle q_1,q_2,q_3\rangle\\ &=\mathbf{q}. \end{aligned} $$

Each coordinate change cancels the corresponding starting coordinate and leaves the endpoint coordinate. This proves the relationship component by component. It also shows why the subtraction check works: subtracting the start from the endpoint must return the displacement that was added.

A Reliable Calculation and Check

For a numerical problem, write down the starting coordinates and displacement separately before combining them. This makes it easier to preserve signs, especially when some components are negative. Then check the result by subtracting the starting coordinates from the endpoint coordinates in the same order.

1
Identify the start.
Write the starting point’s three coordinates in order.
2
Identify the directed displacement.
Confirm that the given vector points from the stated start toward the unknown endpoint.
3
Add corresponding components.
Combine the first coordinates, the second coordinates, and the third coordinates separately.
4
Verify by subtraction.
Subtract the starting position vector from the endpoint position vector and confirm that the given displacement is recovered.

Worked Examples

Worked Example: Finding an Endpoint with Positive Components

Suppose a displacement starts at \(P=(2,-1,4)\) and is \(\mathbf{d}=\langle 3,5,-2\rangle\). Find the endpoint \(Q\).

The starting position vector is \(\mathbf{p}=\langle 2,-1,4\rangle\). Add the displacement component by component:

$$ \begin{aligned} \mathbf{q} &=\mathbf{p}+\mathbf{d}\\ &=\langle 2,-1,4\rangle+\langle 3,5,-2\rangle\\ &=\langle 2+3,\ -1+5,\ 4+(-2)\rangle\\ &=\langle 5,4,2\rangle. \end{aligned} $$

Therefore, the endpoint is \(Q=(5,4,2)\). Check by subtracting the start from the endpoint:

$$ \overrightarrow{PQ} = \langle 5-2,\ 4-(-1),\ 2-4\rangle = \langle 3,5,-2\rangle = \mathbf{d}. $$

The subtraction recovers all three components of the given displacement, so the endpoint is consistent with the stated start and motion.

Worked Example: Handling Negative Coordinates and Displacements

A point \(R=(-4,3,1)\) is the starting point. A displacement \(\mathbf{d}=\langle -2,-5,6\rangle\) begins at \(R\). Find its endpoint \(S\).

The start has position vector \(\mathbf{r}=\langle -4,3,1\rangle\). Add each displacement component to the coordinate in the same position:

$$ \begin{aligned} \mathbf{s} &=\mathbf{r}+\mathbf{d}\\ &=\langle -4,3,1\rangle+\langle -2,-5,6\rangle\\ &=\langle -4+(-2),\ 3+(-5),\ 1+6\rangle\\ &=\langle -6,-2,7\rangle. \end{aligned} $$

Thus \(S=(-6,-2,7)\). The first coordinate decreases from \(-4\) to \(-6\); the second decreases from \(3\) to \(-2\); and the third increases from \(1\) to \(7\). Check by subtracting the starting point:

$$ \overrightarrow{RS} = \langle -6-(-4),\ -2-3,\ 7-1\rangle = \langle -2,-5,6\rangle = \mathbf{d}. $$

The calculation confirms that the endpoint produces exactly the given displacement when measured from \(R\).

Worked Example: Applying Units to Coordinates

A position is recorded in metres. An object starts at \(A=(1.5,-2,0.5)\,\text{m}\) and undergoes the displacement \(\mathbf{d}=\langle 0.5,3,-1.5\rangle\,\text{m}\). Find its endpoint \(B\), including units.

Since both the position coordinates and displacement components are in metres, they can be added component by component:

$$ \begin{aligned} \mathbf{b} &=\mathbf{a}+\mathbf{d}\\ &=\langle 1.5,-2,0.5\rangle\,\text{m} +\langle 0.5,3,-1.5\rangle\,\text{m}\\ &=\langle 1.5+0.5,\ -2+3,\ 0.5+(-1.5)\rangle\,\text{m}\\ &=\langle 2,1,-1\rangle\,\text{m}. \end{aligned} $$

Therefore, \(B=(2,1,-1)\,\text{m}\). Verify by taking endpoint minus start:

$$ \overrightarrow{AB} = \langle 2-1.5,\ 1-(-2),\ -1-0.5\rangle\,\text{m} = \langle 0.5,3,-1.5\rangle\,\text{m} = \mathbf{d}. $$

All components retain the unit metre, and the check returns the original displacement. Adding quantities with mismatched units would not be valid; the starting coordinates and displacement components must use compatible units.

Direction and Order Matter

The vector from \(P\) to \(Q\) is generally not the vector from \(Q\) to \(P\). Reversing the endpoints changes the sign of every displacement component. Thus, to recover an endpoint, use the displacement in the direction from the known start toward that endpoint.

$$ \overrightarrow{QP} = \mathbf{p}-\mathbf{q} = -\overrightarrow{PQ}. $$

If a problem instead gives the displacement from the unknown endpoint back to the known starting point, that vector points in the opposite direction. In that case, the endpoint is found by subtracting the given vector from the starting position vector, or equivalently by adding its negative.

Common Pitfall: Reversing the Subtraction

A frequent error is to subtract the displacement from the starting position vector even though the displacement is stated to point from the start to the endpoint. For example, let \(P=(1,2,3)\) and let the displacement from \(P\) to \(Q\) be \(\mathbf{d}=\langle 4,-1,2\rangle\). The incorrect calculation \(\mathbf{p}-\mathbf{d}\) gives \(\langle -3,3,1\rangle\).

That result fails the required check:

$$ \langle -3-1,\ 3-2,\ 1-3\rangle = \langle -4,1,-2\rangle = -\mathbf{d}, $$

so it corresponds to moving in the opposite direction. The correction is to add the displacement to the start:

$$ \mathbf{q} = \langle 1,2,3\rangle+\langle 4,-1,2\rangle = \langle 5,1,5\rangle. $$

Now the check gives \(\langle 5-1,1-2,5-3\rangle=\langle 4,-1,2\rangle=\mathbf{d}\), as required. The subtraction check is especially useful because it detects a reversed direction immediately.

What the Calculation Tells You

Finding an endpoint from a starting point and a displacement is a coordinate description of a simple geometric relationship. The starting position tells where the motion begins, while the displacement tells how the coordinates change. A positive component increases its coordinate, a negative component decreases it, and a zero component leaves that coordinate unchanged.

The method applies whenever the start and displacement are expressed in the same coordinate system and compatible units. It does not depend on the starting point being the origin. If the starting point is the origin, its position vector is the zero vector, so the endpoint coordinates equal the displacement components. For any other start, include its coordinates in the addition.

Takeaway: For a displacement directed from a known start \(P\) to an unknown endpoint \(Q\), calculate \(\mathbf{q}=\mathbf{p}+\overrightarrow{PQ}\). Then verify that \(\mathbf{q}-\mathbf{p}\) equals the stated displacement.

Check Your Understanding

For each question, pay attention to the starting point, the direction of the displacement, and the subtraction check.

  1. If \(P=(3,-2,1)\) and the displacement from \(P\) is \(\langle -1,4,2\rangle\), what is the endpoint?
  2. A point \(A=(-2,0,5)\) moves to \(B=(1,-3,2)\). What is the displacement from \(A\) to \(B\)?
  3. If a given vector describes the displacement from an unknown endpoint back to a known start, should you add it to or subtract it from the start’s position vector? Explain briefly.
  4. Why does subtracting the starting position vector from a calculated endpoint verify the result?
  5. A starting point and displacement are given in different units. What must be done before adding their components?