Tutorials › Calculus III › Why Translating Both Endpoints Preserves Displacement

Vector Components and Operations · Tutorial 49 of 999

Why Translating Both Endpoints Preserves Displacement

Learn to prove and verify that adding the same vector to both endpoints preserves the directed vector from one endpoint to the other.

Beginner 8 min read

What You'll Learn

  • Define the displacement from one point to another using their coordinates.
  • Prove component by component that a common translation cancels when endpoints are subtracted.
  • Calculate and check translated endpoints in three-dimensional examples.
  • Distinguish translating both endpoints from translating only one.
  • Explain what the result preserves and why the same vector must be added to each endpoint.

Why a Shared Translation Does Not Change Displacement

A displacement describes the change from a starting point to an ending point. Its components depend on the differences between the endpoint coordinates, not on where the endpoints sit relative to the coordinate origin. This distinction matters when a route, segment, or pair of points is moved to a different location: if both endpoints move by the same vector, the directed displacement between them remains unchanged.

We will establish that fact directly from component subtraction. The proof is short, but it identifies exactly why the result holds: each coordinate added to the starting point is also added to the ending point, so the equal additions cancel when the new displacement is calculated.

Points, Translation, and Directed Displacement

Let \(P=(x_1,y_1,z_1)\) and \(Q=(x_2,y_2,z_2)\) be points in three-dimensional space. The directed vector from \(P\) to \(Q\), written \(\overrightarrow{PQ}\), is found by subtracting the coordinates of the starting point \(P\) from the corresponding coordinates of the ending point \(Q\).

$$ \overrightarrow{PQ} = \langle x_2-x_1,\ y_2-y_1,\ z_2-z_1\rangle. $$

Here, “directed” means that the order of the endpoints matters. In general, reversing the endpoints reverses the vector: \(\overrightarrow{QP}=-\overrightarrow{PQ}\).

Now let \(\mathbf{t}=\langle a,b,c\rangle\) be a vector. Translating a point by \(\mathbf{t}\) means adding \(a\) to its \(x\)-coordinate, \(b\) to its \(y\)-coordinate, and \(c\) to its \(z\)-coordinate. The translated point is the original point shifted by the same coordinate amounts.

$$ P'=P+\mathbf{t}=(x_1+a,\ y_1+b,\ z_1+c), \qquad Q'=Q+\mathbf{t}=(x_2+a,\ y_2+b,\ z_2+c). $$
Definition: A common translation of two points \(P\) and \(Q\) by a vector \(\mathbf{t}\) replaces them with \(P+\mathbf{t}\) and \(Q+\mathbf{t}\). The directed displacement after translation is the vector from \(P+\mathbf{t}\) to \(Q+\mathbf{t}\).

The word common is essential: the same vector is added to both points. The result does not claim that changing just one endpoint leaves the displacement unchanged.

The Component Proof

To calculate the displacement after translation, subtract the coordinates of \(P'\) from the corresponding coordinates of \(Q'\). In the \(x\)-coordinate, the difference is \((x_2+a)-(x_1+a)\). The \(a\) cancels. The same cancellation occurs for \(b\) and \(c\) in the other two coordinates.

$$ \begin{aligned} \overrightarrow{P'Q'} &= \langle (x_2+a)-(x_1+a),\ (y_2+b)-(y_1+b),\ (z_2+c)-(z_1+c)\rangle\\ &= \langle x_2-x_1,\ y_2-y_1,\ z_2-z_1\rangle\\ &= \overrightarrow{PQ}. \end{aligned} $$
Theorem: For any points \(P,Q\) in three-dimensional space and any vector \(\mathbf{t}\), translating both points by \(\mathbf{t}\) preserves their directed displacement: \(\overrightarrow{(P+\mathbf{t})(Q+\mathbf{t})}=\overrightarrow{PQ}\). This follows because the same added amount appears in the corresponding coordinate of both endpoints and cancels in subtraction.

The component calculation proves the claim for arbitrary coordinates and an arbitrary translation vector. It is not based on a special numerical example. The only condition needed is that both endpoints receive the same translation. If the added vectors differ, the cancellation used in the proof no longer applies in general.

Worked Examples

Worked Example: Translating Two Given Endpoints

Let \(P=(1,-2,3)\), \(Q=(4,0,-1)\), and translate both points by \(\mathbf{t}=\langle 5,-3,2\rangle\). Find the original and translated directed vectors, then check whether they agree.

First subtract the coordinates of \(P\) from those of \(Q\):

$$ \overrightarrow{PQ} = \langle 4-1,\ 0-(-2),\ -1-3\rangle = \langle 3,\ 2,\ -4\rangle. $$

Next add \(\mathbf{t}\) to each endpoint, one coordinate at a time:

$$ \begin{aligned} P'&=(1+5,\ -2+(-3),\ 3+2)=(6,-5,5),\\ Q'&=(4+5,\ 0+(-3),\ -1+2)=(9,-3,1). \end{aligned} $$

Subtract the translated starting point from the translated ending point:

$$ \overrightarrow{P'Q'} = \langle 9-6,\ -3-(-5),\ 1-5\rangle = \langle 3,\ 2,\ -4\rangle. $$

Check: The result is exactly \(\overrightarrow{PQ}=\langle 3,2,-4\rangle\). Each coordinate difference is unchanged.

Worked Example: A Translation with Negative Components

Let \(P=(-2,4,1)\), \(Q=(3,-1,6)\), and \(\mathbf{t}=\langle -4,2,-3\rangle\). Find the displacement before and after the translation.

The original displacement is:

$$ \overrightarrow{PQ} = \langle 3-(-2),\ -1-4,\ 6-1\rangle = \langle 5,\ -5,\ 5\rangle. $$

Add the same translation vector to both endpoints:

$$ \begin{aligned} P'&=(-2+(-4),\ 4+2,\ 1+(-3))=(-6,6,-2),\\ Q'&=(3+(-4),\ -1+2,\ 6+(-3))=(-1,1,3). \end{aligned} $$

Now subtract \(P'\) from \(Q'\):

$$ \overrightarrow{P'Q'} = \langle -1-(-6),\ 1-6,\ 3-(-2)\rangle = \langle 5,\ -5,\ 5\rangle. $$

Check: The translated displacement equals the original displacement in all three components. Negative translation components shift coordinates downward where appropriate, but they do not alter the differences between corresponding endpoints.

Worked Example: Constructing the Translated Endpoint from a Displacement

Suppose \(P=(2,1,-3)\), and the displacement from \(P\) to \(Q\) is \(\mathbf{d}=\langle -4,3,2\rangle\). Translate both endpoints by \(\mathbf{t}=\langle 7,-2,5\rangle\). Find the translated endpoints and verify their displacement.

Since the displacement is ending point minus starting point, recover \(Q\) by adding \(\mathbf{d}\) to \(P\):

$$ Q=(2+(-4),\ 1+3,\ -3+2)=(-2,4,-1). $$

Translate both points:

$$ \begin{aligned} P'&=(2+7,\ 1+(-2),\ -3+5)=(9,-1,2),\\ Q'&=(-2+7,\ 4+(-2),\ -1+5)=(5,2,4). \end{aligned} $$

Subtract the translated starting point from the translated ending point:

$$ \overrightarrow{P'Q'} = \langle 5-9,\ 2-(-1),\ 4-2\rangle = \langle -4,\ 3,\ 2\rangle. $$

Check: This equals the given displacement \(\mathbf{d}\). The endpoint \(Q\) was first reconstructed using the definition of displacement, and the translated coordinates confirm that the common shift does not change it.

Why the Same Translation Matters

The result depends on matching the translation at both endpoints. If \(P\) is shifted by \(\mathbf{t}_P\) and \(Q\) by a different vector \(\mathbf{t}_Q\), then subtracting the new coordinates gives the original displacement plus the difference between the shifts:

$$ \overrightarrow{(P+\mathbf{t}_P)(Q+\mathbf{t}_Q)} = \overrightarrow{PQ}+(\mathbf{t}_Q-\mathbf{t}_P). $$

This identity follows by expanding the coordinate differences, just as in the common-translation proof. When \(\mathbf{t}_P=\mathbf{t}_Q\), the added difference is the zero vector, so the original displacement remains. When the shifts differ, an extra vector generally remains.

This also explains a useful interpretation: displacement records the relative position of the endpoints, not their absolute location. Moving both endpoints together changes their coordinates, but preserves their coordinate differences. Since the directed vector is identical after translation, any quantity determined only by that vector, such as its magnitude, is unchanged as well.

Common Mistake: Translating Only One Endpoint

A frequent error is to translate one endpoint and then expect the old displacement to remain. For example, take \(P=(1,2,0)\), \(Q=(4,-1,5)\), and \(\mathbf{t}=\langle 2,3,-4\rangle\). The original displacement is:

$$ \overrightarrow{PQ} = \langle 4-1,\ -1-2,\ 5-0\rangle = \langle 3,-3,5\rangle. $$

If only \(Q\) is translated, then \(Q'=(6,2,1)\), while \(P\) remains \((1,2,0)\). The new displacement is:

$$ \overrightarrow{PQ'} = \langle 6-1,\ 2-2,\ 1-0\rangle = \langle 5,0,1\rangle. $$

This is not \(\langle 3,-3,5\rangle\). The translation vector was added to the ending point but not the starting point, so there is no matching addition to cancel. To preserve the displacement, translate \(P\) as well: \(P'=P+\mathbf{t}=(3,5,-4)\). Then:

$$ \overrightarrow{P'Q'} = \langle 6-3,\ 2-5,\ 1-(-4)\rangle = \langle 3,-3,5\rangle. $$

The correction is not to alter the subtraction rule. It is to apply the same translation to both endpoints before subtracting.

Using the Result Carefully

When checking whether a calculation preserves a displacement, identify the starting point, the ending point, and the vector added to each. Then subtract endpoint coordinates in the correct order. This makes it easier to detect two separate errors: changing only one endpoint, or reversing the order of subtraction.

1
Record the original direction.
For a vector from \(P\) to \(Q\), subtract \(P\)'s coordinates from \(Q\)'s.
2
Apply one common translation.
Add the same vector to each endpoint, coordinate by coordinate.
3
Recalculate the directed vector.
Subtract the translated starting point from the translated ending point and check that each component matches the original.

This method applies to any two points in three-dimensional space and any translation vector. It does not depend on particular coordinate values, on the direction of the translation, or on whether some coordinates are negative. Its condition is simply that both endpoints receive the identical vector.

Check Your Understanding

Use component subtraction and the order of the endpoints in each question.

  1. In one sentence, what is the directed displacement from \(P\) to \(Q\) in terms of their coordinates?
  2. Let \(P=(0,3,-2)\), \(Q=(5,1,4)\), and \(\mathbf{t}=\langle -2,4,1\rangle\). Find \(\overrightarrow{PQ}\), \(P+\mathbf{t}\), and \(Q+\mathbf{t}\), then verify the translated displacement.
  3. Why does adding the same vector to both endpoints leave each displacement component unchanged?
  4. If \(P\) is shifted by \(\mathbf{u}\) and \(Q\) by \(\mathbf{v}\), what additional vector appears in the new displacement compared with \(\overrightarrow{PQ}\)?
  5. Give a specific reason that translating only the ending point does not, in general, preserve the displacement.