Can we combine movements before making them?

A robot makes two moves: $\mathbf u=(2,1)$ , then $\mathbf v=(-1,2)$ . Could you give it one instruction that reaches the same endpoint?

Use what you learned about displacement: count the total horizontal change and the total vertical change. Then predict what happens if the robot makes the moves in the opposite order.

Keep a record of your investigation

Use $\mathbf u=(2,1)$ , $\mathbf v=(-1,2)$ , and $k=1$ as the reference. Write down the predicted endpoint and intermediate corner. For the order experiment, keep both vectors and $k$ fixed. For the scaling experiment, restore the original order and change only $k$ . Record the endpoint before and after each change; explain which change affected the route and which affected the net movement.

Two paths, one endpoint?

Start with multiplier $k=1$ . Compare the path with its dashed net-change arrow. Before selecting Reverse order, sketch where you expect the robot to finish.

Restore the original order with Reverse order. Before selecting Double u, write your predicted endpoint. This trial doubles only the first vector, keeping the second vector and the move order fixed.

Two moves, one net change

u = (2, 1), v = (−1, 2). The path follows ku, then v.

kuvNet change (dashed)Axes rescale to keep the vectors visible.

Multiplier k

ku = (2, 1); v = (-1, 2); net change = (1, 3). Compare the net change after reversing the order.

Compare the two experiments

With $k=1$ , either order ends at $(1,3)$ : two right and one left give one right; one up and two up give three up. The intermediate corner changes.

With $k=2$ , the scaled move is $(4,2)$ and the net change is $(3,4)$ . That is not twice the original endpoint, which would be $(2,6)$ .

Addition combines corresponding changes

To combine vectors, add corresponding components:

\mathbf u+\mathbf v=(2+(-1),\;1+2)=(1,3).

For any two vectors in the same coordinate space,

(u_1,u_2)+(v_1,v_2)=(u_1+v_1,\;u_2+v_2).

Ordinary number addition is commutative, so the component sums are the same in either order. That explains why $\mathbf u+\mathbf v=\mathbf v+\mathbf u$ for every pair, not just our drawing.

The vectors must have matching coordinates. Adding a two-component movement to a three-component record needs a definition of how their slots correspond; we cannot silently drop or invent an entry.

A scalar changes every component

A scalar is a number used to scale a vector. Multiplying $\mathbf u$ by $k$ gives

k\mathbf u=(2k,k).

Predict the path for $k=-1$ and then select Negate u. What does a negative multiplier do? Set Multiplier k to zero next.

Explain the negative and zero cases

Negation reverses the movement: $-\mathbf u=(-2,-1)$ . Adding $\mathbf v$ then gives $(-3,1)$ . At $k=0$ , the scaled move is $(0,0)$ , so only $\mathbf v$ contributes and the endpoint is $(-1,2)$ .

The vector $(0,0)$ is the zero vector. It represents no change. Subtracting a vector means adding its reverse:

\mathbf u-\mathbf v=\mathbf u+(-\mathbf v)=(3,-1).

Where the parentheses matter

Compare these two instructions on paper:

2\mathbf u+\mathbf v=(3,4),\qquad 2(\mathbf u+\mathbf v)=(2,6).

The second expression doubles both moves because scaling distributes across addition:

2(\mathbf u+\mathbf v)=2\mathbf u+2\mathbf v.

The first expression leaves $\mathbf v$ unchanged. In a model update, this is the difference between scaling one contribution and scaling the whole update.

Try unfamiliar moves

Let $\mathbf p=(1,-2)$ and $\mathbf q=(3,1)$ . Predict $\mathbf p-2\mathbf q$ and explain it as a path before calculating.

Check your path

First move $(1,-2)$ , then move twice opposite to $\mathbf q$ , which contributes $(-6,-2)$ . The result is $(-5,-4)$ . Reversing the order of these two already-specified moves keeps that net change.

Design your own comparison

Choose a multiplier $k$ different from the worked values, within the explorer’s range. Predict and test the endpoint with the original move order. Keep that $k$ fixed, reverse the order, and compare the endpoints.

On paper, compare $k\mathbf u+\mathbf v$ with $k(\mathbf u+\mathbf v)$ for your chosen $k$ . Can you find a value of $k$ for which they agree? Explain what made that case special.

Write a short conclusion: what you predicted, what you observed, and how the evidence changed or supported your explanation.

Check the reasoning in your investigation

Reversing the two specified moves preserves their sum for any $k$ . The paper expressions differ by $(k-1)\mathbf v$ , so for this nonzero $\mathbf v$ they agree exactly at $k=1$ . Your explanation should distinguish changing order from changing which contributions are scaled.

Question 1 of 30 checked · 0 correct

Let u=(1, 2) and v=(3, −1). What is 2u+v?(8, 2)(5, 3)(4, 1)