Which targets can two vector controls reach?

Suppose two controls move a point by scaled copies of $\mathbf u=(2,1)$ and $\mathbf v=(-1,2)$ . Can you reach the target $(3,4)$ ? Could the same controls reach any point in the plane?

A setting $(a,b)$ produces $a\mathbf u+b\mathbf v$ . Negative coefficients reverse a movement, so the controls can move backward as well as forward.

Keep a record of your investigation

Use $a=2$ , $b=1$ and the original two directions as your reference; record the resulting endpoint. Before replacing $\mathbf v$ , write what endpoint you expect with those same coefficients. Keep $\mathbf u$ , the target, and both coefficients fixed for that comparison. Only afterward search for new coefficients. Record tested settings and endpoints, and distinguish evidence from a few trials from a reason a target is impossible.

Reach a target, then change one direction

Before using the controls, sketch where $2\mathbf u+\mathbf v$ should land. Then try coefficients $a=2$ , $b=1$ .

Before selecting Parallel controls, predict the endpoint if $\mathbf v$ becomes $(4,2)$ with $a=2$ , $b=1$ unchanged. Make that change and record the endpoint. The target remains $(3,4)$ . Now predict whether some other coefficient settings can reach it, and test your prediction.

Can your controls reach the target?

u = (2, 1); v = (-1, 2). Target = (3, 4).

aubvau + bv (dashed)× target (3, 4)Axes rescale to keep the vectors visible.

Coefficient a

Coefficient b

au = (0, 0); bv = (0, 0); sum = (0, 0). Target not reached by these coefficients.

Controls sample coefficients between −6 and 6; number fields also accept fractions as decimals. Span allows every real coefficient. A failed slider search does not prove a target is unreachable.

Try several settings, including a negative coefficient. What do all the new endpoints have in common?

Compare the two arrangements

Initially, $2(2,1)+(-1,2)=(3,4)$ reaches the target. With parallel controls, $\mathbf v=2\mathbf u$ , so every result is $(a+2b)\mathbf u$ . Each endpoint lies on the line $y=x/2$ . The target $(3,4)$ does not lie on that line.

A failed search is not the proof. The coordinate relation is: every reachable point must satisfy $y=x/2$ , but $4\ne3/2$ .

Name the whole set of possibilities

A sum of scaled vectors is a linear combination. The span of a list of vectors is the set of every linear combination of them, allowing every real coefficient.

\operatorname{span}(\mathbf u,\mathbf v) =\{a\mathbf u+b\mathbf v:\ a,b\in\mathbb R\}.

Read the braces as “the set of,” and the colon as “where.” The explorer samples only some coefficients. Span includes large values, fractions, and all real values between the slider steps.

With the original directions, any target $(r,s)$ requires

2a-b=r,\qquad a+2b=s.

Solving gives

a=\frac{2r+s}{5},\qquad b=\frac{2s-r}{5}.

Substitute these expressions into the two equations to check them. They work for every real $r,s$ , so these two vectors span the entire plane. We will develop a systematic way to solve equations like these in the next group of lessons.

One target, different settings

Keep the parallel controls and select Target (6, 3). Predict whether the target is reachable now. Find two different coefficient settings that reach it.

Compare your settings

Both $(a,b)=(3,0)$ and $(1,1)$ reach $(6,3)$ . In fact, every pair with $a+2b=3$ works. There is one reachable line but many descriptions of each point on it.

Adding more controls does not necessarily add more reachable targets. If a new vector is already a combination of the others, its effect can be reproduced using the existing controls. For example, adding $3\mathbf u$ to a list already containing $\mathbf u$ does not enlarge its span.

The zero vector is always reachable: choose all coefficients zero. If every available vector is zero, the span contains only that one point.

Prove a new case without searching

Using the parallel controls $(2,1)$ and $(4,2)$ , decide which targets are reachable: $(-2,-1)$ and $(2,-1)$ .

Check your reasoning

The first is reachable with $a=-1,b=0$ . The second violates $y=x/2$ and is impossible. Negative targets are not automatically unreachable; the direction relation is what matters.

In a model, several features may contribute the same underlying direction. More coefficients can give more ways to describe an output without giving the model more possible outputs.

Design your own comparison

Keep the parallel directions and choose your own target on paper, different from the worked targets. Predict whether it is reachable. If it is, find two coefficient pairs; if it is not, identify a coordinate relation that rules it out. Use arithmetic when the target is outside the explorer’s presets.

Now change only one target coordinate to reverse your reachability verdict. Keep both available directions fixed. Explain why the changed coordinate makes the difference.

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

Every reachable target has the form $(2c,c)$ . Such a target has multiple descriptions because only $a+2b=c$ matters. Changing one coordinate to break that relation makes it unreachable; repairing the relation makes it reachable. Slider search alone cannot establish impossibility.

Question 1 of 30 checked · 0 correct

Using u=(1, 3) and v=(2, 6), can a linear combination reach (2, 5)?Yes, because two controls reach the whole plane.No; every reachable point satisfies y=3x.No, because negative coefficients are forbidden.

Connect the first six discoveries

Use the coordinate directions $\mathbf e_1=(1,0)$ and $\mathbf e_2=(0,1)$ to reach $\mathbf w=(-3,4)$ . Without the explorer, answer these questions:

Which coefficients reach $\mathbf w$ ? Could $\mathbf e_1$ alone reach it?
What is the length of $\mathbf w$ , and what unit vector points in its direction?
What are $\mathbf w\cdot\mathbf e_1$ and the vector projection onto the horizontal line? Explain why one is a number and the other is a vector.

Compare the connected explanation

The coefficients are $-3$ and $4$ . The first direction alone always gives a zero second coordinate, so it cannot reach $\mathbf w$ . Its length is $5$ , and its unit direction is $(-3/5,4/5)$ . The dot product is the signed number $-3$ , while the projection vector is $(-3,0)$ . The reference vector has unit length, so the dot product equals the signed shadow length in this case.