Unit 5 · Lesson 3 of 7

Components describe direction and magnitude.

A displacement has both length and direction. Change its components and follow the arrow and its magnitude.

Change the input. Follow the mathematics.

What you'll be able to do

Set up the model and calculate its vector magnitude.
Connect the equation to the changing visual.
Explain the assumptions and identify a misleading interpretation.

Predict before exploring

What will change if you increase horizontal component? Predict the direction, then explain the response.

A model you can changeMove one control at a time.

Vector

Horizontal component3 mVertical component2 m

vector magnitude

|\mathbf v|=\sqrt{3^2+2^2}=3.60555\,\mathrm m

Components depend on coordinate axes, while vector magnitude does not. Vector addition is componentwise.

Assumptions, units, and limits

A two-dimensional displacement vector. Length is nonnegative; zero vector has no direction.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

From the situation to an equation

Resolve a vector along coordinate axes and use the Pythagorean norm for length. Add and scale vectors componentwise. A unit direction vector divides by magnitude, which requires a nonzero vector.

\mathbf v=\langle v_x,v_y\rangle,\quad|\mathbf v|=\sqrt{v_x^2+v_y^2}

Keep the reasoning visible

\hat{\mathbf v}=\mathbf v/|\mathbf v|\quad(\mathbf v\ne\mathbf0)

Components depend on coordinate axes, while vector magnitude does not. Vector addition is componentwise.

What makes this model useful?

A two-dimensional displacement vector. Length is nonnegative; zero vector has no direction.

Magnitude is a scalar, not another direction vector. A zero vector has no unique unit direction.

A worked example

Calculate the starting scenario

Use horizontal component = 3 m; vertical component = 2 m.
$|\mathbf v|=\sqrt{3^2+2^2}=3.60555\,\mathrm m$
Components depend on coordinate axes, while vector magnitude does not. Vector addition is componentwise. The checked quantity is vector magnitude.

Make the idea your own

Calculate. Compare. Explain.

Numeric checks allow 1% rounding error. Values near zero use a small absolute tolerance. Work through the setup before opening a solution.

Calculate and explain

Rebuild the starting calculation

Calculate the vector magnitude using horizontal component 3 m, vertical component 2 m. State the meaning of its sign and units.

Checked quantity: vector magnitude · m. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Resolve a vector along coordinate axes and use the Pythagorean norm for length. Add and scale vectors componentwise. A unit direction vector divides by magnitude, which requires a nonzero vector.

Explore the visual solution

Vector

Horizontal component3 mVertical component2 m

vector magnitude

|\mathbf v|=\sqrt{3^2+2^2}=3.60555\,\mathrm m

Components depend on coordinate axes, while vector magnitude does not. Vector addition is componentwise.

Assumptions, units, and limits

A two-dimensional displacement vector. Length is nonnegative; zero vector has no direction.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

Resolve a vector along coordinate axes and use the Pythagorean norm for length. Add and scale vectors componentwise. A unit direction vector divides by magnitude, which requires a nonzero vector.

\mathbf v=\langle v_x,v_y\rangle,\quad|\mathbf v|=\sqrt{v_x^2+v_y^2}

Calculate and explain

Apply the idea to a new scenario

For components (−3,4) m, find displacement magnitude.

Checked quantity: vector magnitude · m. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Resolve a vector along coordinate axes and use the Pythagorean norm for length. Add and scale vectors componentwise. A unit direction vector divides by magnitude, which requires a nonzero vector.

Explore the visual solution

Vector

Horizontal component-3 mVertical component4 m

vector magnitude

|\mathbf v|=\sqrt{-3^2+4^2}=5\,\mathrm m

Components depend on coordinate axes, while vector magnitude does not. Vector addition is componentwise.

Assumptions, units, and limits

A two-dimensional displacement vector. Length is nonnegative; zero vector has no direction.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

Resolve a vector along coordinate axes and use the Pythagorean norm for length. Add and scale vectors componentwise. A unit direction vector divides by magnitude, which requires a nonzero vector.

\mathbf v=\langle v_x,v_y\rangle,\quad|\mathbf v|=\sqrt{v_x^2+v_y^2}

Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual

Vector

Horizontal component3 mVertical component2 m

vector magnitude

|\mathbf v|=\sqrt{3^2+2^2}=3.60555\,\mathrm m

Components depend on coordinate axes, while vector magnitude does not. Vector addition is componentwise.

Assumptions, units, and limits

A two-dimensional displacement vector. Length is nonnegative; zero vector has no direction.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. Resolve a vector along coordinate axes and use the Pythagorean norm for length. Add and scale vectors componentwise. A unit direction vector divides by magnitude, which requires a nonzero vector.

Your learning path

Can you explain the connection?

Mark complete when you can set up the calculation and explain what the visual is showing. Opening a page never completes it automatically.

Completion is saved on this browser. This is separate from account practice progress.