CLASSICAL-MECHANICS · Interactive Practice | Unit 2 · Video 3

Three Numbers Tell You Everything: Vectors in Cartesian Coordinates

IKey Formulas

Formula	Name	Description
$\vec{A} = A_x\,\hat{i} + A_y\,\hat{j} + A_z\,\hat{k}$	Decomposition	Three scalar components, each with a sign
$A = \sqrt{A_x^2 + A_y^2 + A_z^2}$	Magnitude	Pythagoras in 3D (always $\geq 0$ )
$\theta = \tan^{-1}\!\left(\frac{A_y}{A_x}\right)$	Direction (in $xy$ -plane)	Verify the quadrant from signs of $A_x,\ A_y$
$C_x = A_x + B_x,\quad C_y = A_y + B_y$	Component-wise addition	Add vectors arithmetically

IIInteractive Visualizations

Visualization 1 — Components, Magnitude, and Direction

Two signed components fix a vector completely: its length by Pythagoras, its angle by the inverse tangent.

Aₓ 3 Aᵧ 4

Visualization 2 — Why the Inverse Tangent Is Ambiguous

The ratio $A_y/A_x$ cannot name a direction: $\vec{A}$ and $-\vec{A}$ share it yet point exactly opposite.

Aₓ 3 Aᵧ 2

Visualization 3 — Adding Vectors Component by Component

Component addition turns the tip-to-tail picture into arithmetic: $C_x = A_x + B_x,\ C_y = A_y + B_y$ .

Aₓ 3 Aᵧ 4 Bₓ -1 Bᵧ 2

💡 Challenge: set $\vec{B}$ so the resultant vanishes, $\vec{C} = \vec{0}$ — what must $\vec{B}$ equal?

IIIQuiz Questions

Question 1

A vector lies in the $xy$ -plane with components $A_x = 6$ and $A_y = 8$ .

What is the magnitude $|\vec{A}|$ ?

✅ Correct! $\sqrt{36 + 64} = \sqrt{100} = 10$ .

❌ Not quite. Square each component, add, then take the square root — don't just add the components.

Show solution

Solution:

Use the Pythagorean magnitude formula: $|\vec{A}| = \sqrt{A_x^2 + A_y^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$

The answer is |A| = 10.

Question 2

A vector has components $A_x = -3$ and $A_y = 3$ .

A student computes $\tan^{-1}(A_y/A_x) = \tan^{-1}(-1) = -45\degree$ and reports the direction as $-45\degree$ . What is the correct direction angle $\theta$ measured counterclockwise from the positive $x$ -axis?

✅ Correct! Negative $A_x$ , positive $A_y$ means quadrant II, so $\theta = 135\degree$ .

❌ Not quite. Check the signs: $A_x < 0$ , $A_y > 0$ puts the vector in quadrant II. The calculator's $-45\degree$ must be corrected.

Show solution

Solution:

The signs are $A_x < 0$ and $A_y > 0$ , which places the vector in the second quadrant ( $90\degree < \theta < 180\degree$ ).

The inverse tangent only returns a value between $-90\degree$ and $90\degree$ , so the raw $-45\degree$ is off by $180\degree$ . Add $180\degree$ to land in the correct quadrant: $\theta = -45\degree + 180\degree = 135\degree$

The answer is $\theta = 135\degree$ .

Question 3

Given $\vec{A} = 3\hat{i} + 4\hat{j}$ and $\vec{B} = -1\hat{i} + 2\hat{j}$ , compute the resultant $\vec{C} = \vec{A} + \vec{B}$ .

Which expression is correct?

✅ Correct! $C_x = 3 + (-1) = 2$ and $C_y = 4 + 2 = 6$ .

❌ Not quite. Add the $\hat{i}$ components together and the $\hat{j}$ components together — keep track of the negative sign in $B_x$ .

Show solution

Solution:

Add the components separately: $C_x = A_x + B_x = 3 + (-1) = 2$ $C_y = A_y + B_y = 4 + 2 = 6$

So $\vec{C} = 2\hat{i} + 6\hat{j}$

The answer is $\vec{C} = 2\hat{i} + 6\hat{j}$ .

Question 4

True or False: For a vector $\vec{A} = A_x\hat{i} + A_y\hat{j}$ , the scalar component $A_x$ is always equal to the magnitude of the component vector $\vec{A}_x = A_x\hat{i}$ .

✅ Correct! The component $A_x$ carries a sign, while the magnitude $|\vec{A}_x| = |A_x|$ is never negative.

❌ Not quite. Consider a negative component such as $A_x = -3$ : its magnitude is $3$ , not $-3$ .

Show solution

Solution:

The scalar component $A_x$ can be positive, zero, or negative. The magnitude of the component vector is $|\vec{A}_x| = \sqrt{A_x^2} = |A_x| \geq 0,$ which is always non-negative.

When $A_x < 0$ , the magnitude $|A_x|$ differs from $A_x$ . For example, if $A_x = -3$ , then $A_x = -3$ but $|\vec{A}_x| = 3$ . They are equal only when $A_x \geq 0$ , so the blanket statement is False.

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