CLASSICAL-MECHANICS · Interactive Practice

From Components to Coordinates: Vectors

IKey Formulas

Formula	Name	Description
$\vec{A} = A_x\,\hat{i} + A_y\,\hat{j} + A_z\,\hat{k}$	Components	Vector by its components
$\lvert\vec{A}\rvert = \sqrt{A_x^2 + A_y^2 + A_z^2}$	Magnitude	Square, add, square root
$\vec{A}\pm\vec{B} = (A_x\pm B_x,\ A_y\pm B_y,\ A_z\pm B_z)$	Add / Subtract	Component by component
$\hat{A} = \dfrac{\vec{A}}{\lvert\vec{A}\rvert}$	Unit vector	Direction, length 1
$(x, y) = (r\cos\theta,\ r\sin\theta)$	Polar form	Angle $\theta$ from $+x$ axis

IIVisualization 1 — Components, Magnitude & Direction

Two perpendicular components fix both how long a vector is and which way it points.

Ax 3 Ay 4

IIIVisualization 2 — Bearings to Standard Angles

Why must a compass bearing become a standard angle from $+x$ before it goes into $r\cos\theta$ and $r\sin\theta$ ?

Bearing Angle b 52 Distance r 5

💡 Plug the raw bearing straight into $\cos$ and $\sin$ and you land in the wrong quadrant — convert to $\theta$ from $+x$ first.

IVVisualization 3 — Point on a Segment

As a point slides from $P_1$ to $P_2$ , its position vector is a weighted average of the two endpoints.

a / d 0.40

VQuiz Questions

Question 1

Given the vector $\vec{A} = 2\,\hat{i} - 3\,\hat{j} + 7\,\hat{k}$ , what is its magnitude $\lvert\vec{A}\rvert$ ?

✅ Correct! $\sqrt{4+9+49} = \sqrt{62}$ .

❌ Not quite. Remember to square each component (including the negative one) before adding.

Show solution

Solution:

Square each component, add, then take the square root: $\lvert\vec{A}\rvert = \sqrt{2^2 + (-3)^2 + 7^2} = \sqrt{4 + 9 + 49} = \sqrt{62} \approx 7.87$

The negative component squares to a positive value — magnitudes are never negative.

Question 2

A vector points 18° south of east at a distance of 24 km from the origin (east is $+x$ , north is $+y$ ). What standard angle $\theta$ , measured from the positive x-axis, should you use in $r\cos\theta$ and $r\sin\theta$ ?

✅ Correct! South of east is below the +x axis, so θ = -18°.

❌ Not quite. 'South of east' rotates below the positive x-axis — that is a negative angle.

Show solution

Solution:

"South of east" means start on the positive x-axis (east) and rotate downward (toward south) by 18°. Measuring from the positive x-axis, a downward rotation is negative: $\theta = -18°$

A common mistake is to plug the compass number $+18°$ straight in, which would point above the x-axis instead of below it.

Question 3

A Coast Guard ship is at position $\vec{r_1} = -21.5\,\hat{i} + 27.6\,\hat{j}$ km and a sailboat is at $\vec{r_2} = 22.8\,\hat{i} - 7.4\,\hat{j}$ km. What is the displacement vector $\vec{r_2} - \vec{r_1}$ the ship must travel?

✅ Correct! Subtracting components gives 44.3 i - 35.0 j km.

❌ Not quite. Subtract component by component: be careful with the double negative in x (22.8 − (−21.5)).

Show solution

Solution:

Subtract component by component, $\vec{r_2} - \vec{r_1}$ :

$x:\quad 22.8 - (-21.5) = 44.3 \text{ km}$ $y:\quad -7.4 - 27.6 = -35.0 \text{ km}$

So $\vec{r_2} - \vec{r_1} = 44.3\,\hat{i} - 35.0\,\hat{j}$ km.

The magnitude is $\sqrt{44.3^2 + 35.0^2} \approx 56.5$ km, and the heading is $\tan^{-1}(-35.0/44.3) \approx -38.3°$ , i.e. 38.3° south of east.

Question 4

The position vector of a point on the segment from $P_1$ to $P_2$ is $\vec{A} = \left(1 - \tfrac{a}{d}\right)\vec{r_1} + \tfrac{a}{d}\,\vec{r_2}$ where $a$ is the distance from $P_1$ and $d$ is the segment length.

True or False: When $a = d$ , the formula gives $\vec{A} = \vec{r_2}$ (the point sits exactly at $P_2$ ).

✅ Correct! With a/d = 1 the first term vanishes, leaving A = r₂.

❌ Not quite. Plug in a/d = 1: the coefficient of r₁ becomes 0 and r₂'s becomes 1.

Show solution

Solution:

Substitute $a = d$ , so $a/d = 1$ : $\vec{A} = (1 - 1)\,\vec{r_1} + (1)\,\vec{r_2} = \vec{r_2}$

The point sits exactly at $P_2$ . (Similarly, $a = 0$ gives $\vec{A} = \vec{r_1}$ .) This confirms the formula is a weighted average that slides smoothly between the two endpoints — so the statement is True.

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