Statics and Dynamics Quiz: Vector Representation

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Vector RepresentationQuestion 1 of 7

A position vector $\mathbf{r}$ in 3D space makes equal angles with all three positive coordinate axes. A force of magnitude 150 N acts along the direction defined by $\mathbf{r}$ , but in the direction opposite to $\mathbf{r}$ 's orientation (i.e., toward the origin). Which expression correctly gives this force in Cartesian unit-vector notation?

\mathbf{F} = -\frac{150}{\sqrt{2}}\hat{i} - \frac{150}{\sqrt{2}}\hat{j} - \frac{150}{\sqrt{2}}\hat{k} \ \text{N}

\mathbf{F} = 86.6\hat{i} + 86.6\hat{j} + 86.6\hat{k} \ \text{N}

\mathbf{F} = -50\hat{i} - 50\hat{j} - 50\hat{k} \ \text{N}

\mathbf{F} = -86.6\hat{i} - 86.6\hat{j} - 86.6\hat{k} \ \text{N}

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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Vector Representation

Practice Vector Representation in Statics and Dynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Vector Representation, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$\mathbf{F} = -\frac{150}{\sqrt{2}}\hat{i} - \frac{150}{\sqrt{2}}\hat{j} - \frac{150}{\sqrt{2}}\hat{k} \ \text{N}$
$\mathbf{F} = 86.6\hat{i} + 86.6\hat{j} + 86.6\hat{k} \ \text{N}$
$\mathbf{F} = -50\hat{i} - 50\hat{j} - 50\hat{k} \ \text{N}$
$\mathbf{F} = -86.6\hat{i} - 86.6\hat{j} - 86.6\hat{k} \ \text{N}$ (correct answer)

Explanation: Whenever you see a force described by a direction in space, your first move should always be to find the unit vector along that direction, then scale it by the magnitude. If a vector makes equal angles with all three positive coordinate axes, its direction cosines are all equal:

\cos\alpha = \cos\beta = \cos\gamma

. Since direction cosines must satisfy

\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1

, you get

3\cos^2\alpha = 1

, so

\cos\alpha = \frac{1}{\sqrt{3}}

. The unit vector along r is therefore

\hat{u} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}

. Because the force acts opposite to this direction, you negate the unit vector and multiply by 150 N:

\mathbf{F} = -\frac{150}{\sqrt{3}}\hat{i} - \frac{150}{\sqrt{3}}\hat{j} - \frac{150}{\sqrt{3}}\hat{k} \approx -86.6\hat{i} - 86.6\hat{j} - 86.6\hat{k} \ \text{N}

This confirms D is correct. A uses

\sqrt{2}

instead of

\sqrt{3}

in the denominator — a classic error from forgetting there are three axes contributing, not two. B has the right magnitude per component but the wrong sign; it points away from the origin, not toward it. C uses 50 N per component, which would only be correct if the components summed to 150 N directly (i.e.,

50 \times 3 = 150

) — but force magnitude is found using the Pythagorean theorem, not simple addition. Study tip: Always build your unit vector from the constraint

\sum\cos^2\theta_i = 1

before scaling by magnitude — and never confuse "sum of components" with "magnitude of the vector."

Question 2

Three forces act on a particle: $\mathbf{F}_1 = 10\hat{i} + 0\hat{j} + 0\hat{k}$ N, $\mathbf{F}_2 = 0\hat{i} + 10\hat{j} + 0\hat{k}$ N, and $\mathbf{F}_3 = 0\hat{i} + 0\hat{j} + 10\hat{k}$ N.

A resultant force $\mathbf{R}$ is the sum of two forces: $\mathbf{F}_1$ , which has magnitude 60 N and acts along the unit vector $\hat{u}_1 = 0.5\hat{i} + 0.5\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$ , and $\mathbf{F}_2$ , which has magnitude 80 N and acts along the unit vector $\hat{u}_2 = -0.5\hat{i} + 0.5\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$ . Before computing the resultant, a student checks whether $\hat{u}_1$ and $\hat{u}_2$ are valid unit vectors. Which conclusion is correct, and what is the Cartesian form of $\mathbf{R}$ ?

$\hat{u}_1$ and $\hat{u}_2$ are not valid unit vectors because the sum of their components equals $0.5+0.5+\frac{1}{\sqrt{2}} \approx 1.707 \neq 1$ ; consequently, the resultant cannot be determined without re-normalizing.
Both are valid unit vectors; $\mathbf{R} = -10\hat{i} + 70\hat{j} + 98.99\hat{k} \ \text{N}$ (correct answer)
Both are valid unit vectors; $\mathbf{R} = 10\hat{i} + 70\hat{j} + 98.99\hat{k} \ \text{N}$
Both are valid unit vectors; $\mathbf{R} = -10\hat{i} + 70\hat{j} + 84.85\hat{k} \ \text{N}$

Explanation: Whenever you see a question involving force vectors along specified directions, your first instinct should be to verify the unit vectors before computing anything. A unit vector must satisfy

|\hat{u}| = \sqrt{u_x^2 + u_y^2 + u_z^2} = 1

— it's the sum of squares, not the sum of components, that must equal 1. Check

\hat{u}_1 = 0.5\hat{i} + 0.5\hat{j} + \frac{1}{\sqrt{2}}\hat{k}

|\hat{u}_1| = \sqrt{(0.5)^2 + (0.5)^2 + \left(\frac{1}{\sqrt{2}}\right)^2} = \sqrt{0.25 + 0.25 + 0.5} = \sqrt{1} = 1 \checkmark

The same calculation confirms

|\hat{u}_2| = 1

. Both are valid. Now compute the resultant by multiplying each magnitude by its unit vector and adding:

\mathbf{F}_1 = 60\hat{u}_1 = 30\hat{i} + 30\hat{j} + \frac{60}{\sqrt{2}}\hat{k}

\mathbf{F}_2 = 80\hat{u}_2 = -40\hat{i} + 40\hat{j} + \frac{80}{\sqrt{2}}\hat{k}

\mathbf{R} = (30-40)\hat{i} + (30+40)\hat{j} + \frac{140}{\sqrt{2}}\hat{k} = -10\hat{i} + 70\hat{j} + 98.99\hat{k} \ \text{N}

This confirms B is correct. Choice A commits the classic blunder of summing components instead of computing the magnitude — a fundamental error in vector mathematics. Choice C gets the

\hat{k}

component right but incorrectly makes the

\hat{i}

component positive, likely by dropping the negative sign from

\hat{u}_2

. Choice D correctly identifies the

\hat{i}

and

\hat{j}

components but miscalculates

\frac{140}{\sqrt{2}}

, perhaps using

60 + 80 = 140

divided by 2 instead of

\sqrt{2}

. Study tip: Always validate unit vectors using the sum-of-squares rule, and when multiplying a magnitude by a unit vector, carefully carry through every sign — sign errors are the most common source of wrong answers in vector addition problems.

Question 3

Two forces act on a particle: $\mathbf{F}_1 = 3\hat{i} - 4\hat{j} + 0\hat{k}$ kN and $\mathbf{F}_2 = -6\hat{i} + 0\hat{j} + 8\hat{k}$ kN.

A third force $\mathbf{F}_3$ is to be added such that the unit vector in the direction of the resultant $\mathbf{R} = \mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3$ is $\hat{u}_R = 0.6\hat{i} + 0.8\hat{j} + 0\hat{k}$ . If $|\mathbf{R}| = 10$ kN, what is $\mathbf{F}_3$ ?

$\mathbf{F}_3 = 9\hat{i} + 12\hat{j} - 8\hat{k} \ \text{kN}$ (correct answer)
$\mathbf{F}_3 = 9\hat{i} + 12\hat{j} + 8\hat{k} \ \text{kN}$
$\mathbf{F}_3 = 3\hat{i} + 12\hat{j} - 8\hat{k} \ \text{kN}$
$\mathbf{F}_3 = 9\hat{i} + 8\hat{j} - 8\hat{k} \ \text{kN}$

Explanation: When a question gives you a unit vector and a magnitude for the resultant, your first move should always be to reconstruct the resultant vector itself, then work backwards to find the unknown force. This is a classic vector decomposition problem. Since

\hat{u}_R = 0.6\hat{i} + 0.8\hat{j} + 0\hat{k}

and

|\mathbf{R}| = 10

kN, the resultant vector is simply their product:

\mathbf{R} = 10(0.6\hat{i} + 0.8\hat{j} + 0\hat{k}) = 6\hat{i} + 8\hat{j} + 0\hat{k}

kN. Next, add the two known forces:

\mathbf{F}_1 + \mathbf{F}_2 = (3-6)\hat{i} + (-4+0)\hat{j} + (0+8)\hat{k} = -3\hat{i} - 4\hat{j} + 8\hat{k}

kN. Now solve for

\mathbf{F}_3 = \mathbf{R} - (\mathbf{F}_1 + \mathbf{F}_2) = (6-(-3))\hat{i} + (8-(-4))\hat{j} + (0-8)\hat{k} = 9\hat{i} + 12\hat{j} - 8\hat{k}

kN. That confirms A is correct. Choice B flips the sign on the

\hat{k}

component to

+8

, which would mean

\mathbf{F}_3

doesn't cancel the

+8\hat{k}

contribution from

\mathbf{F}_2

— the resultant would have a nonzero

z

-component, contradicting

\hat{u}_R

. Choice C uses

3\hat{i}

instead of

9\hat{i}

, likely from forgetting to account for the negative

x

-components in

\mathbf{F}_1 + \mathbf{F}_2

. Choice D uses

8\hat{j}

instead of

12\hat{j}

, a sign error when subtracting the

-4\hat{j}

term. A reliable strategy: always reconstruct

\mathbf{R}

from the unit vector and magnitude first, then subtract the known forces component by component. Track signs carefully — that's where most errors on this type of problem occur.

Question 4

A force $\mathbf{F}$ is expressed in Cartesian form as $\mathbf{F} = F_0(2\hat{i} - 1\hat{j} + 2\hat{k})$ , where $F_0$ is a scalar constant with units of Newtons.

An engineer states: 'The expression $2\hat{i} - 1\hat{j} + 2\hat{k}$ is a unit vector, so $F_0$ directly equals the magnitude of $\mathbf{F}$ .' Which response most precisely and correctly evaluates this claim, and what is the true magnitude of $\mathbf{F}$ ?

The claim is false; $|2\hat{i}-\hat{j}+2\hat{k}|=\sqrt{3}$ , so the true magnitude is $|\mathbf{F}|=F_0\sqrt{3}$ N, and $F_0$ must be divided by $\sqrt{3}$ to obtain the unit-vector form.
The claim is true; $2\hat{i}-\hat{j}+2\hat{k}$ has components that sum to $2-1+2=3$ , but since the sum of squares equals $9$ , the magnitude of the vector $\mathbf{F}$ is $3F_0$ N, confirming $F_0$ is the magnitude per unit length.
The claim is false; $|2\hat{i}-\hat{j}+2\hat{k}|=3$ , so the true magnitude is $|\mathbf{F}|=3F_0$ N, meaning $F_0$ is one-third the force magnitude, not equal to it. (correct answer)
The claim is false; the vector $2\hat{i}-\hat{j}+2\hat{k}$ is not a unit vector because it has three components rather than one, so the magnitude of $\mathbf{F}$ cannot be determined without knowing the coordinate system orientation.

Explanation: Whenever you see a force written as a scalar multiplied by a vector expression, your first instinct should be to check whether that vector expression is actually a unit vector — because only then does the scalar directly equal the force's magnitude. To test this, compute the magnitude of

2\hat{i} - \hat{j} + 2\hat{k}

using the standard formula:

\sqrt{(2)^2 + (-1)^2 + (2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3

. Since this magnitude equals 3, not 1, the vector is not a unit vector. Therefore

|\mathbf{F}| = F_0 \times 3 = 3F_0

N, and

F_0

represents only one-third of the force's true magnitude. This confirms C is correct — the engineer's claim is false, and the magnitude is

3F_0

. Choice A gets the logic right (correctly identifying the claim as false) but makes an arithmetic error, computing the magnitude as

\sqrt{3}

instead of the correct value of 3. Double-check your arithmetic under exam pressure. Choice B is doubly wrong: it calls the claim true while simultaneously acknowledging the magnitude is

3F_0

— a direct contradiction — and the phrase "magnitude per unit length" is meaningless in this context. Choice D invents a false rule; the number of components in a vector has nothing to do with whether it's a unit vector, and magnitude is independent of coordinate system orientation. Your takeaway: always verify a purported unit vector by computing

\sqrt{F_x^2 + F_y^2 + F_z^2}

. If the result isn't exactly 1, the scalar prefactor is not the magnitude — multiply by the actual magnitude of the direction vector to find

|\mathbf{F}|

Question 5

Three forces act on a particle: $\mathbf{F}_1 = 10\hat{i} + 0\hat{j} + 0\hat{k}$ N, $\mathbf{F}_2 = 0\hat{i} + 10\hat{j} + 0\hat{k}$ N, and $\mathbf{F}_3 = 0\hat{i} + 0\hat{j} + 10\hat{k}$ N.

What is the unit vector in the direction of the resultant $\mathbf{R} = \mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3$ , and which direction angle (rounded to one decimal place) does R make with the negative z-axis?

$\hat{u}_R = \frac{1}{3}(\hat{i}+\hat{j}+\hat{k})$ ; angle with negative z-axis $\approx 125.3°$
$\hat{u}_R = \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$ ; angle with negative z-axis $\approx 54.7°$
$\hat{u}_R = \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$ ; angle with negative z-axis $\approx 125.3°$ (correct answer)
$\hat{u}_R = \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$ ; angle with negative z-axis $\approx 35.3°$

Explanation: When working with resultant forces and direction angles in 3D, your two key tasks are: (1) finding the unit vector by dividing R by its magnitude, and (2) carefully identifying which axis the problem references for the direction angle. Start by adding the three forces component-wise:

\mathbf{R} = 10\hat{i} + 10\hat{j} + 10\hat{k}

N. The magnitude is

|\mathbf{R}| = \sqrt{10^2 + 10^2 + 10^2} = \sqrt{300} = 10\sqrt{3}

N. Dividing R by its magnitude gives the unit vector

\hat{u}_R = \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})

, confirming the scalar is

\frac{1}{\sqrt{3}}

, not

\frac{1}{3}

. Now for the direction angle. The angle R makes with the positive z-axis uses

\cos\gamma = \frac{R_z}{|\mathbf{R}|} = \frac{1}{\sqrt{3}}

, giving

\gamma \approx 54.7°

. The negative z-axis points opposite to

+z

, so the angle with the negative z-axis is

180° - 54.7° \approx 125.3°

. That confirms C as correct. Choice A uses

\frac{1}{3}

as the scalar — a classic error where students forget to take the square root when computing magnitude, effectively using

\sqrt{3}

instead of

\sqrt{3}

... or simply squaring rather than rooting. Choice B gets the unit vector right but reports the angle with the positive z-axis (

54.7°

), not the negative one — a careful reading failure. Choice D gives yet another incorrect angle (

35.3°

), which has no geometric basis here. A reliable tip: whenever a direction angle question specifies a negative axis, compute the angle with the positive axis first using the dot product, then subtract from

180°

to flip the reference direction.

Question 6

A force is described by its direction angles: $\alpha = 60°$ (angle with the positive x-axis), $\beta = 45°$ (angle with the positive y-axis), and an unknown angle $\gamma$ with the positive z-axis. If the z-component of the force is negative and the force magnitude is $F = 200$ N, what is the Cartesian representation of the force?

$\mathbf{F} = 100\hat{i} + 141.4\hat{j} - 100\hat{k} \ \text{N}$ (correct answer)
$\mathbf{F} = 100\hat{i} + 141.4\hat{j} + 100\hat{k} \ \text{N}$
$\mathbf{F} = 100\hat{i} + 141.4\hat{j} - 122.5\hat{k} \ \text{N}$
$\mathbf{F} = 141.4\hat{i} + 100\hat{j} - 100\hat{k} \ \text{N}$

Explanation: When a force is defined by direction angles, the key relationship to remember is the direction cosine identity:

\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1

. This constraint links all three angles, so if two are known, the third is determined — up to a sign. Here,

\alpha = 60°

and

\beta = 45°

, so:

\cos^2(60°) + \cos^2(45°) + \cos^2\gamma = 1

0.25 + 0.5 + \cos^2\gamma = 1 \implies \cos^2\gamma = 0.25 \implies \cos\gamma = \pm 0.5

. Since the z-component is negative, you take

\cos\gamma = -0.5

, giving

\gamma = 120°

. The Cartesian components are then:

F_x = 200\cos(60°) = 100 \ \text{N}

F_y = 200\cos(45°) = 141.4 \ \text{N}

F_z = 200(-0.5) = -100 \ \text{N}

. This confirms Answer A is correct. Answer B ignores the problem's sign condition and uses

\cos\gamma = +0.5

, giving a positive z-component — a direct contradiction of the given information. Answer C uses an incorrect z-component of

-122.5

N. You can verify this fails the identity:

100^2 + 141.4^2 + 122.5^2 \neq 200^2

, meaning the magnitude would no longer be 200 N. Answer D swaps the x- and y-components (141.4 vs. 100), confusing which angle corresponds to which axis —

\alpha

goes with x,

\beta

with y, always. Your go-to strategy: always verify your components satisfy

F_x^2 + F_y^2 + F_z^2 = F^2

. This quick magnitude check catches sign errors and swapped components before they cost you points.

Question 7

A cable exerts a tension of 500 N on an anchor point. The cable runs from the anchor to a point described in cylindrical coordinates as $(r, \theta, z) = (3, 120°, 4)$ m, where the anchor is at the origin. Expressing the tension force as a Cartesian vector (from anchor toward the cable attachment point), which answer is correct?

$\mathbf{T} = -150\hat{i} + 259.8\hat{j} - 400\hat{k} \ \text{N}$
$\mathbf{T} = 150\hat{i} + 259.8\hat{j} + 400\hat{k} \ \text{N}$
$\mathbf{T} = -259.8\hat{i} + 150\hat{j} + 400\hat{k} \ \text{N}$
$\mathbf{T} = -150\hat{i} + 259.8\hat{j} + 400\hat{k} \ \text{N}$ (correct answer)

Explanation: When working with cylindrical coordinates

(r, \theta, z)

, your first task is always to convert the position to Cartesian form before computing a unit vector. The conversion formulas are

x = r\cos\theta

y = r\sin\theta

, and

z = z

. With

(r, \theta, z) = (3, 120°, 4)

, the attachment point in Cartesian coordinates is:

x = 3\cos(120°) = 3(-0.5) = -1.5 \text{ m}

y = 3\sin(120°) = 3(0.866) = 2.598 \text{ m}

z = 4 \text{ m}

So the position vector from the anchor to the attachment point is

\mathbf{r} = -1.5\hat{i} + 2.598\hat{j} + 4\hat{k}

. Its magnitude is

|\mathbf{r}| = \sqrt{(-1.5)^2 + (2.598)^2 + 4^2} = \sqrt{2.25 + 6.75 + 16} = \sqrt{25} = 5 \text{ m}

. The unit vector is

\hat{u} = \frac{\mathbf{r}}{5}

, and multiplying by 500 N gives

\mathbf{T} = -150\hat{i} + 259.8\hat{j} + 400\hat{k} \text{ N}

, confirming D. Choice A flips the sign on the

z

-component, which would point the cable downward — a sign error with no geometric justification. Choice B uses

\cos(120°) = +0.5

instead of

-0.5

, the classic mistake of forgetting that 120° lies in the second quadrant where

x

is negative. Choice C swaps the

x

and

y

components, suggesting confusion between sine and cosine in the conversion formulas. Your study tip: always sketch the angle's quadrant before plugging into trig. Second-quadrant angles (90°–180°) always have negative

x

(cosine) and positive

y

(sine) — a sign error here cascades through the entire problem.