Vector Representation - Statics and Dynamics

1. 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?
   Answer: $\hat{u}_R = \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$; angle with negative z-axis $\approx 125.3°$ 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.
2. 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?
   Answer: $\mathbf{F} = -86.6\hat{i} - 86.6\hat{j} - 86.6\hat{k} \ \text{N}$ 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."
3. 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?
   Answer: $\mathbf{T} = -150\hat{i} + 259.8\hat{j} + 400\hat{k} \ \text{N}$ 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.
4. 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 $\mathbf{F}$?
   Answer: 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. 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 $|\mathbf{F}|$.
5. 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 $\mathbf{R}$?
   Answer: Both are valid unit vectors; $\mathbf{R} = -10\hat{i} + 70\hat{j} + 98.99\hat{k} \ \text{N}$ 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.
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?
   Answer: $\mathbf{F} = 100\hat{i} + 141.4\hat{j} - 100\hat{k} \ \text{N}$ 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.
7. 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 $\mathbf{F}_3$?
   Answer: $\mathbf{F}_3 = 9\hat{i} + 12\hat{j} - 8\hat{k} \ \text{kN}$ 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.

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