Vector Addition: Analytical Methods Quiz

Graphical vector addition relies on rulers and protractors, and it carries an irreducible reading error. Analytical methods replace that ritual with arithmetic: resolve each vector into perpendicular components, sum those components independently, then reassemble the resultant using the Pythagorean theorem and an inverse tangent.

This quiz tests the three-step rhythm — resolve, sum, reassemble — along with the trigonometry, sign conventions, and quadrant reasoning that make it work.

What you will practise

Resolving vectors with ##A_x = A\cos\theta## and ##A_y = A\sin\theta##
Adding components along ##x## and ##y## independently
Finding magnitude with ##R = \sqrt{R_x^2 + R_y^2}##
Finding direction with ##\theta = \tan^{-1}\!\left(\dfrac{R_y}{R_x}\right)## and correcting for quadrants
Writing vectors in unit vector notation ##\hat{i}, \hat{j}##
Recognising why analytical addition beats graphical construction in precision

Difficulty: Medium. Twenty questions, answerable from the lesson alone.

Q1. A vector ##\vec{A}## of magnitude ##A## makes an angle ##\theta## measured from the positive x-axis. Which pair of expressions gives its horizontal and vertical components?

##A_x = A\sin\theta,\ A_y = A\cos\theta##

##A_x = A\cos\theta,\ A_y = A\sin\theta##

##A_x = A\tan\theta,\ A_y = A\cos\theta##

##A_x = A\cos\theta,\ A_y = A\tan\theta##

The lesson states that a vector at angle theta from the positive x-axis splits into a horizontal component ##A\cos\theta## and a vertical component ##A\sin\theta##.

Q2. Which statement best captures the central insight of analytical vector addition?

Once decomposed, the geometry of arrows dissolves into arithmetic of scalars

Vectors can only be added if they are perpendicular

The resultant always has magnitude equal to the sum of magnitudes

Components must be found with a protractor before adding

The lesson says that once a vector is decomposed, the geometry of arrows dissolves into arithmetic of scalars, and the components behave like ordinary signed numbers.

Q3. Two vectors are written as ##\vec{A} = A_x\hat{i} + A_y\hat{j}## and ##\vec{B} = B_x\hat{i} + B_y\hat{j}##. What is ##\vec{R} = \vec{A} + \vec{B}##?

##\vec{R} = (A_x B_x)\hat{i} + (A_y B_y)\hat{j}##

##\vec{R} = (A_x - B_x)\hat{i} + (A_y - B_y)\hat{j}##

##\vec{R} = (A_x + B_x)\hat{i} + (A_y + B_y)\hat{j}##

##\vec{R} = (A_x + B_y)\hat{i} + (A_y + B_x)\hat{j}##

The sum collects like unit vectors: the ##\hat{i}## terms add together and the ##\hat{j}## terms add together, giving ##(A_x + B_x)\hat{i} + (A_y + B_y)\hat{j}##.

Q4. Why may the x and y components of a set of vectors be summed completely independently of each other?

Because components are always positive numbers

Because perpendicular directions are independent, an exact consequence of Euclidean geometry

Because the resultant is always horizontal

Because sine and cosine are equal for all angles

The orthogonality principle: perpendicular directions are independent. A displacement eastward has no effect on northward position, which licenses splitting a 2D problem into two 1D problems.

Q5. A vector has components ##R_x = 3## and ##R_y = 4##. What is the magnitude of the resultant?

##7##

##5##

##1##

##25##

Magnitude is reassembled with the Pythagorean theorem: ##R = \sqrt{R_x^2 + R_y^2} = \sqrt{9 + 16} = \sqrt{25} = 5##.

Q6. A resultant has components ##R_x = 1## and ##R_y = 1##. What angle does it make with the positive x-axis?

##30^{\circ}##

##45^{\circ}##

##60^{\circ}##

##90^{\circ}##

##\theta = \tan^{-1}\!\left(\dfrac{R_y}{R_x}\right) = \tan^{-1}(1) = 45^{\circ}##, and since both components are positive the vector lies in the first quadrant.

Q7. A resultant has components ##R_x = -3## and ##R_y = 3##. In which quadrant does the resultant lie?

First quadrant

Second quadrant

Third quadrant

Fourth quadrant

With ##R_x 0## the vector points up and to the left, which is the second quadrant.

Q8. A resultant has components ##R_x = -4## and ##R_y = -4##. A calculator returns ##\tan^{-1}\!\left(\dfrac{-4}{-4}\right) = 45^{\circ}##. What is the correct direction measured from the positive x-axis?

##45^{\circ}##

##135^{\circ}##

##225^{\circ}##

##315^{\circ}##

The inverse tangent cannot distinguish opposite quadrants. Both components are negative, so the vector lies in the third quadrant: ##180^{\circ} + 45^{\circ} = 225^{\circ}##.

Q9. Which sequence correctly describes the three-step rhythm of analytical vector addition?

Resolve, sum, reassemble

Sum, resolve, reassemble

Reassemble, resolve, sum

Resolve, reassemble, sum

The lesson names the rhythm explicitly: resolve each vector into perpendicular parts, sum those parts independently, then reassemble the result through the Pythagorean theorem and an inverse tangent.

Q10. What is a principal advantage of analytical vector addition over graphical construction?

It requires no trigonometry

It can provide greater precision than graphical construction, subject to the precision of the inputs and calculations

It works only for perpendicular vectors

It avoids the need for components

Analytical addition avoids the measurement uncertainty inherent in drawing and reading a graphical construction. Its accuracy still depends on the precision of the input values and calculations.

Q11. According to the lesson, roughly how finely can a protractor resolve an angle in graphical vector addition?

To about ##0.01^{\circ}##

To about half a degree

To about ##5^{\circ}##

To about ##10^{\circ}##

The lesson states that a protractor cannot resolve an angle finer than about half a degree, whereas analytical methods have no such ceiling.

Q12. A pencil line is drawn to the nearest millimetre using a scale of ##1## unit per centimetre. Which pair of resultant magnitudes would differ by less than ##1## millimetre on the drawing and therefore could not reliably be distinguished by that measurement?

##12.4## units and ##12.45## units

##12## units and ##20## units

##5## units and ##15## units

##1## unit and ##9## units

At ##1## unit per centimetre, the difference of ##0.05## units corresponds to ##0.5## millimetres on the drawing, less than the stated ##1## millimetre measurement resolution. The other pairs have much larger differences.

Q13. A vector of magnitude ##10## makes an angle of ##30^{\circ}## with the positive x-axis. What are its components?

##A_x = 5,\ A_y = 5##

##A_x = 5\sqrt{3},\ A_y = 5##

##A_x = 5,\ A_y = 5\sqrt{3}##

##A_x = 10,\ A_y = 10##

##A_x = 10\cos 30^{\circ} = 10 \times \dfrac{\sqrt{3}}{2} = 5\sqrt{3}## and ##A_y = 10\sin 30^{\circ} = 10 \times \dfrac{1}{2} = 5##.

Q14. Two vectors have components ##\vec{A} = 2\hat{i} + 3\hat{j}## and ##\vec{B} = 4\hat{i} + 1\hat{j}##. What is the resultant in unit vector notation?

##6\hat{i} + 4\hat{j}##

##2\hat{i} + 2\hat{j}##

##8\hat{i} + 3\hat{j}##

##6\hat{i} + 2\hat{j}##

Add like unit vectors: ##(2 + 4)\hat{i} + (3 + 1)\hat{j} = 6\hat{i} + 4\hat{j}##.

Q15. A resultant has components ##R_x = 3## and ##R_y = 3\sqrt{3}##. What angle does it make with the positive x-axis?

##30^{\circ}##

##45^{\circ}##

##60^{\circ}##

##120^{\circ}##

##\theta = \tan^{-1}\!\left(\dfrac{3\sqrt{3}}{3}\right) = \tan^{-1}(\sqrt{3}) = 60^{\circ}##, and both components are positive so the vector is in the first quadrant.

Q16. Which statement about the unit vectors ##\hat{i}## and ##\hat{j}## is correct according to the lesson?

They are parallel to each other

They are orthogonal, so no terms mix between them

They have magnitudes equal to the vector's magnitude

They can only be used for horizontal vectors

The lesson states that ##\hat{i}## and ##\hat{j}## are orthogonal, so no terms mix between them; this is why like unit vectors are collected separately.

Q17. A vector points along the negative y-axis. What are the signs of its components?

##A_x > 0,\ A_y > 0##

##A_x 0##

##A_x = 0,\ A_y < 0##

##A_x < 0,\ A_y = 0##

A vector pointing straight down has no horizontal part, so ##A_x = 0##, and its vertical part is negative, so ##A_y < 0##.

Q18. Why are components a rigorous method for vector addition rather than merely a shortcut?

Because components eliminate the need for the Pythagorean theorem

Because they resolve vectors into perpendicular coordinates that can be summed independently, preserving the precision supported by the input data

Because graphical methods cannot add more than two vectors

Because components make all resultants horizontal

Resolving vectors into perpendicular components allows each coordinate to be summed independently. The calculation can preserve the precision supported by its inputs; for example, small differences between launch angles such as ##37^{\circ}## and ##37.5^{\circ}## can produce measurably different projectile ranges.

Q19. A resultant has components ##R_x = 0## and ##R_y = -5##. What is its direction measured from the positive x-axis?

##0^{\circ}##

##90^{\circ}##

##180^{\circ}##

##270^{\circ}##

With no horizontal component and a negative vertical component, the vector points straight down, which is ##270^{\circ}## from the positive x-axis.

Q20. A vector of magnitude ##8## makes an angle of ##30^{\circ}## above the negative x-axis. What are its components?

##A_x = 4\sqrt{3},\ A_y = 4##

##A_x = -4\sqrt{3},\ A_y = 4##

##A_x = -4,\ A_y = 4\sqrt{3}##

##A_x = 4\sqrt{3},\ A_y = -4##

The horizontal part points left, so ##A_x = -8\cos 30^{\circ} = -4\sqrt{3}##, while the vertical part points up, so ##A_y = 8\sin 30^{\circ} = 4##. Sign conventions must follow the quadrant.