Introduction solutions

A = (6 cos 45°, 6 sin 45°) = (4.2426, 4.2426) m/s. B = (4 cos 150°, 4 sin 150°) = (−3.4641, 2) m/s.

Why: The angles start at +x, so cosine gives each x projection and sine gives each y projection. B points left, which makes its x component negative.

Add components: R = (0.7785, 6.2426) m/s.

Why: Add x to x and y to y: each axis records its own part of the resultant velocity.

|R| = √(0.7785² + 6.2426²) = 6.2910 m/s. atan2(6.2426, 0.7785) = 82.89° from +x, or 7.11° east of north.

Why: The x and y components are perpendicular, so Pythagoras gives the magnitude. atan2 keeps the direction in the correct quadrant; subtracting from 90° gives east of north.

Divide the components by 6.2910 to get the dimensionless unit vector (0.1238, 0.9923).

Why: Dividing R by its own magnitude removes the m/s units and leaves a vector of length one pointing the same way.

Add all path lengths: 2 + 4 + 5 + 4 + 5 + 4 + 5 + 4 + 10 + 4 + 5 + 8 + 5 + 4 + 2 = 71 km.

Why: Distance counts every leg you actually walk. Opposite directions still add to the path length, so none of these lengths cancel.

Add signed components, choosing east and north as positive: Δx = 9 km and Δy = −8 km.

Why: Displacement cares only about the endpoints. With east and north positive, west and south legs subtract from the matching components.

Magnitude = √(9² + 8²) = 12.0416 km. The bearing is atan(8/9) = 41.63° south of east.

Why: The net east and south components form a right triangle. Pythagoras gives its length, and the component ratio gives the angle below east.

On a flat plane, the north and south legs cancel, leaving a 3 km eastward displacement.

Why: On a flat map, equal north and south moves undo each other, but neither undoes the 3 km eastward move.

Starting at the North Pole, walking south reaches a latitude circle. Eastward travel stays on that circle, and walking 1 km north reaches the pole again.

Why: Every meridian meets at the North Pole. After the eastward leg changes longitude, a 1 km trip north still returns to that same pole.

There are also families of starts near the South Pole: begin 1 km north of a latitude circle whose circumference is 3/n km. The eastward leg then makes n complete circuits.

Why: Choose a small southern latitude circle so 3 km is an integer number of laps. The eastward leg then ends where it began, allowing the north leg to retrace the south leg.

The angle from A to B is −45°, so the z component has a negative sign.

Why: Turning from A at +30° to B at −15° is clockwise. In a right-handed plane, that rotation gives a cross product along −z.

(A × B)𝓏 = |A||B| sin(−45°) = 50(13)(−√2/2).

Why: A cross product's signed z component is the product of the two magnitudes and the sine of their directed angle.

Multiplying two vectors measured in metres gives square metres.

Why: The sine has no units, so the two metre factors multiply to m². This is a product of lengths, not another displacement.

Use î × ĵ = k̂ as the test. Curl the right-hand fingers from +x toward +y and compare the thumb with +z.

Why: Handedness is about axis orientation, not the labels' positions on the page. The thumb must point along +z when the fingers curl from +x toward +y.

A matches the right-hand rule. B reverses the y and z roles relative to A and is left-handed.

Why: Swapping two axes reverses the orientation. B's interchange of y and z flips the right-handed arrangement in A.

C is the familiar arrangement with +x toward the viewer, +y right and +z up. D cyclically permutes those axes, retaining handedness.

Why: A cyclic permutation moves all three axes around without swapping just one pair, so D keeps C's right-handed orientation.

The normal force is N = mg cos 30° = 84.87 N. Kinetic friction is f = 0.4N = 33.948 N, directed up the ramp.

Why: There is no acceleration through the ramp, so N balances the perpendicular component of weight. Sliding friction has magnitude μkN and opposes the downhill motion.

Weight points downward. The angle between uphill friction and weight is 120°.

Why: Friction points 30° above horizontal, while weight points straight down. Their tail-to-tail angle is therefore 120°, not the ramp's 30°.

f · W = (33.948)(98) cos 120° ≈ −1.6635 × 10³ N².

Why: A dot product uses the cosine of the angle between the forces. cos 120° is negative, and multiplying two forces gives N² rather than joules.

If α is nonzero, α × β = 0 implies β = cα for some real c.

Why: For nonzero α, a zero cross product says β is parallel or antiparallel to α. Writing β = cα also includes the zero vector.

Then β · γ = c(α · γ) = 0. This also covers β = 0.

Why: Substitute cα into the dot product. Scaling α cannot change its zero dot product with γ, so β stays perpendicular to γ unless it is zero.

If α = 0, both stated conditions hold for every β and γ, so β · γ can be any real value.

Why: The zero vector makes both original products vanish automatically. Those conditions then place no restriction on the angle or magnitudes of β and γ.

Distance = 2 + 8 + 5 + 4 + 10 + 8 + 5 + 4 + 3 = 49 km.

Why: Route distance is the sum of all nine travelled lengths, regardless of whether a later leg goes back the other way.

Net components are (−5, −8) km, so displacement magnitude = √89 = 9.434 km.

Why: Cancel opposite directions within each axis to get the endpoint change. The remaining west and south components are perpendicular, so their squared lengths add.

η = √89/49 = 0.19253. The units cancel.

Why: Efficiency compares the direct endpoint distance with the actual path length. Both are in kilometres, so the ratio is dimensionless; multiply by 100 for percent.

A full lap returns to the initial position, so displacement is zero.

Why: Displacement is final position minus initial position. A complete lap has the same endpoint as its start, even though the runner travelled a long way.

Each semicircle has arc length πr. The four arcs total π(30 + 10 + 10 + 10) = 60π m.

Why: A semicircle is half a circumference, so its length is πr. Add the four separate radii before multiplying by π.

The four straight sections total 2(250) + 2(150) = 800 m.

Why: There are two straight pieces of each stated length. Count both copies to get the full straight-line part of the lap.

Total distance = 800 + 60π = 988.496 m.

Why: The arcs and straight sections are different pieces of the same path. Adding their lengths gives the distance for the whole lap.

The cross products point along z. α × β = [4(3) − 3(1)] k̂ = 9 k̂.

Why: With both vectors in the x–y plane, their cross product has only a z component. The determinant 4·3 − 3·1 sets its signed size.

γ × δ = [(−1)(0) − 4(8)] k̂ = −32 k̂.

Why: Apply the same x–y determinant to γ and δ. The negative result puts this cross product along −z.

Dot the two resulting vectors: 9(−32)(k̂ · k̂) = −288.

Why: The two cross products share the z axis, and k̂·k̂ = 1. Their opposite directions make the final dot product negative.

A vector requires components or a magnitude and a direction. A scalar does not require a spatial direction.

Why: Spatial direction is the deciding feature: displacement, velocity and force need it; mass, temperature and kinetic energy can be specified by one number and its unit.

A negative sign on temperature expresses a position on the temperature scale, not a direction in space.

Why: A minus sign can tell you which side of a chosen scale zero you are on. It does not supply temperature with a spatial direction.

Take the magnitude: |F| = |−12| = 12 N.

Why: Magnitude is a nonnegative size. Taking the absolute value removes the directional minus sign without changing the force's units.

The negative component points west. Divide the vector by its magnitude to obtain the dimensionless direction vector.

Why: East is +x, so −x is west. Dividing (−12, 0) N by 12 N leaves the unit-length direction (−1, 0).

Magnitude = √(36 + 64) = 10 m.

Why: The two components lie along perpendicular axes, so use Pythagoras to turn their signed components into a positive length.

The vector lies in quadrant II. atan2(8, −6) gives 126.87°, or 53.13° north of west.

Why: Negative x and positive y put the vector in quadrant II. atan2 uses both signs, avoiding the wrong quadrant that a bare tangent ratio could give.

vₓ = 12 cos(120°); vᵧ = 12 sin(120°).

Why: The angle is measured from +x. The adjacent projection is 12 cos 120° and the upward projection is 12 sin 120°.

cos(120°) = −1/2 and sin(120°) = √3/2.

Why: At 120° the vector points left and up. The exact trig values give a negative x component and a positive y component with the original m/s units.

Add components: (3 − 3, −4 + 4) = (0, 0) m.

Why: Each component is the opposite of the other vector's matching component, so the two displacements cancel exactly.

Its magnitude is zero. Normalizing would require division by zero.

Why: A zero resultant has no direction to preserve. Dividing it by its zero magnitude is undefined, so it has no unit vector.

Magnitude = √(4 + 9 + 36) = 7.

Why: In three perpendicular directions, the squared magnitude is the sum of all three squared components. The negative y component contributes positively when squared.

Divide every component by 7. The squared magnitude is (4 + 9 + 36)/49 = 1.

Why: Dividing every component by the same magnitude keeps the direction. Checking the squared components confirms the new vector has length one.

A · B = 3(8) + 4(−6) = 24 − 24 = 0.

Why: The dot product adds products along matching axes. Here the positive x contribution exactly cancels the negative y contribution.

Both vectors are nonzero, so a zero dot product means the angle between them is 90°.

Why: For nonzero vectors, A·B = |A||B| cos θ. Since neither magnitude is zero, the vanishing product requires cos θ = 0.

A · B = 4; |A| = |B| = √5.

Why: Find the dot product and both lengths so you can compare directional overlap with the vectors' full sizes.

cos θ = 4/5, so θ = arccos(0.8).

Why: Divide A·B by |A||B| to isolate cos θ. arccos returns the smaller angle between these two nonzero vectors.

Use Fₓ = F cos θ and Fᵧ = F sin θ.

Why: The 37° angle starts at the horizontal. Cosine gives the horizontal projection, while sine gives the upward projection.

Fₓ = 50 cos 37°; Fᵧ = 50 sin 37°.

Why: Insert the rope's 50 N magnitude in both projections. Both components are positive because the force points right and above the floor.

The right-hand rule gives î × ĵ = k̂.

Why: The axes are right-handed, so curling from +x to +y points the thumb along +z.

A × B = 6 k̂ m². Reversing the order changes the sign.

Why: Multiply the perpendicular magnitudes 2 and 3 for the cross-product size. Reversing the order reverses its direction, and the metre units multiply to m².

The z component of A × B is 4(3) − 1(−2) = 14 m².

Why: For vectors in the x–y plane, the signed cross-product component is AxBy − AyBx. The negative Bx makes the second contribution positive here.

Area is the magnitude of the cross product.

Why: A parallelogram's area is base times perpendicular height, which equals |A×B|. Area uses the magnitude, so it cannot be negative.

W = F · Δr, not |F| times the path length.

Why: Only the part of the constant force along the displacement transfers energy. The dot product includes that directional projection automatically.

W = 6(3) − 2(4) = 10 N·m.

Why: Multiply force and displacement along each matching axis. The negative y force removes 8 J from the positive 18 J x contribution.

The first path length is 300 + 400 = 700 m.

Why: The first walker travels both legs, so the path length adds even though the directions differ.

The direct distance is √(300² + 400²) = 500 m.

Why: The direct path is the hypotenuse of a right triangle with 300 m and 400 m legs, so Pythagoras gives 500 m.

Both changes in position are (300, 400) m.

Why: Displacement depends on the shared start and finish, not the route. Both walkers end 300 m east and 400 m north of the start.

Total distance = 200 m; total time = 50 s.

Why: The runner covers 100 m twice, while both travel times belong in the whole-trip time interval.

Average speed = 200/50. Total displacement is zero.

Why: Average speed uses total distance divided by total time. Average velocity uses displacement instead, and returning to the start makes that displacement zero.

Add the relative velocities: (4, 0) + (0, 3) = (4, 3) m/s.

Why: The boat's motion through the water and the water's motion past the bank add as vectors. Their east and north components describe the boat in the bank's frame.

Magnitude = 5 m/s; direction = atan2(3, 4).

Why: The bank-frame components form a 3–4–5 right triangle. atan2(3, 4) gives the angle from east toward north.

Add forces to obtain Fnet = (6, 4) N.

Why: Newton's second law uses the net force, so combine all forces on the cart before calculating acceleration.

Apply Newton’s second law separately to each component: a = Fnet/m.

Why: Divide each net-force component by the same cart mass. This keeps the acceleration parallel to the net force and gives N/kg = m/s².

The person exerts a force on the wall; the wall exerts an equal opposite force on the person.

Why: A third-law pair links the same interaction in opposite directions: person on wall and wall on person, with equal magnitudes.

Only forces acting on the person belong in that person’s free-body diagram.

Why: A free-body diagram follows one chosen body. The person's push acts on the wall, so it cannot cancel the wall's reaction acting on the person.

Take upward as positive: N − mg = ma.

Why: For the passenger, the floor force points up and weight points down. The upward-positive force sum must equal the positive upward ma.

N = m(g + a) = 60(9.8 + 1.5).

Why: Solve N − mg = ma for N. The floor must supply both the force that balances weight and the extra force that accelerates the passenger upward.

Horizontal components cancel by symmetry.

Why: Equal rope angles and equal tensions give opposite horizontal components. They cancel, leaving only vertical components to support the lamp.

Vertical balance requires 2T sin 30° = 100 N.

Why: Each rope supplies T sin 30° upward because the angle is measured from horizontal. Their two upward contributions must total the 100 N weight.

The largest sum occurs when both point in the same direction: 5 + 7.

Why: Aligning the forces makes every part of one reinforce the other, giving the largest allowed resultant.

The smallest occurs when they point in opposite directions: |7 − 5|.

Why: Putting the forces opposite each other gives the greatest cancellation. The stronger 7 N force leaves a 2 N resultant in its direction.