Vectors in the Plane · applications

Applications: Vectors in the Plane

12 question types · Secondary 4 · each worked step by step with a figure that follows the steps

O-Level · GCSE Higher · H2

01

A Delivery Drone's Two Legs and the Flight Back to Base

methodWrite Each Leg as a Column Vector, Add the Components to Get the Displacement from Base, Then Find Its Magnitude by Pythagoras for the Straight Flight Back

A delivery drone leaves its base O. Its first leg is a = 25 and its second leg is b = 43, in kilometers east and north. (a) Find the displacement of the drone from its base as a column vector. (b) The drone has enough charge left for 12 km. Can it fly straight back to base, and how many kilometers of charge will it have to spare?

24682468km eastkm northabOa =25, b =43east on top, north underneath
Each leg is a column vector, kilometers east on top and north underneath: a = 25 and b = 43.
Each leg is a column vector: the top number is kilometers east and the bottom number is kilometers north. The first leg goes 2 km east and 5 km north, and the second goes 4 km east and 3 km north.
step 1 of 5

A displacement is a vector: it has a size and a direction. Two legs are added by adding their components, and the sum is the single journey from the base to the drone. The distance straight back is the magnitude of that vector, found by Pythagoras.

  1. Each leg is a column vector: the top number is kilometers east and the bottom number is kilometers north. The first leg goes 2 km east and 5 km north, and the second goes 4 km east and 3 km north.
  2. Add the components: a + b = 25 + 43 = 2 + 45 + 3 = 68.
  3. (a) The drone is 6 km east and 8 km north of its base, so its displacement is 68.
  4. The straight flight back is the magnitude of this vector: √62 + 82 = √36 + 64 = √100 = 10 km.
  5. (b) 10 km is less than 12 km, so the drone can fly straight back, with 12 − 10 = 2 km of charge to spare. Check: the two legs are √29 ≈ 5.4 km and 5 km, and the straight line back, 10 km, is shorter than the 10.4 km flown, as it must be.

answer(a) 68 km; (b) yes: the flight back is 10 km, leaving 2 km to spare

techniqueAdding Vectors · Column Vectors · Magnitude of a Vector

examsO-Level · GCSE Higher · H2

Common pitfalls

  • Adding the lengths of the two legs, 5.4 + 5 = 10.4 km, and calling it the distance from base. The legs point in different directions, so their lengths do not add; only their components do.
  • Working out √6 + 8 instead of √62 + 82. The magnitude is the hypotenuse of a right-angled triangle with sides 6 and 8, so each component is squared first.
02

A Boat Crossing a River While the Current Carries It Downstream

methodAdd the Boat's Own Velocity and the Current's to Get the Resultant, Take the Crossing Time from the Velocity Straight Across Alone, and Multiply the Current by That Time for the Drift

A river is 60 m wide and flows due east at 3 m/s. A boat sets off from a point O on the south bank and steers due north, straight across, at 4 m/s through the water. (a) Find the resultant velocity of the boat as a column vector, and its speed. (b) How long does the crossing take, and how far downstream of the point opposite O does the boat land?

204060153045meters east of Ometers northriver4 m/s north3 m/s eastOboat04, current30each arrow drawn is 15 seconds of that velocity
With east and north as the components, the boat's own velocity is 04 m/s and the current's is 30 m/s. Each arrow is drawn as 15 seconds of its velocity.
Take the components east and north. The boat's own velocity is 04 m/s and the current's velocity is 30 m/s.
step 1 of 5

The boat is moved by its engine and by the current at the same time, so its velocity over the ground is the sum of the two velocity vectors. The crossing is decided by the part of the velocity straight across the river; the current decides how far the boat drifts in that time.

  1. Take the components east and north. The boat's own velocity is 04 m/s and the current's velocity is 30 m/s.
  2. The boat moves with both at once, so its resultant velocity is the sum: 04 + 30 = 34 m/s.
  3. (a) The speed is the magnitude of the resultant: √32 + 42 = √25 = 5 m/s.
  4. Only the north component carries the boat across. The river is 60 m wide, so the crossing takes 604 = 15 s.
  5. (b) In those 15 s the current carries the boat 3 × 15 = 45 m east, so it lands 45 m downstream after 15 s. Check: the boat travels 5 × 15 = 75 m along its path, and 452 + 602 = 2025 + 3600 = 5625 = 752.

answer(a) 34 m/s, a speed of 5 m/s; (b) 15 s, landing 45 m downstream

techniqueAdding Vectors · Vectors · Magnitude of a Vector

examsO-Level · GCSE Higher · H2

Common pitfalls

  • Dividing the width by the speed of 5 m/s to get 12 s. The 5 m/s is along the slanting path, which is longer than 60 m; only the 4 m/s straight across brings the far bank nearer.
  • Expecting the boat to land opposite O because it steers due north. The current acts for the whole crossing, so the boat drifts east all the way over.
03

A Warehouse Robot Sent to Its Charger

methodAdd the Moves Written in i and j to Find Where the Robot Is, Then Subtract Its Position from the Charger's to Get the Move Still to Make

A robot on a warehouse floor starts at its dock D. Distances are in meters, with i one meter east and j one meter north. The robot makes the move 4i + j and then the move 3i + 5j. Its charger is at 10i + 2j from the dock. (a) Find the robot's position from the dock in terms of i and j. (b) Find the move that takes it straight to the charger, and the distance it must travel.

246246810meters east (i)meters north (j)4i + j3i + 5jD(4i + j) + (3i + 5j)= 7i + 6j
Add the i parts and the j parts separately: (4i + j) + (3i + 5j) = 7i + 6j.
Add the i parts and the j parts separately: (4i + j) + (3i + 5j) = 7i + 6j.
step 1 of 5

In i and j notation a vector is written as so many steps east plus so many steps north, and vectors are added by adding the i parts and the j parts separately. The move from one point to another is the second position minus the first.

  1. Add the i parts and the j parts separately: (4i + j) + (3i + 5j) = 7i + 6j.
  2. (a) The robot R is at 7i + 6j from the dock, that is, 7 m east and 6 m north of it.
  3. The move from the robot R to the charger C is the charger's position minus the robot's: RC = (10i + 2j) − (7i + 6j) = 3i − 4j.
  4. Its length is √32 + (−4)2 = √9 + 16 = √25 = 5 m.
  5. (b) The robot must move 3i − 4j, which is 3 m east and 4 m south, a distance of 5 m. Check: (7i + 6j) + (3i − 4j) = 10i + 2j, which is the charger.

answer(a) 7i + 6j; (b) 3i − 4j, a distance of 5 m

techniqueThe Unit Vectors i and j · Subtracting Vectors · Magnitude of a Vector

examsO-Level · GCSE Higher · H2

Common pitfalls

  • Subtracting the wrong way round, the robot minus the charger, to get −3i + 4j. That is the move from the charger back to the robot; the move from R to C is C minus R.
  • Squaring −4 as −16 and getting √9 − 16, which has no value. The square of a negative number is positive: (−4)2 = 16.
04

Two Ropes Pulling a Crate, and the Third Rope That Holds It Still

methodAdd the Two Forces as Column Vectors, Which Is the Diagonal of Their Parallelogram, Find the Size of the Resultant, and Take Its Negative for the Force That Balances It

Two workers pull a crate across a yard with ropes. With components east and north in newtons, the ropes pull with forces F = 4070 and G = 80−20. (a) Find the resultant force R on the crate as a column vector, and its magnitude. (b) A third worker holds the crate still with a single rope. What force must that rope exert?

−5050−100100newtons eastnewtons northFGcrateF =4070, G =80−20the resultant is the diagonal of the parallelogram
Both forces act on the crate at once: F = 4070 N and G = 80−20 N. Their resultant is the diagonal of the parallelogram with the two ropes as sides.
Both forces act on the crate at once, so the resultant is their sum. On the diagram it is the diagonal of the parallelogram whose sides are the two ropes.
step 1 of 5

Forces acting on one object add as vectors: the resultant is the sum of the column vectors, and on a diagram it is the diagonal of the parallelogram with the two forces as sides. An object stays still when all the forces on it add to the zero vector.

  1. Both forces act on the crate at once, so the resultant is their sum. On the diagram it is the diagonal of the parallelogram whose sides are the two ropes.
  2. Add the components: R = 4070 + 80−20 = 40 + 8070 − 20 = 12050 N.
  3. (a) The magnitude is √1202 + 502 = √14400 + 2500 = √16900 = 130 N.
  4. The crate stays still when the three forces add to zero, so the third force is the negative of the resultant: −R = −120−50 N.
  5. (b) The third rope must pull with −120−50 N, a force of 130 N directly opposite the resultant. Check: 4070 + 80−20 + −120−50 = 00.

answer(a) R = 12050 N, of magnitude 130 N; (b) −120−50 N

techniqueAdding Vectors · Magnitude of a Vector · Scalar Multiplication

examsO-Level · GCSE Higher · H2

Common pitfalls

  • Adding the sizes of the two forces: √402 + 702 ≈ 80.6 N and √802 + 202 ≈ 82.5 N make about 163 N. The ropes pull in different directions, so part of each pull is canceled by the other.
  • Giving the third force as 12050 N, the same as the resultant. That would double the pull; to hold the crate still the force must point the opposite way.
05

Pointing a Camera at a Basket Along a Unit Vector

methodFind the Vector from the Camera to the Target by Subtracting Position Vectors, Divide It by Its Magnitude for the Unit Vector, and Multiply the Unit Vector by a Distance to Reach a Point Along the Line

On a plan of a sports hall, with distances in meters from one corner, a camera is mounted at C(1, 2) and must point at the basket G(7, 10). Its motor is set by the unit vector in the direction it points. (a) Find the unit vector in the direction from C to G. (b) The camera is to focus on a point F that is 5 m from C in that direction. Find the coordinates of F.

2468102468meters eastmeters northCGC(1, 2)G(7, 10)CG =710−12=68
The vector from C to G is the position vector of G minus that of C: CG = 710 − 12 = 68.
The vector from C to G is the position vector of G minus that of C: CG = 710 − 12 = 68.
step 1 of 5

A unit vector has length 1 and keeps only the direction. It is found by dividing a vector by its own magnitude. A point a given distance along the line is then the starting position plus that distance times the unit vector.

  1. The vector from C to G is the position vector of G minus that of C: CG = 710 − 12 = 68.
  2. Its magnitude is √62 + 82 = √100 = 10 m.
  3. (a) Divide the vector by its magnitude: the unit vector is 11068 = 0.60.8. Check: 0.62 + 0.82 = 0.36 + 0.64 = 1.
  4. A step of 5 m in that direction is 5 times the unit vector: 50.60.8 = 34.
  5. (b) Add this step to the position vector of C: 12 + 34 = 46, so F is at (4, 6). Check: 5 m is half of 10 m, and the midpoint of CG is (4, 6).

answer(a) 0.60.8; (b) F(4, 6)

techniqueUnit Vectors · Magnitude of a Vector · Subtracting Vectors

examsH2

Common pitfalls

  • Setting the motor with 68. It points the right way, but its length is 10, not 1; a unit vector is the vector divided by its magnitude.
  • Giving F as (3, 4). That is the step from C to F; the position of F is the position of C plus that step.
06

A Phone Mast Halfway Between Two Villages, and a Second a Third of the Way Along

methodWrite the Midpoint's Position Vector as Half the Sum of the Two Position Vectors, Then Reach a Point a Third of the Way Along by Adding a Third of the Vector Between the Villages

On a map with its origin O at a railway station and distances in kilometers, village A has position vector a = 29 and village B has position vector b = 143. (a) A phone mast M is built halfway between the villages. Find its position vector and its distance from the station. (b) A second mast R is built on the straight road from A to B, one third of the way from A. Find its position vector.

51051015km eastkm northabmABM(8, 6)Oa + b =29+143=1612m = half of1612=86
The midpoint has half the sum of the two position vectors: m = 121612 = 86.
The midpoint's position vector is half the sum of the two: m = 12(a + b) = 121612 = 86.
step 1 of 5

A position vector is the journey from the origin to a point. The midpoint of two points has half the sum of their position vectors. To reach a point part of the way along a road, start at one end and add that fraction of the vector along the road.

  1. The midpoint's position vector is half the sum of the two: m = 12(a + b) = 121612 = 86.
  2. (a) M has position vector 86, and its distance from the station is √82 + 62 = √100 = 10 km.
  3. The vector along the road from A to B is AB = b − a = 143 − 29 = 12−6.
  4. One third of the way from A is one third of that vector: 1312−6 = 4−2.
  5. (b) r = a + 13AB = 29 + 4−2 = 67. Check: from R to B is 8−4, twice 4−2, so R is twice as far from B as from A.

answer(a) m = 86, 10 km from the station; (b) r = 67

techniquePosition Vectors · Scalar Multiplication · Subtracting Vectors

examsO-Level · H2

Common pitfalls

  • Writing the midpoint as 12(b − a) = 6−3. That is half the step from A to B; it has to be added to a before it is a position.
  • Taking 13b for the second mast. A third of the position vector of B is a third of the way from the station to B, not a third of the way from A to B.
07

Three Survey Posts on One Straight Fence Line

methodShow One Vector Is a Scalar Multiple of the Other, So the Two Are Parallel, and Use the Point They Share to Put All Three Posts on One Line

A surveyor marks three posts on flat ground. Measured in meters east and north of a corner post O, they are at P(1, 1), Q(5, 4) and R(13, 10). (a) Show that P, Q and R lie on one straight line. (b) Find the ratio PQ : QR and the length of fence from P to R.

2468102468101214meters eastmeters northPQQRP(1, 1)Q(5, 4)R(13, 10)PQ =43, QR =86
PQ = 5 − 14 − 1 = 43 and QR = 13 − 510 − 4 = 86.
PQ = 5 − 14 − 1 = 43 and QR = 13 − 510 − 4 = 86.
step 1 of 5

Two vectors are parallel when one is a scalar multiple of the other. Parallel alone does not put three points on one line; the two vectors must also share a point. The scalar then gives the ratio of the lengths.

  1. PQ = 5 − 14 − 1 = 43 and QR = 13 − 510 − 4 = 86.
  2. 86 = 243, so QR = 2PQ. One vector is a scalar multiple of the other, so the two are parallel.
  3. (a) PQ and QR are parallel and share the point Q. Two parallel lines through one point are the same line, so P, Q and R lie on one straight line.
  4. PQ = √42 + 32 = √25 = 5 m, and QR is twice as long, so QR = 10 m.
  5. (b) PQ : QR = 5 : 10 = 1 : 2, and the fence from P to R is 5 + 10 = 15 m. Check: PR = 129 and √122 + 92 = √225 = 15 m.

answer(a) QR = 2PQ and Q is shared, so the posts are in one line; (b) 1 : 2, and 15 m

techniqueCollinear Vectors · Scalar Multiplication · Subtracting Vectors

examsO-Level · GCSE Higher · H2

Common pitfalls

  • Stopping at parallel. Two parallel vectors can lie on two different lines, like the opposite sides of a rectangle; it is the shared point Q that makes them one line.
  • Writing the ratio as 2 : 1 because QR = 2PQ. The ratio PQ : QR puts PQ first, and PQ is the shorter part, so the ratio is 1 : 2.
08

A Tie Beam Joining the Midpoints of Two Rafters

methodWrite Every Side in Terms of Two Vectors from the Apex, Halve the Rafters to Reach the Midpoints, and Recognize the Tie as Half the Base Vector

A roof truss is a triangle ABC with its apex at A. A tie beam joins M, the midpoint of the rafter AB, to N, the midpoint of the rafter AC. Let AB = p and AC = q. (a) Prove that the tie MN is parallel to the base BC and half as long. (b) The base BC is 10 m long. A second tie joins P, the midpoint of AM, to Q, the midpoint of AN. How long is the second tie?

pqq − pABCBC = BA + AC = −p + q = q − p
Go from B to C by way of A: BC = BA + AC = −p + q.
Go from B to C by way of A: BC = BA + AC = −p + q = q − p.
step 1 of 5

A vector proof writes every line in the figure in terms of the same two vectors. A route from one point to another can go by any path, and a vector that is a scalar multiple of another is parallel to it, with the scalar giving the ratio of their lengths.

  1. Go from B to C by way of A: BC = BA + AC = −p + q = q − p.
  2. M is halfway along AB and N is halfway along AC, so AM = 12p and AN = 12q.
  3. Go from M to N by way of A: MN = MA + AN = −12p + 12q = 12(q − p).
  4. (a) So MN = 12BC. A scalar multiple of BC is parallel to it, and the scalar 12 makes the tie half as long as the base.
  5. The tie MN is 12 × 10 = 5 m long. In triangle AMN, P and Q are the midpoints of the sides AM and AN, so part (a) applies to that triangle too: PQ is parallel to MN and half as long. (b) The second tie is 12 × 5 = 2.5 m long. Check with a truss whose corners are B(0, 0), C(10, 0) and A(4, 6) in meters: M is (2, 3), N is (7, 3), P is (3, 4.5) and Q is (5.5, 4.5), so PQ = 2.50, a quarter of 100.

answer(a) MN = 12BC, so MN is parallel to BC and half as long; (b) 2.5 m

techniqueVector Proofs with Midpoints · Scalar Multiplication · Subtracting Vectors

examsO-Level · GCSE Higher · H2

Common pitfalls

  • Writing BC = p − q. Going from B to C means going back along AB, which is −p, and then along AC, which is +q.
  • Measuring one drawing with a ruler and calling the result proved. A measurement shows one triangle; the vector working holds for every triangle, because it uses nothing but the two midpoints.
09

The Work Done Pulling a Trolley Up a Loading Ramp

methodMultiply Matching Components and Add Them, the Dot Product of the Force and the Displacement, and Read a Negative Result as Work Done Against the Force

A trolley is pulled 10 m up a straight loading ramp. With components along the ground and straight up, in meters, its displacement is d = 86. The rope pulls with a force F = 4060 N, and the trolley's weight is W = 0−100 N. The work done by a force, in joules, is the dot product of the force and the displacement. (a) Find the work done by the rope. (b) Find the work done by the weight, and say what its sign means.

246246810meters along the groundmeters updFWF · d = 40 × 8 + 60 × 6
The forces are drawn at 1 m for every 50 N. The dot product multiplies matching components and adds: F · d = 40 × 8 + 60 × 6.
The dot product multiplies matching components and adds the products: F · d = 40 × 8 + 60 × 6.
step 1 of 5

The dot product of two vectors multiplies their matching components and adds the products. For a force and a displacement it counts only the part of the force along the motion, which is the work done. A negative result means that the force acts against the motion.

  1. The dot product multiplies matching components and adds the products: F · d = 40 × 8 + 60 × 6.
  2. (a) F · d = 320 + 360 = 680, so the rope does 680 J of work.
  3. For the weight: W · d = 0 × 8 + (−100) × 6 = −600.
  4. (b) The weight does −600 J of work. The sign is negative because the weight points down while the trolley rises: the weight acts against the motion. Only the rise of 6 m counts, and 100 × 6 = 600.
  5. Check: √82 + 62 = 10 m, the length of the ramp. The part of the pull along the ramp is 68010 = 68 N, less than the full pull √402 + 602 ≈ 72.1 N, because the rope is angled above the ramp.

answer(a) 680 J; (b) −600 J: the weight acts against the motion

techniqueThe Dot Product · Column Vectors

examsH2

Common pitfalls

  • Multiplying the magnitudes, 72.1 × 10 = 721 J. That counts the whole pull as if the rope were in line with the ramp; the dot product counts only the part along the motion.
  • Dropping the minus sign and giving 600 J for the weight. The weight acts against the motion up the ramp, and the negative sign is how the dot product says so.
10

The Angle Between Two Flight Paths from an Airport

methodDivide the Dot Product by the Product of the Magnitudes for the Cosine of the Angle, and Read a Dot Product of Zero as a Right Angle

Two aircraft take off from an airport O. With components in kilometers east and north, the first flies along u = 71 and the second along v = 34. (a) Find the angle between the two flight paths. (b) A training flight is to cross the second path at right angles, flying along w = −43. Show that w is perpendicular to v.

24−448km eastkm northuvOu · v = 7 × 3 + 1 × 4 = 21 + 4 = 25
u · v = 7 × 3 + 1 × 4 = 25.
u · v = 7 × 3 + 1 × 4 = 21 + 4 = 25.
step 1 of 5

The dot product of two vectors equals the product of their magnitudes and the cosine of the angle between them. So the cosine of the angle is the dot product divided by the two magnitudes, and a dot product of zero between two non-zero vectors means a right angle.

  1. u · v = 7 × 3 + 1 × 4 = 21 + 4 = 25.
  2. |u| = √72 + 12 = √50 = 5√2 and |v| = √32 + 42 = 5.
  3. cosθ = u · v|u||v| = 2525√2 = 1√2.
  4. (a) cosθ = 1√2, so the angle between the flight paths is θ = 45°.
  5. (b) w · v = −4 × 3 + 3 × 4 = −12 + 12 = 0. Neither vector is zero, so cosθ = 0 and w is at 90° to v.

answer(a) 45°; (b) w · v = 0, so the paths are at 90°

techniqueThe Angle Between Vectors · The Dot Product · Magnitude of a Vector

examsH2

Common pitfalls

  • Leaving out the magnitudes and taking cosθ = 25. A cosine is never more than 1; the dot product has to be divided by the product of the lengths.
  • Reading a dot product of zero as meaning that one of the vectors is zero. Neither w nor v is zero; a zero dot product between two non-zero vectors means that they are perpendicular.
11

A Cable Across a Warehouse from a Floor Corner to the Far Top Corner

methodSet Up Three Axes Along the Edges of the Room, Find the Length by Pythagoras in Three Dimensions, and Take the Midpoint as the Mean of Each Coordinate

A warehouse is a cuboid 12 m long, 4 m wide and 3 m high. With axes along its edges from the corner O on the floor, a cable runs straight from O to a light L at (12, 4, 3) in the opposite top corner. A sensor S is at (0, 4, 3). (a) Find the length of the cable. (b) A hook holds the cable at its midpoint M. Find the coordinates of M and the distance from the sensor to the hook.

12 m4 m3 mOLSOL =1243along the length, across the width, up the height
The cable is the vector OL = 1243: along the length, across the width and up the height of the room.
The cable is the vector OL = 1243: along the length, across the width, and up the height of the room.
step 1 of 5

In three dimensions a vector has three components, and its length is the square root of the sum of their three squares. The midpoint of two points has the mean of each of their coordinates.

  1. The cable is the vector OL = 1243: along the length, across the width, and up the height of the room.
  2. (a) In three dimensions the length is √122 + 42 + 32 = √144 + 16 + 9 = √169 = 13 m.
  3. The midpoint has the mean of each coordinate: M = (0 + 122, 0 + 42, 0 + 32) = (6, 2, 1.5).
  4. From the sensor to the hook: SM = 6 − 02 − 41.5 − 3 = 6−2−1.5.
  5. (b) M is at (6, 2, 1.5), and SM = √62 + 22 + 1.52 = √36 + 4 + 2.25 = √42.25 = 6.5 m. Check: 6.52 = 42.25.

answer(a) 13 m; (b) M(6, 2, 1.5), and 6.5 m from the sensor

techniqueDistance and Midpoint in Three Dimensions · Vectors in Three Dimensions

Common pitfalls

  • Using only the floor: √122 + 42 = √160 ≈ 12.6 m. That is the diagonal of the floor; the cable also climbs 3 m, so the height enters the sum as a third square.
  • Putting the hook at a height of 3 m because the light is at the ceiling. The cable starts on the floor, so halfway along it is halfway up, at 1.5 m.
12

A Ship on a Straight Course, and a Buoy and a Reef on Its Chart

methodRead the Start as the Position Vector and the Hourly Step as the Direction Vector, Substitute a Time for the Position, and Solve Both Components for One Time to Test a Point

A ship's course is r = 21 + λ34, where r is its position in kilometers east and north of a harbor and λ is the time in hours since noon. (a) Where is the ship at 2 pm, and what is its speed? (b) At what time does the ship pass the buoy B at (14, 17)? Does it pass over a reef at (11, 12)?

481216481216km eastkm northdB(14, 17)reefnoon (2, 1)r =21+ h34h hours after noon
At noon λ = 0 and the ship is at (2, 1). Each hour adds the direction vector 34.
At noon λ = 0 and the ship is at (2, 1). Each hour adds the direction vector 34.
step 1 of 5

In the vector equation of a line r = a + λd, the vector a is a point on the line and d is its direction. Here λ counts hours, so d is the ship's move in one hour. A point is on the line only when one value of λ gives both of its coordinates.

  1. At noon λ = 0 and the ship is at (2, 1). Each hour adds the direction vector 34.
  2. At 2 pm, λ = 2: r = 21 + 234 = 89.
  3. (a) At 2 pm the ship is at (8, 9). Each hour it moves √32 + 42 = 5 km, so its speed is 5 km/h.
  4. For the buoy: 2 + 3λ = 14 gives λ = 4, and 1 + 4λ = 17 gives λ = 4 as well. Both components agree, so the ship passes the buoy at 4 pm.
  5. (b) For the reef: 2 + 3λ = 11 gives λ = 3, but then 1 + 4 × 3 = 13, not 12. No single time fits both, so the ship does not pass over the reef; at 3 pm it is at (11, 13), 1 km due north of it.

answer(a) (8, 9), at 5 km/h; (b) it passes the buoy at 4 pm, and it misses the reef

techniqueThe Vector Equation of a Line · Position Vectors · Magnitude of a Vector

examsH2

Common pitfalls

  • Solving only the east component, 2 + 3λ = 11, and deciding that the ship reaches the reef at 3 pm. A point is on the line only when one value of λ fits both components.
  • Giving the speed as 3 + 4 = 7 km/h, or as the length of 21. The start says where the ship is, not how fast it goes; the speed is the magnitude of the direction vector, 5 km/h.
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