01 · From one point to another
A = (−2,1,4), B = (4,−1,1). Find AB and the distance AB.
Hint
Subtract A from B component by component.
Worked solution
AB = (6,−2,−3). Distance = √(36 + 4 + 9) = 7.
Understand · explore · practise
3D vectors: mixed practice
Twenty original mixed vector questions with separate hints and worked solutions, covering 3D geometry, ratios, proofs, forces and motion.
Before you startThe preceding Pure 2 vector lessons. Revisit a topic when its first step is unclear.
01 / Choose the mathematical relationship before calculating
A useful sketch and a named vector often reveal the first step.
Work on paper before opening a hint. Keep exact forms where possible, label units and check that your final result answers the requested question. These questions combine the chapter methods without copying textbook exercises.
Position → displacement → magnitude
For geometry, begin by distinguishing where a point is from how to get there.
Choose a first stepExplore
Choose a method and check it when you are ready.
This checks strategy, not a numerical answer. The independent questions below include hints and worked solutions you can open separately.
02 / Distance and displacement
Subtract end minus start before taking a magnitude.
Use the signs in the coordinates, then square the displacement components. A straight-line distance is nonnegative; a displacement retains direction.
02 · Unknown coordinate
P = (x,2,−1) is distance 5 from Q = (1,−1,−1). Find both possible x values.
Hint
Square the distance equation.
Worked solution
(x − 1)² + 3² = 25, so (x − 1)² = 16. Therefore x = 5 or −3.
03 / Magnitude, direction and parallelism
A single scalar must relate every component.
A unit vector keeps direction but has magnitude one. To test parallelism, account for zero components as well as nonzero ones.
03 · Unit direction
Find a unit vector opposite to v = (−2,3,6).
Hint
|v| = 7, then negate v/7.
Worked solution
The unit vector is (2/7,−3/7,−6/7). Its squared magnitude is (4 + 9 + 36)/49 = 1.
04 · Parallel parameters
(p,−6,9) is parallel to (2,−2,3). Find p and say whether orientations agree.
Hint
The second component gives the scale factor.
Worked solution
The factor is 3, also giving 9 = 3 × 3. Hence p = 6. The positive factor means orientations agree.
04 / Axis angles and plane angles
The sign of an axis component matters.
An angle with a positive coordinate axis lies between 0° and 180°. The conventional angle of a line to a plane lies between 0° and 90°; use the absolute perpendicular component.
05 · Negative vertical component
v = (3,4,−12). Find the angle with the positive z axis and the acute angle with the xy plane, to 1 decimal place.
Hint
|v| = 13; horizontal projection length = 5.
Worked solution
Positive-z angle = arccos(−12/13) ≈ 157.4°. Plane angle = arctan(12/5) ≈ 67.4°. They describe different geometric relationships.
06 · Direction-cosine check
A nonzero vector makes 60° with both positive x and positive y axes and has positive z component. Find its positive-z angle.
Hint
cos²α + cos²β + cos²γ = 1.
Worked solution
cos²γ = 1 − 1/4 − 1/4 = 1/2. Positive z gives cosγ = 1/√2, so γ = 45°.
05 / Triangles and areas
Use directed sides from the angle vertex.
You can find 3D side lengths and then apply ordinary triangle rules. Pythagoras is a useful check for a right angle.
07 · Right triangle in space
A = (1,1,0), B = (4,1,4), C = (1,7,0). Find the triangle area.
Hint
Find AB, AC and BC; test the side-length squares.
Worked solution
AB = (3,0,4), AC = (0,6,0), BC = (−3,6,−4). Their squared lengths are 25, 36 and 61. Since 25 + 36 = 61, angle A is right. Area = ½ × 5 × 6 = 15.
08 · Non-right triangle
A = (0,0,0), B = (2,0,0), C = (1,2,2). Find angle A exactly in inverse-trig form and the triangle area.
Hint
AB = 2 and AC = BC = 3. Use the cosine rule, then ½ab sinC.
Worked solution
cos A = (2² + 3² − 3²)/(2 × 2 × 3) = 1/3, so A = arccos(1/3). sin A = 2√2/3. Area = ½ × 2 × 3 × 2√2/3 = 2√2.
06 / Parallelograms and segment ratios
Boundary order and the ratio order are part of the data.
Write the point formula first; it helps avoid exchanging the weights on the endpoints.
09 · Parallelogram vertex
A = (1,−1,2), B = (4,0,1), C = (3,3,5) are consecutive vertices of ABCD. Find D and the shared diagonal midpoint.
Hint
D = A + C − B.
Worked solution
D = (0,2,6). The midpoint of AC is (2,1,3.5); the midpoint of BD gives the same coordinates.
10 · Internal and external points
A = (−2,0,1), B = (4,3,−2). Find P with AP : PB = 2 : 1 internally, and Q = A + 2(B − A).
Hint
Use t = 2/3 for P and t = 2 for Q.
Worked solution
B − A = (6,3,−3). P = (2,2,−1); Q = (10,6,−5). Q lies beyond B; its positive length ratio AQ : QB is 2 : 1.
07 / Prove a general relationship
Use arbitrary vectors and state the assumptions.
For line intersections, parameters belong to different lines and need not have the same value. For general shape proofs, special numerical coordinates alone are insufficient.
11 · Midpoint segment
Triangle ABC has position vectors a, b, c. M and N are midpoints of AB and AC. Prove MN is parallel to BC and half its length.
Hint
Subtract the midpoint positions.
Worked solution
m = (a + b)/2 and n = (a + c)/2, so MN = n − m = (c − b)/2 = BC/2. For B ≠ C this proves the direction and length claims.
12 · Check a line intersection
One line has positions (s,2s,s), another (2−t,t,2t). Determine whether they intersect.
Hint
First solve s = 2 − t and 2s = t; then check z.
Worked solution
The first two equations give s = 2/3, t = 4/3. The third would require 2/3 = 8/3, so there is no intersection.
08 / Resultants and equilibrium
Add every force before using R = ma.
State whether weight is already included in a quoted resultant. Do not add it twice, or omit it when only applied forces are quoted.
13 · Include weight
A 2 kg particle is acted on by applied force (6,−8,23.6) N and weight. Use g = 9.8. Find a and |a|.
Hint
Weight is (0,0,−19.6) N.
Worked solution
R = (6,−8,4) N; a = (3,−4,2) m s⁻² and |a| = √29 m s⁻².
14 · Remove an equilibrating force
A 3 kg particle is in equilibrium. A force (−6,3,−9) N is removed. Find the new acceleration.
Hint
The remaining resultant is the negative of the removed force.
Worked solution
R = (6,−3,9) N, so a = (2,−1,3) m s⁻².
09 / Motion and distance
Speed is |v|; distance may require a turning-point split.
Check the constant-acceleration assumption, and use vertical velocity to interpret vertical motion.
15 · Starting from rest
A particle starts from rest at the origin with constant a = (2,−2,1) m s⁻². Find position, speed and distance travelled at 4 s.
Hint
|a| = 3; this motion follows a fixed direction without reversal.
Worked solution
r = 8a = (16,−16,8) m. v = 4a, so speed = 12 m s⁻¹. Distance = |r| = 24 m.
16 · Return journey
A particle moves vertically with u = (0,0,6) m s⁻¹ and a = (0,0,−3) m s⁻². Find displacement and distance after 4 s.
Hint
Vertical velocity vanishes at 2 s.
Worked solution
z = 6t − 1.5t², so z(2) = 6 and z(4) = 0. Displacement is zero; distance = 6 + 6 = 12 m.
10 / Find the flaw before fixing the arithmetic
Correct algebra still needs valid assumptions.
Watch: candidate, check, conclusion
Pause, replay or seek freely. The notes explain the same idea and stay in view.
17 · Independence claim
A student says three pairwise nonparallel vectors must be independent. Give a counterexample.
Hint
Choose three different directions in one plane.
Worked solution
(1,0,0), (0,1,0), (1,1,0) are pairwise nonparallel, but the third is the sum of the first two. Hence they are dependent.
18 · Acceleration claim
At one instant v = (2,0,−3) and a = (0,0,1), in SI units. A student says the particle is ascending because az > 0. Correct the claim.
Hint
Inspect the vertical velocity.
Worked solution
It is descending because vz = −3 < 0. Its downward vertical speed is decreasing at that instant.
11 / Finish with a complete component check
Record the method you would revisit after each question.
If your result seems plausible, still check every coordinate, required sign and unit. Reattempt a question without its solution before treating the method as secure.
19 · Reflection check
P = (2,−3,4) is reflected in the x axis. Find its image and compare with reflection in the xy plane.
Hint
The perpendicular foot on the x axis is (2,0,0).
Worked solution
Reflection in the x axis gives (2,3,−4), since the foot is the midpoint. Reflection in the xy plane changes only z and gives (2,−3,−4).
20 · All three components
For noncoplanar a, b, c, solve (p−1)a + 2pb + qc = 3a + rb − 2c.
Hint
Compare coefficients in the independent directions.
Worked solution
p − 1 = 3 gives p = 4; then r = 2p = 8, and q = −2. Substitution checks all three coefficients.
Section 1 of 11 · Choose the mathematical relationship before calculating