Vectors

Find the magnitude of the vector 5i − 12j.

• A17
• B7
• C13✓
• D169

Answer: The magnitude is √(5² + (−12)²) = √(25 + 144) = √169 = 13. Adding the components gives 5 + 12 = 17, and omitting the final square root leaves 169.

Find the unit vector in the direction of 6i + 8j.

• A0.6i + 0.8j✓
• B3i + 4j
• C6i + 8j
• D0.6i − 0.8j

Answer: The magnitude is √(6² + 8²) = √100 = 10, so the unit vector is (6i + 8j)/10 = 0.6i + 0.8j. Dividing by 2 instead of by the magnitude 10 gives 3i + 4j, which is not a unit vector.

Given a = 2i + 3j and b = 4i − j, find 3a − 2b.

• A14i + 7j
• B−2i + 7j
• C−2i − 11j
• D−2i + 11j✓

Answer: 3a = 6i + 9j and 2b = 8i − 2j, so 3a − 2b = (6 − 8)i + (9 − (−2))j = −2i + 11j. (Mishandling the sign on −2b's j-term gives the +7j distractor.)

The points A and B have position vectors a = 3i − 2j and b = 7i + 5j. Find the vector AB.

• A−4i − 7j
• B4i + 7j✓
• C4i + 3j
• D10i + 3j

Answer: AB = b − a = (7 − 3)i + (5 − (−2))j = 4i + 7j. Working out a − b instead reverses the direction and gives −4i − 7j, and missing the double negative in 5 − (−2) gives 4i + 3j.

Which of the following vectors is parallel to 3i − 4j?

• A3i + 4j
• B6i − 8j✓
• C4i − 3j
• D6i + 8j

Answer: A parallel vector is a scalar multiple: 6i − 8j = 2(3i − 4j). The others are not scalar multiples (e.g. 4i − 3j swaps the components).

The points A and B have position vectors 2i + 5j and 8i − 3j. Find the position vector of the midpoint M of AB.

• A10i + 2j
• B3i + 4j
• C5i + j✓
• D6i − 4j

Answer: The midpoint is (a + b)/2 = ((2 + 8)/2)i + ((5 + (−3))/2)j = 5i + j. (The value 10i + 2j is the sum a + b without halving; the other values do not follow from the midpoint formula.)

Find the magnitude of the vector 2i − 3j + 6k.

• A11
• B√13
• C49
• D7✓

Answer: The magnitude is √(2² + (−3)² + 6²) = √(4 + 9 + 36) = √49 = 7. Adding the components gives 2 + 3 + 6 = 11; ignoring the k-term gives √(4 + 9) = √13; and omitting the square root leaves 49.

The points A and B have coordinates A(2, 1) and B(8, 13). The point P lies on AB such that AP : PB = 1 : 2. Find the coordinates of P.

• A(10, 14)
• B(4, 5)✓
• C(5, 7)
• D(6, 9)

Answer: AB = (8 − 2)i + (13 − 1)j = 6i + 12j. Since AP : PB = 1 : 2, AP = (1/3)AB = 2i + 4j, so P = A + AP = (2 + 2, 1 + 4) = (4, 5). The point (5, 7) is the midpoint of AB, and (6, 9) comes from reversing the ratio to 2 : 1.

Find the magnitude of the vector 3i + 4j.

• A25
• B7
• C5✓
• D3/4

Answer: |v| = √(3² + 4²) = √25 = 5. The value before square-rooting is the answer most often given, because that final step is the one left out.

Find the magnitude of the vector 5i + 12j.

• A13✓
• B17
• C169
• D5/12

Answer: |v| = √(5² + 12²) = √169 = 13. The value before square-rooting is the answer most often given, because that final step is the one left out.

Find the magnitude of the vector 8i + 6j.

• A100
• B14
• C10✓
• D4/3

Answer: |v| = √(8² + 6²) = √100 = 10. The value before square-rooting is the answer most often given, because that final step is the one left out.

Find the scalar product of (3i + 4j) and (2i − j).

• A8
• B6
• C5
• D2✓

Answer: Multiply matching components and add: (3)(2) + (4)(−1) = 6 − 4 = 2. A scalar product returns a NUMBER, not a vector — that is what 'scalar' is telling you.

Find the scalar product of (i + 2j) and (4i + 3j).

• A4
• B10✓
• C11
• D6

Answer: Multiply matching components and add: (1)(4) + (2)(3) = 4 + 6 = 10. A scalar product returns a NUMBER, not a vector — that is what 'scalar' is telling you.

Find the scalar product of (2i + 5j) and (3i + j).

• A11✓
• B6
• C17
• D5

Answer: Multiply matching components and add: (2)(3) + (5)(1) = 6 + 5 = 11. A scalar product returns a NUMBER, not a vector — that is what 'scalar' is telling you.

Points A and B have position vectors 2i + 3j and 7i − 9j. Find the vector AB.

• A5i − 12j✓
• B9i − 6j
• C−5i + 12j
• D5i + 12j

Answer: The vector from A to B is b − a = (7 − 2)i + (−9 − 3)j = 5i − 12j. Adding the position vectors instead gives 9i − 6j, and working out a − b reverses the direction to −5i + 12j.

Find the unit vector in the direction of 9i + 12j.

• A(3/7)i + (4/7)j
• B(9)i + (12)j
• C(3/5)i + (4/5)j✓
• D(1/25)i + (4/75)j

Answer: Divide the vector by its magnitude: |v| = √(9² + 12²) = 15, so the unit vector is (9/15)i + (12/15)j. Dividing by the sum of components, or by |v|², are the two usual errors — a unit vector must have magnitude exactly 1.

Vectors a = 2i + 3j and b = 6i + 9j. What is the relationship between them?

• AThey are equal in magnitude
• BThey are perpendicular, since a·b = 0
• Cb is parallel to a, since b = 3a✓
• DThey are neither parallel nor perpendicular

Answer: Check whether one is a scalar multiple of the other: 3(2i + 3j) = 6i + 9j = b, so they are parallel and point the same way. Parallel vectors are scalar multiples; perpendicular ones have a scalar product of zero.

What does it mean if two vectors have a scalar product of zero?

• AAt least one of the two vectors must be the zero vector
• BThey are parallel and point in exactly the same direction
• CThey have equal magnitude
• DThey are perpendicular✓

Answer: a·b = |a||b|cosθ, so a zero result with non-zero vectors forces cosθ = 0 and θ = 90°. It is the standard test for perpendicularity, and much quicker than comparing gradients.

A particle starts at the point with position vector (i + 4j) m and moves with constant velocity (5i − 12j) m s⁻¹. Find its distance from its starting point after 2 seconds.

• A22.8 m
• B34 m
• C26 m✓
• D13 m

Answer: The displacement after 2 s is 2(5i − 12j) = 10i − 24j, with magnitude √(100 + 576) = √676 = 26 m. The value 22.8 m is the new position's distance from the origin, √(11² + 20²), not from the start; 34 m adds the components instead of using Pythagoras, and 13 m is the speed, the distance covered in 1 second.

A constant force F = (3i − 4j) N acts on a particle of mass 0.5 kg, which is initially at rest. Find the speed of the particle after 3 seconds.

• A15 m s⁻¹
• B7.5 m s⁻¹
• C30 m s⁻¹✓
• D45 m s⁻¹

Answer: a = F/m = (6i − 8j) m s⁻², so v = 3a = 18i − 24j and the speed is √(324 + 576) = 30 m s⁻¹. 15 m s⁻¹ uses a = F, forgetting to divide by the mass; 7.5 m s⁻¹ multiplies by the mass instead; 45 is the distance travelled in metres, ½|a|t².

A particle moves so that its position vector at time t seconds is r = (1 + 2t)i + (3 − t)j metres, for t ≥ 0. Find the least distance of the particle from the origin O.

• A7√5/5 m✓
• B√10 m
• C49/5 m
• D7/5 m

Answer: |r|² = (1 + 2t)² + (3 − t)² = 5t² − 2t + 10. This is least when 10t − 2 = 0, i.e. t = 1/5, giving |r|² = 10 − 1/5 = 49/5 and |r| = 7/√5 = 7√5/5 ≈ 3.13 m. √10 is the distance at t = 0, which is not the minimum because |r|² is still decreasing there; 49/5 forgets the square root, and 7/5 roots only the numerator.

Find the acute angle between the vectors a = i + 2j + 2k and b = 2i − j + 2k, to 1 decimal place.

• A26.4°
• B116.4°
• C63.6°✓
• D48.2°

Answer: a · b = 2 − 2 + 4 = 4 and |a| = |b| = 3, so cos θ = 4/9 and θ = 63.6°. The value 26.4° is 90° − 63.6°, from using sine instead of cosine; 116.4° is the obtuse angle, and 48.2° divides the dot product by |a| + |b| = 6 instead of |a||b| = 9.