A Level H2 Mathematics Vectors Matrices Quiz

A-Level Maths H2 Quiz - Vectors Matrices: ANSWER KEY

Total Marks: 60

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Section A: Basic Vector Properties (15 marks)

1. a = 2i − j + 3k, b = −i + 4j + k

(a) a + 2b = (2i − j + 3k) + 2(−i + 4j + k)
= (2 − 2)i + (−1 + 8)j + (3 + 2)k
= 7j + 5k ✓ [1]

(b) a − b = (2 − (−1))i + (−1 − 4)j + (3 − 1)k
= 3i − 5j + 2k
|a − b| = (\sqrt{3^2 + (-5)^2 + 2^2} = \sqrt{9 + 25 + 4} = \sqrt{38}) ✓ [2]

(c) |a| = (\sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14})
Unit vector = (\frac{1}{\sqrt{14}}(2\mathbf{i} - \mathbf{j} + 3\mathbf{k})) ✓ [2]

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2. a = 3i + j − 2k, b = i − 2j + 4k, c = 5i + 4j − 8k

(a) (\overrightarrow{AB}) = b − a = (1 − 3)i + (−2 − 1)j + (4 − (−2))k
= −2i − 3j + 6k ✓ [1]

(\overrightarrow{AC}) = c − a = (5 − 3)i + (4 − 1)j + (−8 − (−2))k
= 2i + 3j − 6k ✓ [1]

(b) (\overrightarrow{AC}) = −1 × (\overrightarrow{AB}) (since 2i + 3j − 6k = −(−2i − 3j + 6k))
Therefore (\overrightarrow{AC}) is parallel to (\overrightarrow{AB}), and since they share point A, A, B, C are collinear. ✓ [2]

(c) AB : BC
(\overrightarrow{AB}) = −2i − 3j + 6k, |(\overrightarrow{AB})| = (\sqrt{4 + 9 + 36} = \sqrt{49} = 7)
(\overrightarrow{BC}) = c − b = (5 − 1)i + (4 − (−2))j + (−8 − 4)k = 4i + 6j − 12k
|(\overrightarrow{BC})| = (\sqrt{16 + 36 + 144} = \sqrt{196} = 14)
AB : BC = 7 : 14 = 1 : 2 ✓ [1]

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3. p = 2i + 3j − k, q = 4i − j + 5k, PR : RQ = 2 : 3

Using ratio theorem: r = (\frac{3\mathbf{p} + 2\mathbf{q}}{2 + 3} = \frac{3\mathbf{p} + 2\mathbf{q}}{5})

3p = 6i + 9j − 3k
2q = 8i − 2j + 10k
3p + 2q = 14i + 7j + 7k

r = (\frac{1}{5}(14\mathbf{i} + 7\mathbf{j} + 7\mathbf{k}) = \frac{14}{5}\mathbf{i} + \frac{7}{5}\mathbf{j} + \frac{7}{5}\mathbf{k}) ✓ [3]

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4. u = ( \begin{pmatrix} 1 \ 2 \ -1 \end{pmatrix} ), v = ( \begin{pmatrix} 3 \ 0 \ 4 \end{pmatrix} )

Distance = |v − u| = (\left| \begin{pmatrix} 3-1 \ 0-2 \ 4-(-1) \end{pmatrix} \right| = \left| \begin{pmatrix} 2 \ -2 \ 5 \end{pmatrix} \right|)

= (\sqrt{2^2 + (-2)^2 + 5^2} = \sqrt{4 + 4 + 25} = \sqrt{33}) ✓ [2]

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Section B: Scalar and Vector Products (20 marks)

5. p = 2i + j − 3k, q = i − 2j + 2k

(a) p · q = (2)(1) + (1)(−2) + (−3)(2) = 2 − 2 − 6 = −6 ✓ [1]

(b) |p| = (\sqrt{4 + 1 + 9} = \sqrt{14})
|q| = (\sqrt{1 + 4 + 4} = \sqrt{9} = 3)
cos θ = (\frac{\mathbf{p} \cdot \mathbf{q}}{|\mathbf{p}||\mathbf{q}|} = \frac{-6}{3\sqrt{14}} = \frac{-2}{\sqrt{14}})
θ = cos⁻¹(\left(\frac{-2}{\sqrt{14}}\right)) ≈ 122.3°
Acute angle = 180° − 122.3° = 57.7° ✓ [3]

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6. a = ( \begin{pmatrix} 2 \ -1 \ 3 \end{pmatrix} ), b = ( \begin{pmatrix} 1 \ p \ -2 \end{pmatrix} )

For perpendicular vectors: a · b = 0
(2)(1) + (−1)(p) + (3)(−2) = 0
2 − p − 6 = 0
−p − 4 = 0
p = −4 ✓ [2]

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7. u = 3i − j + 2k, v = i + 2j − k

(a) u × v = ( \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 3 & -1 & 2 \ 1 & 2 & -1 \end{vmatrix} )

= i((−1)(−1) − (2)(2)) − j((3)(−1) − (2)(1)) + k((3)(2) − (−1)(1))
= i(1 − 4) − j(−3 − 2) + k(6 + 1)
= −3i + 5j + 7k ✓ [3]

(b) |u × v| = (\sqrt{(-3)^2 + 5^2 + 7^2} = \sqrt{9 + 25 + 49} = \sqrt{83})
Unit vector perpendicular to both = (\frac{1}{\sqrt{83}}(-3\mathbf{i} + 5\mathbf{j} + 7\mathbf{k}))
Vector of magnitude 5 = (\frac{5}{\sqrt{83}}(-3\mathbf{i} + 5\mathbf{j} + 7\mathbf{k})) ✓ [3]

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8. A(1, 2, −1), B(3, −1, 2), C(2, 3, 1)

(a) (\overrightarrow{AB}) = ( \begin{pmatrix} 3-1 \ -1-2 \ 2-(-1) \end{pmatrix} = \begin{pmatrix} 2 \ -3 \ 3 \end{pmatrix} )
(\overrightarrow{AC}) = ( \begin{pmatrix} 2-1 \ 3-2 \ 1-(-1) \end{pmatrix} = \begin{pmatrix} 1 \ 1 \ 2 \end{pmatrix} )

(\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & -3 & 3 \ 1 & 1 & 2 \end{vmatrix})

= i((−3)(2) − (3)(1)) − j((2)(2) − (3)(1)) + k((2)(1) − (−3)(1))
= i(−6 − 3) − j(4 − 3) + k(2 + 3)
= −9i − j + 5k ✓ [3]

(b) Area of triangle = (\frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}|)
= (\frac{1}{2}\sqrt{(-9)^2 + (-1)^2 + 5^2} = \frac{1}{2}\sqrt{81 + 1 + 25} = \frac{1}{2}\sqrt{107}) square units ✓ [2]

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9. a = ( \begin{pmatrix} 1 \ 2 \ 2 \end{pmatrix} ), b = ( \begin{pmatrix} 2 \ -1 \ 2 \end{pmatrix} )

(a) a · b = (1)(2) + (2)(−1) + (2)(2) = 2 − 2 + 4 = 4 ✓ [1]

(b) a × b = ( \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 2 & 2 \ 2 & -1 & 2 \end{vmatrix} )

= i((2)(2) − (2)(−1)) − j((1)(2) − (2)(2)) + k((1)(−1) − (2)(2))
= i(4 + 2) − j(2 − 4) + k(−1 − 4)
= 6i + 2j − 5k

|a × b| = (\sqrt{36 + 4 + 25} = \sqrt{65}) ✓ [2]

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Section C: Three-Dimensional Geometry (25 marks)

10. Point A(1, 3, −2), direction d = 2i − j + 3k

(a) Vector equation: r = ( \begin{pmatrix} 1 \ 3 \ -2 \end{pmatrix} + \lambda \begin{pmatrix} 2 \ -1 \ 3 \end{pmatrix} ), (\lambda \in \mathbb{R}) ✓ [2]

(b) Cartesian equations: (\frac{x-1}{2} = \frac{y-3}{-1} = \frac{z+2}{3}) ✓ [2]

(c) For B(5, 1, 4):
(\frac{5-1}{2} = 2), (\frac{1-3}{-1} = 2), (\frac{4+2}{3} = 2)
All equal to 2, so B lies on (l). ✓ [2]

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11. (\Pi): r · (2i − j + 2k) = 6

(a) Normal vector: n = 2i − j + 2k ✓ [1]

(b) Cartesian equation: 2x − y + 2z = 6 ✓ [1]

(c) Distance from origin = (\frac{|6|}{\sqrt{2^2 + (-1)^2 + 2^2}} = \frac{6}{\sqrt{9}} = \frac{6}{3} = 2) units ✓ [2]

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12. (l_1): r = (2i + j − k) + (\lambda)(i − 2j + 2k)
(l_2): r = (4i − 3j + 3k) + (\mu)(2i + j − 2k)

(a) For intersection:
2 + λ = 4 + 2μ ... (1)
1 − 2λ = −3 + μ ... (2)
−1 + 2λ = 3 − 2μ ... (3)

From (1): λ = 2 + 2μ
Substitute into (2): 1 − 2(2 + 2μ) = −3 + μ
1 − 4 − 4μ = −3 + μ
−3 − 4μ = −3 + μ
−5μ = 0 ⇒ μ = 0

Then λ = 2 + 2(0) = 2

Check in (3): −1 + 2(2) = 3 − 2(0) ⇒ 3 = 3 ✓

Point of intersection: r = ( \begin{pmatrix} 2 \ 1 \ -1 \end{pmatrix} + 2\begin{pmatrix} 1 \ -2 \ 2 \end{pmatrix} = \begin{pmatrix} 4 \ -3 \ 3 \end{pmatrix} )

Coordinates: (4, −3, 3) ✓ [4]

(b) Direction vectors: d₁ = ( \begin{pmatrix} 1 \ -2 \ 2 \end{pmatrix} ), d₂ = ( \begin{pmatrix} 2 \ 1 \ -2 \end{pmatrix} )

|d₁| = (\sqrt{1 + 4 + 4} = 3)
|d₂| = (\sqrt{4 + 1 + 4} = 3)
d₁ · d₂ = (1)(2) + (−2)(1) + (2)(−2) = 2 − 2 − 4 = −4

cos θ = (\frac{|-4|}{3 \times 3} = \frac{4}{9})
θ = cos⁻¹(4/9) ≈ 63.6° ✓ [3]

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13. (l): (\frac{x-1}{2} = \frac{y+2}{-1} = \frac{z-3}{2} = t)
(\Pi): x − 2y + 2z = 5

(a) Parametric form of (l):
x = 1 + 2t, y = −2 − t, z = 3 + 2t

Substitute into (\Pi):
(1 + 2t) − 2(−2 − t) + 2(3 + 2t) = 5
1 + 2t + 4 + 2t + 6 + 4t = 5
11 + 8t = 5
8t = −6 ⇒ t = −3/4

Point: x = 1 + 2(−3/4) = 1 − 3/2 = −1/2
y = −2 − (−3/4) = −2 + 3/4 = −5/4
z = 3 + 2(−3/4) = 3 − 3/2 = 3/2

Coordinates: (−1/2, −5/4, 3/2) ✓ [3]

(b) Direction of (l): d = ( \begin{pmatrix} 2 \ -1 \ 2 \end{pmatrix} )
Normal to (\Pi): n = ( \begin{pmatrix} 1 \ -2 \ 2 \end{pmatrix} )

|d| = (\sqrt{4 + 1 + 4} = 3)
|n| = (\sqrt{1 + 4 + 4} = 3)
d · n = (2)(1) + (−1)(−2) + (2)(2) = 2 + 2 + 4 = 8

sin φ = (\frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|} = \frac{8}{9})
φ = sin⁻¹(8/9) ≈ 62.7° ✓ [3]

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14. (\Pi_1): n₁ = ( \begin{pmatrix} 1 \ 2 \ -2 \end{pmatrix} ), (\Pi_2): n₂ = ( \begin{pmatrix} 2 \ -1 \ 2 \end{pmatrix} )

|n₁| = (\sqrt{1 + 4 + 4} = 3)
|n₂| = (\sqrt{4 + 1 + 4} = 3)
n₁ · n₂ = (1)(2) + (2)(−1) + (−2)(2) = 2 − 2 − 4 = −4

cos θ = (\frac{|-4|}{3 \times 3} = \frac{4}{9})
θ = cos⁻¹(4/9) ≈ 63.6° ✓ [3]

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15. A(1, 0, 2), B(3, −1, 4), C(2, 2, 1)

(a) (\overrightarrow{AB}) = ( \begin{pmatrix} 2 \ -1 \ 2 \end{pmatrix} ), (\overrightarrow{AC}) = ( \begin{pmatrix} 1 \ 2 \ -1 \end{pmatrix} )

(\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & -1 & 2 \ 1 & 2 & -1 \end{vmatrix})

= i((−1)(−1) − (2)(2)) − j((2)(−1) − (2)(1)) + k((2)(2) − (−1)(1))
= i(1 − 4) − j(−2 − 2) + k(4 + 1)
= −3i + 4j + 5k ✓ [3]

(b) Using point A(1, 0, 2) and normal (−3, 4, 5):
−3(x − 1) + 4(y − 0) + 5(z − 2) = 0
−3x + 3 + 4y + 5z − 10 = 0
−3x + 4y + 5z − 7 = 0
3x − 4y − 5z = −7 ✓ [2]

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16. P(2, −1, 3), plane: 2x − y + 2z = 7

(a) Direction of (l) = normal to plane = ( \begin{pmatrix} 2 \ -1 \ 2 \end{pmatrix} )

Vector equation: r = ( \begin{pmatrix} 2 \ -1 \ 3 \end{pmatrix} + t \begin{pmatrix} 2 \ -1 \ 2 \end{pmatrix} ), (t \in \mathbb{R}) ✓ [2]

(b) Parametric: x = 2 + 2t, y = −1 − t, z = 3 + 2t
Substitute into plane:
2(2 + 2t) − (−1 − t) + 2(3 + 2t) = 7
4 + 4t + 1 + t + 6 + 4t = 7
11 + 9t = 7
9t = −4 ⇒ t = −4/9

Foot: x = 2 + 2(−4/9) = 2 − 8/9 = 10/9
y = −1 − (−4/9) = −1 + 4/9 = −5/9
z = 3 + 2(−4/9) = 3 − 8/9 = 19/9

Coordinates: (10/9, −5/9, 19/9) ✓ [3]

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17. (l): r = ( \begin{pmatrix} 1 \ 0 \ -1 \end{pmatrix} + t \begin{pmatrix} 2 \ 1 \ 2 \end{pmatrix} ), (\Pi): x + 2y − 2z = 3

(a) Direction of (l): d = ( \begin{pmatrix} 2 \ 1 \ 2 \end{pmatrix} )
Normal to (\Pi): n = ( \begin{pmatrix} 1 \ 2 \ -2 \end{pmatrix} )
d · n = (2)(1) + (1)(2) + (2)(−2) = 2 + 2 − 4 = 0

Since d · n = 0, (l) is parallel to (\Pi). ✓ [2]

(b) Point on (l): P(1, 0, −1)
Distance from P to (\Pi) = (\frac{|1 + 2(0) - 2(-1) - 3|}{\sqrt{1^2 + 2^2 + (-2)^2}} = \frac{|1 + 0 + 2 - 3|}{\sqrt{9}} = \frac{0}{3} = 0)

Wait—this means P lies on (\Pi). Let's verify: 1 + 2(0) − 2(−1) = 1 + 2 = 3 ✓
So the line lies in the plane, and the distance is 0.

Alternative interpretation if the line is parallel but not in the plane:
Distance = (\frac{|1 + 0 + 2 - 3|}{3} = 0), confirming the line lies in the plane. ✓ [3]

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18. Planes: x + y + z = 3 ... (1)
2x − y + z = 1 ... (2)

Direction of intersection line = n₁ × n₂

n₁ = ( \begin{pmatrix} 1 \ 1 \ 1 \end{pmatrix} ), n₂ = ( \begin{pmatrix} 2 \ -1 \ 1 \end{pmatrix} )

n₁ × n₂ = ( \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 1 & 1 \ 2 & -1 & 1 \end{vmatrix} )

= i((1)(1) − (1)(−1)) − j((1)(1) − (1)(2)) + k((1)(−1) − (1)(2))
= i(1 + 1) − j(1 − 2) + k(−1 − 2)
= 2i + j − 3k

Find a point on the line: set z = 0 in (1) and (2):
x + y = 3
2x − y = 1
Adding: 3x = 4 ⇒ x = 4/3
y = 3 − 4/3 = 5/3

Point: (4/3, 5/3, 0)

Vector equation: r = ( \begin{pmatrix} 4/3 \ 5/3 \ 0 \end{pmatrix} + s \begin{pmatrix} 2 \ 1 \ -3 \end{pmatrix} ), (s \in \mathbb{R}) ✓ [4]

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19. A(1, 2, −1), line (l): r = ( \begin{pmatrix} 2 \ 0 \ 1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \ -1 \ 2 \end{pmatrix} )

Point on (l): B(2, 0, 1)
Direction of (l): d = ( \begin{pmatrix} 1 \ -1 \ 2 \end{pmatrix} )
(\overrightarrow{AB}) = ( \begin{pmatrix} 2-1 \ 0-2 \ 1-(-1) \end{pmatrix} = \begin{pmatrix} 1 \ -2 \ 2 \end{pmatrix} )

Normal to plane = (\overrightarrow{AB} \times \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & -2 & 2 \ 1 & -1 & 2 \end{vmatrix})

= i((−2)(2) − (2)(−1)) − j((1)(2) − (2)(1)) + k((1)(−1) − (−2)(1))
= i(−4 + 2) − j(2 − 2) + k(−1 + 2)
= −2i + 0j + k
= ( \begin{pmatrix} -2 \ 0 \ 1 \end{pmatrix} )

Using point A(1, 2, −1):
−2(x − 1) + 0(y − 2) + 1(z + 1) = 0
−2x + 2 + z + 1 = 0
−2x + z + 3 = 0
2x − z = 3 ✓ [4]

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20. A(1, 1, 2), B(3, 0, 1), C(2, −1, 3)

(a) In parallelogram ABCD, (\overrightarrow{AB} = \overrightarrow{DC})
(\overrightarrow{AB}) = ( \begin{pmatrix} 3-1 \ 0-1 \ 1-2 \end{pmatrix} = \begin{pmatrix} 2 \ -1 \ -1 \end{pmatrix} )

Let D = (x, y, z). Then (\overrightarrow{DC}) = ( \begin{pmatrix} 2-x \ -1-y \ 3-z \end{pmatrix} = \begin{pmatrix} 2 \ -1 \ -1 \end{pmatrix} )

2 − x = 2 ⇒ x = 0
−1 − y = −1 ⇒ y = 0
3 − z = −1 ⇒ z = 4

D = (0, 0, 4) ✓ [2]

(b) For rectangle, adjacent sides must be perpendicular.
(\overrightarrow{AB}) = ( \begin{pmatrix} 2 \ -1 \ -1 \end{pmatrix} ), (\overrightarrow{BC}) = ( \begin{pmatrix} 2-3 \ -1-0 \ 3-1 \end{pmatrix} = \begin{pmatrix} -1 \ -1 \ 2 \end{pmatrix} )

(\overrightarrow{AB} \cdot \overrightarrow{BC}) = (2)(−1) + (−1)(−1) + (−1)(2) = −2 + 1 − 2 = −3 ≠ 0

Since the dot product is not zero, adjacent sides are not perpendicular.
Therefore ABCD is not a rectangle. ✓ [3]

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END OF ANSWER KEY