A Level H2 Mathematics Vectors Matrices Quiz

A-Level Maths H2 Quiz - Vectors Matrices: Answer Key

Total Marks: 80

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Section A: Basic Vector Properties and Operations (20 marks)

1. a = 3i − 2j + k, b = −i + 4j − 2k

(a) 2a − 3b = 2(3i − 2j + k) − 3(−i + 4j − 2k)
= (6i − 4j + 2k) − (−3i + 12j − 6k)
= (6 + 3)i + (−4 − 12)j + (2 + 6)k
= 9i − 16j + 8k [2 marks: M1 for correct scalar multiplication, A1 for correct result]

(b) a + b = (3 − 1)i + (−2 + 4)j + (1 − 2)k = 2i + 2j − k
|a + b| = (\sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3) [2 marks: M1 for correct sum, A1 for magnitude]

(c) |a| = (\sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14})
Unit vector = (\frac{1}{\sqrt{14}}(3\mathbf{i} - 2\mathbf{j} + \mathbf{k})) [2 marks: M1 for magnitude, A1 for unit vector]

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

(a) (\overrightarrow{AB}) = b − a = (3−1)i + (−1−2)j + (2−(−1))k = 2i − 3j + 3k
(\overrightarrow{AC}) = c − a = (5−1)i + (1−2)j + (4−(−1))k = 4i − j + 5k [2 marks: A1 each]

(b) For collinearity, (\overrightarrow{AC}) must be a scalar multiple of (\overrightarrow{AB}).
Check: (\overrightarrow{AC}) = k(\overrightarrow{AB}) ⇒ (4, −1, 5) = k(2, −3, 3)
From x-component: 4 = 2k ⇒ k = 2
Check y: −1 = 2(−3) = −6 ✗
Wait—recheck: (\overrightarrow{BC}) = c − b = (5−3)i + (1−(−1))j + (4−2)k = 2i + 2j + 2k
(\overrightarrow{AB}) = 2i − 3j + 3k
These are not scalar multiples. Let's verify collinearity properly:
Points are collinear if (\overrightarrow{AB}) and (\overrightarrow{BC}) are parallel.
(\overrightarrow{BC}) = 2i + 2j + 2k
Is (\overrightarrow{BC}) = k(\overrightarrow{AB})? 2 = 2k ⇒ k = 1; but 2 ≠ −3. Not collinear.
Correction: Let's re-examine. (\overrightarrow{AB}) = 2i − 3j + 3k, (\overrightarrow{BC}) = 2i + 2j + 2k. These are not parallel. Points are not collinear.
Alternative approach: Check if area of triangle ABC is zero.
(\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & -3 & 3 \ 4 & -1 & 5 \end{vmatrix} = \mathbf{i}((-3)(5) - (3)(-1)) - \mathbf{j}((2)(5) - (3)(4)) + \mathbf{k}((2)(-1) - (-3)(4)))
= i(−15 + 3) − j(10 − 12) + k(−2 + 12) = −12i + 2j + 10k ≠ 0. Not collinear.
Note: The question as written has an error. For the answer key, we will show the method and note that the points are not collinear.
[2 marks: M1 for method (checking if one vector is scalar multiple of another), A1 for correct conclusion that they are not collinear]

(c) Since points are not collinear, ratio AB : BC is not defined in the collinear sense.
AB = |(\overrightarrow{AB})| = (\sqrt{4 + 9 + 9} = \sqrt{22})
BC = |(\overrightarrow{BC})| = (\sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3})
Ratio AB : BC = (\sqrt{22} : 2\sqrt{3}) [2 marks: M1 for finding lengths, A1 for ratio]

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3. p = 2i + j − 3k, q = 4i − j + k, ratio 2:3 internally.

Position vector of R = (\frac{3\mathbf{p} + 2\mathbf{q}}{2+3} = \frac{3(2\mathbf{i} + \mathbf{j} - 3\mathbf{k}) + 2(4\mathbf{i} - \mathbf{j} + \mathbf{k})}{5})
= (\frac{(6\mathbf{i} + 3\mathbf{j} - 9\mathbf{k}) + (8\mathbf{i} - 2\mathbf{j} + 2\mathbf{k})}{5})
= (\frac{14\mathbf{i} + \mathbf{j} - 7\mathbf{k}}{5} = \frac{14}{5}\mathbf{i} + \frac{1}{5}\mathbf{j} - \frac{7}{5}\mathbf{k}) [3 marks: M1 for correct ratio theorem formula, M1 for substitution, A1 for correct result]

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4. u = ( \begin{pmatrix} 2 \ -1 \ 3 \end{pmatrix} ), v = ( \begin{pmatrix} -4 \ 2 \ -6 \end{pmatrix} )

(a) v = −2u, so u and v are parallel (and in opposite directions). [1 mark]

(b) u + λv = ( \begin{pmatrix} 2 \ -1 \ 3 \end{pmatrix} + \lambda \begin{pmatrix} -4 \ 2 \ -6 \end{pmatrix} = \begin{pmatrix} 2-4\lambda \ -1+2\lambda \ 3-6\lambda \end{pmatrix} )
For this to be parallel to ( \begin{pmatrix} 1 \ 0 \ -1 \end{pmatrix} ), we need the y-component to be 0:
−1 + 2λ = 0 ⇒ λ = (\frac{1}{2})
Check: with λ = (\frac{1}{2}), u + λv = ( \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix} ) which is the zero vector (parallel to any vector by convention, though some definitions exclude it).
Alternative: For non-zero parallel vectors, ratios of components must be equal.
(\frac{2-4\lambda}{1} = \frac{3-6\lambda}{-1}) and −1+2λ = 0. From second: λ = (\frac{1}{2}).
[2 marks: M1 for setting up parallel condition, A1 for λ = 1/2]

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Section B: Scalar and Vector Products (20 marks)

5. p = 2i + j − 2k, q = −i + 3j + k

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

(b) cos θ = (\frac{\mathbf{p} \cdot \mathbf{q}}{|\mathbf{p}||\mathbf{q}|})
|p| = (\sqrt{4 + 1 + 4} = \sqrt{9} = 3)
|q| = (\sqrt{1 + 9 + 1} = \sqrt{11})
cos θ = (\frac{-1}{3\sqrt{11}})
θ = cos⁻¹(\left(\frac{-1}{3\sqrt{11}}\right)) ≈ 95.7° (to nearest 0.1°) [3 marks: M1 for formula, M1 for magnitudes, A1 for angle]

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

(a) a × b = ( \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 2 & -1 \ 3 & -1 & 2 \end{vmatrix} )
= i((2)(2) − (−1)(−1)) − j((1)(2) − (−1)(3)) + k((1)(−1) − (2)(3))
= i(4 − 1) − j(2 + 3) + k(−1 − 6)
= 3i − 5j − 7k = ( \begin{pmatrix} 3 \ -5 \ -7 \end{pmatrix} ) [2 marks: M1 for determinant setup, A1 for correct result]

(b) A vector perpendicular to both a and b is a × b = ( \begin{pmatrix} 3 \ -5 \ -7 \end{pmatrix} ).
|a × b| = (\sqrt{9 + 25 + 49} = \sqrt{83})
Unit perpendicular vector = (\frac{1}{\sqrt{83}}\begin{pmatrix} 3 \ -5 \ -7 \end{pmatrix})
Vector of magnitude 5 = (\frac{5}{\sqrt{83}}\begin{pmatrix} 3 \ -5 \ -7 \end{pmatrix}) (or its negative) [3 marks: M1 for identifying cross product as perpendicular, M1 for unit vector, A1 for scaling to magnitude 5]

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7. |u| = 3, |v| = 4, angle = 60°

(a) u · v = |u||v| cos 60° = (3)(4)(0.5) = 6 [1 mark]

(b) |u × v| = |u||v| sin 60° = (3)(4)(\left(\frac{\sqrt{3}}{2}\right)) = 6√3 [2 marks: M1 for formula, A1 for correct value]

(c) |2u − v|² = (2u − v) · (2u − v) = 4|u|² − 4u·v + |v|²
= 4(9) − 4(6) + 16 = 36 − 24 + 16 = 28
|2u − v| = √28 = 2√7 [3 marks: M1 for expanding dot product, M1 for substitution, A1 for result]

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8. a = i + 2j + 2k, b = 2i − j + k

Projection of a onto b = (\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2} \mathbf{b})
a · b = (1)(2) + (2)(−1) + (2)(1) = 2 − 2 + 2 = 2
|b|² = 4 + 1 + 1 = 6
Projection = (\frac{2}{6}\begin{pmatrix} 2 \ -1 \ 1 \end{pmatrix} = \frac{1}{3}\begin{pmatrix} 2 \ -1 \ 1 \end{pmatrix} = \begin{pmatrix} 2/3 \ -1/3 \ 1/3 \end{pmatrix}) [3 marks: M1 for formula, M1 for dot product and magnitude squared, A1 for vector]

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Section C: Lines and Planes in 3D (20 marks)

9. Point A(1, −2, 3), direction ( \begin{pmatrix} 2 \ 1 \ -1 \end{pmatrix} )

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

(b) Parametric: x = 1 + 2λ, y = −2 + λ, z = 3 − λ
Substitute into plane x + 2y − z = 4:
(1 + 2λ) + 2(−2 + λ) − (3 − λ) = 4
1 + 2λ − 4 + 2λ − 3 + λ = 4
5λ − 6 = 4 ⇒ 5λ = 10 ⇒ λ = 2
Point: x = 1 + 4 = 5, y = −2 + 2 = 0, z = 3 − 2 = 1
Coordinates: (5, 0, 1) [3 marks: M1 for parametric form, M1 for substitution and solving, A1 for coordinates]

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10. Points P(2, −1, 4) and Q(4, 3, 0)

Direction vector = ( \begin{pmatrix} 4-2 \ 3-(-1) \ 0-4 \end{pmatrix} = \begin{pmatrix} 2 \ 4 \ -4 \end{pmatrix} ) (or simplified: ( \begin{pmatrix} 1 \ 2 \ -2 \end{pmatrix} ))

Cartesian equation: (\frac{x-2}{1} = \frac{y+1}{2} = \frac{z-4}{-2})
(Using simplified direction vector and point P) [3 marks: M1 for direction vector, M1 for Cartesian form setup, A1 for correct equation]

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11. Π: r · ( \begin{pmatrix} 2 \ -1 \ 3 \end{pmatrix} ) = 6

(a) Normal vector: ( \begin{pmatrix} 2 \ -1 \ 3 \end{pmatrix} ) [1 mark]

(b) Cartesian: 2x − y + 3z = 6 [1 mark]

(c) Check (1, 2, 2): 2(1) − 2 + 3(2) = 2 − 2 + 6 = 6
Since LHS = RHS, the point lies on Π. [2 marks: M1 for substitution, A1 for conclusion]

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12. Line (l): direction d = ( \begin{pmatrix} 2 \ 1 \ -2 \end{pmatrix} ), Plane Π: normal n = ( \begin{pmatrix} 2 \ -1 \ 2 \end{pmatrix} )

Angle θ between line and plane: sin θ = (\frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|})
d · n = (2)(2) + (1)(−1) + (−2)(2) = 4 − 1 − 4 = −1
|d| = (\sqrt{4 + 1 + 4} = \sqrt{9} = 3)
|n| = (\sqrt{4 + 1 + 4} = \sqrt{9} = 3)
sin θ = (\frac{|-1|}{3 \times 3} = \frac{1}{9})
θ = sin⁻¹(1/9) ≈ 6.38° [4 marks: M1 for correct formula (sin, not cos), M1 for dot product, M1 for magnitudes, A1 for angle]

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13. Point P(3, −1, 2), plane 2x + y − 2z = 8

Distance = (\frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}) where plane is ax + by + cz = d
= (\frac{|2(3) + 1(-1) + (-2)(2) - 8|}{\sqrt{4 + 1 + 4}})
= (\frac{|6 - 1 - 4 - 8|}{\sqrt{9}})
= (\frac{|-7|}{3} = \frac{7}{3}) [4 marks: M1 for distance formula, M1 for substitution, M1 for denominator, A1 for result]

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Section D: Applications and Problem Solving (20 marks)

14. (l_1): r = ( \begin{pmatrix} 1 \ 2 \ -1 \end{pmatrix} + s \begin{pmatrix} 2 \ -1 \ 1 \end{pmatrix} ), (l_2): r = ( \begin{pmatrix} 3 \ 0 \ 1 \end{pmatrix} + t \begin{pmatrix} -1 \ 2 \ 3 \end{pmatrix} )

(a) For intersection, equate:
( \begin{pmatrix} 1+2s \ 2-s \ -1+s \end{pmatrix} = \begin{pmatrix} 3-t \ 2t \ 1+3t \end{pmatrix} )
Equations:
1 + 2s = 3 − t → 2s + t = 2 ...(1)
2 − s = 2t → s + 2t = 2 ...(2)
−1 + s = 1 + 3t → s − 3t = 2 ...(3)

From (2): s = 2 − 2t
Sub into (1): 2(2 − 2t) + t = 2 → 4 − 4t + t = 2 → −3t = −2 → t = 2/3
Then s = 2 − 2(2/3) = 2 − 4/3 = 2/3
Check (3): s − 3t = 2/3 − 3(2/3) = 2/3 − 2 = −4/3 ≠ 2.
Lines do not intersect.

Wait—let me recheck the equations carefully.
Line 1: x = 1+2s, y = 2−s, z = −1+s
Line 2: x = 3−t, y = 0+2t = 2t, z = 1+3t

Equation system:
1+2s = 3−t → 2s + t = 2 ...(1)
2−s = 2t → s + 2t = 2 ...(2)
−1+s = 1+3t → s − 3t = 2 ...(3)

From (2): s = 2 − 2t
Sub into (1): 2(2−2t) + t = 2 → 4 − 4t + t = 2 → −3t = −2 → t = 2/3, s = 2 − 4/3 = 2/3
Check (3): 2/3 − 3(2/3) = 2/3 − 2 = −4/3 ≠ 2.

The lines do NOT intersect. They are skew.

However, the question asks to "show that l₁ and l₂ intersect." This appears to be an error in the question design. For the answer key, we will show the working and conclude correctly.

[4 marks: M1 for equating parametric forms, M1 for forming equations, M1 for solving two equations, A1 for checking third equation and concluding they do not intersect]

(b) Angle between lines uses direction vectors: d₁ = ( \begin{pmatrix} 2 \ -1 \ 1 \end{pmatrix} ), d₂ = ( \begin{pmatrix} -1 \ 2 \ 3 \end{pmatrix} )
cos θ = (\frac{|\mathbf{d_1} \cdot \mathbf{d_2}|}{|\mathbf{d_1}||\mathbf{d_2}|})
d₁ · d₂ = (2)(−1) + (−1)(2) + (1)(3) = −2 − 2 + 3 = −1
|d₁| = (\sqrt{4+1+1} = \sqrt{6})
|d₂| = (\sqrt{1+4+9} = \sqrt{14})
cos θ = (\frac{|-1|}{\sqrt{6}\sqrt{14}} = \frac{1}{\sqrt{84}})
θ = cos⁻¹(1/√84) ≈ 83.7° [3 marks: M1 for formula, M1 for dot product and magnitudes, A1 for angle]

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15. Π₁: x + y + z = 3, Π₂: 2x − y + z = 1

(a) Direction of intersection line: d = n₁ × n₂ = ( \begin{pmatrix} 1 \ 1 \ 1 \end{pmatrix} \times \begin{pmatrix} 2 \ -1 \ 1 \end{pmatrix} )
= ( \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 both planes: set z = 0.
x + y = 3, 2x − y = 1. Add: 3x = 4 ⇒ x = 4/3, y = 3 − 4/3 = 5/3.
Point: (4/3, 5/3, 0)

Line equation: r = ( \begin{pmatrix} 4/3 \ 5/3 \ 0 \end{pmatrix} + \lambda \begin{pmatrix} 2 \ 1 \ -3 \end{pmatrix} ), λ ∈ ℝ [4 marks: M1 for cross product, M1 for finding a point, M1 for correct form, A1 for equation]

(b) Angle between planes = angle between normals:
cos θ = (\frac{|\mathbf{n_1} \cdot \mathbf{n_2}|}{|\mathbf{n_1}||\mathbf{n_2}|})
n₁ · n₂ = (1)(2) + (1)(−1) + (1)(1) = 2 − 1 + 1 = 2
|n₁| = √3, |n₂| = (\sqrt{4+1+1} = \sqrt{6})
cos θ = (\frac{2}{\sqrt{3}\sqrt{6}} = \frac{2}{\sqrt{18}} = \frac{2}{3\sqrt{2}} = \frac{\sqrt{2}}{3})
θ = cos⁻¹(√2/3) ≈ 61.9° [3 marks: M1 for formula, M1 for dot product and magnitudes, A1 for angle]

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16. Plane Π contains A(1, 0, −1) and line (l): r = ( \begin{pmatrix} 2 \ 1 \ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \ -1 \ 2 \end{pmatrix} )

Two vectors in the plane:
v₁ = direction of (l) = ( \begin{pmatrix} 1 \ -1 \ 2 \end{pmatrix} )
v₂ = ( \begin{pmatrix} 2 \ 1 \ 0 \end{pmatrix} - \begin{pmatrix} 1 \ 0 \ -1 \end{pmatrix} = \begin{pmatrix} 1 \ 1 \ 1 \end{pmatrix} ) (from A to a point on (l))

Normal n = v₁ × v₂ = ( \begin{pmatrix} 1 \ -1 \ 2 \end{pmatrix} \times \begin{pmatrix} 1 \ 1 \ 1 \end{pmatrix} )
= ( \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & -1 & 2 \ 1 & 1 & 1 \end{vmatrix} )
= i((−1)(1) − (2)(1)) − j((1)(1) − (2)(1)) + k((1)(1) − (−1)(1))
= i(−1 − 2) − j(1 − 2) + k(1 + 1)
= −3i + j + 2k

Plane equation: −3(x − 1) + 1(y − 0) + 2(z + 1) = 0
−3x + 3 + y + 2z + 2 = 0
−3x + y + 2z + 5 = 0
3x − y − 2z = 5 [4 marks: M1 for finding two vectors in plane, M1 for cross product, M1 for plane equation form, A1 for Cartesian equation]

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17. a = i + j, b = 2i + 3j − k, c = −i + 2j + 2k

(a) (\overrightarrow{AB}) = b − a = (2−1)i + (3−1)j + (−1−0)k = i + 2j − k
(\overrightarrow{AC}) = c − a = (−1−1)i + (2−1)j + (2−0)k = −2i + j + 2k [1 mark]

(b) Area = (\frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}|)
(\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 2 & -1 \ -2 & 1 & 2 \end{vmatrix})
= i((2)(2) − (−1)(1)) − j((1)(2) − (−1)(−2)) + k((1)(1) − (2)(−2))
= i(4 + 1) − j(2 − 2) + k(1 + 4)
= 5i + 0j + 5k = ( \begin{pmatrix} 5 \ 0 \ 5 \end{pmatrix} )

|(\overrightarrow{AB} \times \overrightarrow{AC})| = (\sqrt{25 + 0 + 25} = \sqrt{50} = 5\sqrt{2})
Area = (\frac{1}{2} \times 5\sqrt{2} = \frac{5\sqrt{2}}{2}) square units [3 marks: M1 for cross product, M1 for magnitude, A1 for area]

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18. Line (l): (\frac{x-2}{3} = \frac{y+1}{-2} = \frac{z-4}{1} = \lambda)

Parametric: x = 2 + 3λ, y = −1 − 2λ, z = 4 + λ

Let foot of perpendicular from origin O to (l) be F(2+3λ, −1−2λ, 4+λ).
(\overrightarrow{OF}) is perpendicular to direction of (l), so (\overrightarrow{OF} \cdot \begin{pmatrix} 3 \ -2 \ 1 \end{pmatrix} = 0)
( \begin{pmatrix} 2+3\lambda \ -1-2\lambda \ 4+\lambda \end{pmatrix} \cdot \begin{pmatrix} 3 \ -2 \ 1 \end{pmatrix} = 0 )
3(2+3λ) + (−2)(−1−2λ) + 1(4+λ) = 0
6 + 9λ + 2 + 4λ + 4 + λ = 0
14λ + 12 = 0 ⇒ λ = −6/7

F: x = 2 + 3(−6/7) = 2 − 18/7 = −4/7
y = −1 − 2(−6/7) = −1 + 12/7 = 5/7
z = 4 + (−6/7) = 22/7

Foot: (−4/7, 5/7, 22/7) [4 marks: M1 for parametric form, M1 for perpendicular condition, M1 for solving λ, A1 for coordinates]

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19. A(1, 2, 3), B(2, 4, 5), C(3, 1, 0), D(4, 3, 2)

Volume of tetrahedron = (\frac{1}{6}|\overrightarrow{AB} \cdot (\overrightarrow{AC} \times \overrightarrow{AD})|) (absolute value of scalar triple product)

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

(\overrightarrow{AC} \times \overrightarrow{AD} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & -1 & -3 \ 3 & 1 & -1 \end{vmatrix})
= i((−1)(−1) − (−3)(1)) − j((2)(−1) − (−3)(3)) + k((2)(1) − (−1)(3))
= i(1 + 3) − j(−2 + 9) + k(2 + 3)
= 4i − 7j + 5k

(\overrightarrow{AB} \cdot (\overrightarrow{AC} \times \overrightarrow{AD}) = \begin{pmatrix} 1 \ 2 \ 2 \end{pmatrix} \cdot \begin{pmatrix} 4 \ -7 \ 5 \end{pmatrix} = 4 - 14 + 10 = 0)

Volume = (\frac{1}{6}|0| = 0)

The points are coplanar, so the volume is 0. [4 marks: M1 for scalar triple product formula, M1 for vectors, M1 for cross and dot products, A1 for volume]

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20. Plane through (1, 0, 0), (0, 2, 0), (0, 0, 3)

(a) Intercept form: (\frac{x}{1} + \frac{y}{2} + \frac{z}{3} = 1)
Multiply by 6: 6x + 3y + 2z = 6 [2 marks: M1 for intercept form or normal vector method, A1 for equation]

(b) Distance from (1, 1, 1) to 6x + 3y + 2z = 6:
d = (\frac{|6(1) + 3(1) + 2(1) - 6|}{\sqrt{36 + 9 + 4}} = \frac{|6+3+2-6|}{\sqrt{49}} = \frac{5}{7}) [2 marks: M1 for distance formula, A1 for result]

(c) Reflection: point P(1,1,1), foot F, reflected point P′ such that F is midpoint of PP′.
Normal vector n = ( \begin{pmatrix} 6 \ 3 \ 2 \end{pmatrix} ).
Line through P perpendicular to plane: r = ( \begin{pmatrix} 1 \ 1 \ 1 \end{pmatrix} + t \begin{pmatrix} 6 \ 3 \ 2 \end{pmatrix} )
Parametric: x = 1+6t, y = 1+3t, z = 1+2t
Substitute into plane: 6(1+6t) + 3(1+3t) + 2(1+2t) = 6
6 + 36t + 3 + 9t + 2 + 4t = 6
49t + 11 = 6 ⇒ 49t = −5 ⇒ t = −5/49

Foot F: x = 1 + 6(−5/49) = 1 − 30/49 = 19/49
y = 1 + 3(−5/49) = 1 − 15/49 = 34/49
z = 1 + 2(−5/49) = 1 − 10/49 = 39/49

Reflected point P′: F is midpoint, so P′ = 2F − P
x′ = 2(19/49) − 1 = 38/49 − 49/49 = −11/49
y′ = 2(34/49) − 1 = 68/49 − 49/49 = 19/49
z′ = 2(39/49) − 1 = 78/49 − 49/49 = 29/49

P′ = (−11/49, 19/49, 29/49) [2 marks: M1 for method (finding foot or using formula), A1 for coordinates]

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END OF ANSWER KEY