A2 June 2024 Paper 1 Q11
11 The plane \(\Pi\) has equation \(2x - y + 2z = 4\). The point P has coordinates (8, 4, 5).
The line L has equation \(\dfrac{x - 2}{3} = \dfrac{y}{2} = \dfrac{z + 3}{4}\).
| Scheme | Marks | AO |
|---|---|---|
| \(d = \dfrac{|2 \times 8 + (-1) \times 4 + 2 \times 5 - 4|}{\sqrt{2^2 + (-1)^2 + 2^2}}\) | M1 | 1.1 |
| \(= 6\) | A1 | 1.1 |
| [2] |
Notes
M1: Use of distance from point to plane formula, must be correct. Condone missing modulus.
Alternative method
| Scheme | Marks |
|---|---|
| Perpendicular from P to plane intersects plane at \((4, 6, 1)\) \(d = \sqrt{(8 - 4)^2 + (4 - 6)^2 + (5 - 1)^2}\) | M1 |
| \(= 6\) | A1 |
| [2] |
M1: No slips allowed
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{8 - 2}{3} = \dfrac{4}{2} = \dfrac{5 + 3}{4} \quad [= 2]\) | M1 | 2.1 |
| so P(8, 4, 5) lies on the line | A1 | 2.2a |
| [2] |
Notes
M1: Substituting coordinates into line equation
A1: Conclusion
Alternative method
| Scheme | Marks |
|---|---|
| \(x = 2 + 3\lambda, y = 2\lambda, z = -3 + 4\lambda\) \(2 + 3\lambda = 8\) \(\Rightarrow \lambda = 2\) | M1 |
| \(y = 2 \times 2 = 4, z = -3 + 4 \times 2 = 5\) so P(8, 4, 5) lies on the line | A1 |
| [2] |
M1: A method to find \(\lambda\) from one coordinate
A1: Must show substitution for all three coordinates and conclude
| Scheme | Marks | AO |
|---|---|---|
| \(2(2 + 3\lambda) - 2\lambda + 2(-3 + 4\lambda) = 4\) | M1 | 1.1 |
| \(\Rightarrow \lambda = \frac{1}{2}\) | A1 | 1.1 |
| Point of intersection is \(\left(\frac{7}{2}, 1, -1\right)\) | A1 | 1.1 |
| [3] |
Notes
M1: or \((2\mathbf{i} - \mathbf{j} + 2\mathbf{k}).\left((2 + 3\lambda)\mathbf{i} + 2\lambda\mathbf{j} + (-3 + 4\lambda)\mathbf{k}\right) = 4\)
Allow a slip
A1: Accept \(x = \frac{7}{2}, y = 1, z = -1\)
Do not accept a position vector
Alternative method
| Scheme | Marks |
|---|---|
| \(x = \frac{3y + 4}{2}\) and \(z = 2y - 3 \Rightarrow 2\left(\frac{3y + 4}{2}\right) - y + 2(2y - 3) = 4\) | M1 |
| \(y = 1\) | A1 |
| Point of intersection is \(\left(\frac{7}{2}, 1, -1\right)\) | A1 |
| [3] |
M1: For an equation of the plane in only one of \(x\), \(y\) or \(z\). Allow a slip.
A1: Solving for the correct coordinate
A1: Accept \(x = \frac{7}{2}, y = 1, z = -1\)
Do not accept a position vector
| Scheme | Marks | AO |
|---|---|---|
| Let angle between line and normal = \(\theta\) \(\cos\theta = \dfrac{(2\mathbf{i} - \mathbf{j} + 2\mathbf{k}).(3\mathbf{i} + 2\mathbf{j} + 4\mathbf{k})}{\sqrt{2^2 + (-1)^2 + 2^2} \times \sqrt{3^2 + 2^2 + 4^2}}\) | M1 | 1.1 |
| \(= \dfrac{12}{\sqrt{9}\sqrt{29}}\) | 1.1 | |
| \(\theta = 42.03^\circ\) | A1 | |
| angle between line and plane = \(90 - 42.03\) | M1 | 1.1 |
| \(= 48.0^\circ\) | A1 | 1.1 |
| [4] |
Notes
M1: Must see a dot product or sum of three terms
A1: or 0.73[358…] rad
M1: Allow their angle \(\theta - 90\) if their \(\theta\) obtuse. Can be implied by correct final answer if correct \(\theta\) found.
A1: or 0.837… rad. www.
Alternative method 1
| Scheme | Marks |
|---|---|
| Let angle between line and plane \(= \phi\) \(\sin\phi = \dfrac{(2\mathbf{i} - \mathbf{j} + 2\mathbf{k}).(3\mathbf{i} + 2\mathbf{j} + 4\mathbf{k})}{\sqrt{2^2 + (-1)^2 + 2^2} \times \sqrt{3^2 + 2^2 + 4^2}}\) \(= \dfrac{12}{\sqrt{9}\sqrt{29}}\) | M2 |
| \(\phi = 48.0^\circ\) | A2 |
| [4] |
M2: Must see a dot product or sum of three terms
A2: or 0.837… rad
Alternative method 2
| Scheme | Marks |
|---|---|
| Let angle between line and normal \(= \theta\) \(\sin\theta = \dfrac{|(2\mathbf{i} - \mathbf{j} + 2\mathbf{k}) \times (3\mathbf{i} + 2\mathbf{j} + 4\mathbf{k})|}{\sqrt{2^2 + (-1)^2 + 2^2} \times \sqrt{3^2 + 2^2 + 4^2}}\) \(= \dfrac{\sqrt{117}}{\sqrt{9}\sqrt{29}}\) | M1 |
| \(\theta = 42.03^\circ\) | A1 |
| angle between line and plane = \(90 - 42.03\) | M1 |
| \(= 48.0^\circ\) | A1 |
| [4] |
M1: Complete method using vector product. Do not condone missing modulus unless a correct value for \(\sin\theta\) is given
A1: or 0.73[358…] rad
M1: Allow their angle \(\theta - 90\) if their \(\theta\) obtuse. Can be implied by correct final answer if correct \(\theta\) found.
A1: or 0.837… rad. www.
| Scheme | Marks | AO |
|---|---|---|
| P to intersection \(= \sqrt{\left(8 - \frac{7}{2}\right)^2 + (4 - 1)^2 + (5 + 1)^2}\) | M1 | 3.1a |
| \(= \dfrac{3\sqrt{29}}{2}\) | A1 | 1.1 |
| \(\dfrac{3\sqrt{29}}{2} \times \dfrac{12}{3\sqrt{29}} = 6\) | A1 | 3.2a |
| [3] |
Notes
M1: FT their coordinates from (c)
A1: or 8.0777... Do not FT.
A1: Condone using approximate values, e.g \(\sin 48^\circ\) or \(\cos 42^\circ\) for \(\frac{12}{3\sqrt{29}}\), in which case condone \(\approx\) for =
FT their angle labels from previous parts