A2 June 2023 Paper 1 Q16
16 The point P \((4, 1, 0)\) is equidistant from the plane \(2x + y + 2z = 0\) and the line \(\dfrac{x - 3}{2} = \dfrac{y - 1}{b} = \dfrac{z + 5}{3}\), where \(b \gt 0\).
Determine the value of \(b\). [10]
| Scheme | Marks | AO |
|---|---|---|
| \(\text{Distance from point to plane} = \dfrac{|2 \times 4 + 1 \times 1 + 0 \times 2|}{\sqrt{2^2 + 1^2 + 2^2}}\) | M1* | 3.1a |
| \(= 3\) units | A1 | 1.1 |
| Line is \(\mathbf{r} = 3\mathbf{i} + \mathbf{j} - 5\mathbf{k} + \lambda(2\mathbf{i} + b\mathbf{j} + 3\mathbf{k})\) | M1 | 3.1a |
| \(\overrightarrow{AP} = \mathbf{i} + 5\mathbf{k}\) | A1 | 2.1 |
| \(\overrightarrow{AP} \times \mathbf{d} = (\mathbf{i} + 5\mathbf{k}) \times (2\mathbf{i} + b\mathbf{j} + 3\mathbf{k})\) | M1 | 2.1 |
| \(= -5b\mathbf{i} + 7\mathbf{j} + b\mathbf{k}\) | A1 | 2.1 |
| dist from P to line \(\dfrac{|-5b\mathbf{i} + 7\mathbf{j} + b\mathbf{k}|}{|\mathbf{d}|} = \dfrac{\sqrt{26b^2 + 49}}{\sqrt{13 + b^2}}\) | M1* A1 | 3.1a 1.1 |
| so \(\frac{26b^2 + 49}{13 + b^2} = 9 \Rightarrow 26b^2 + 49 = 117 + 9b^2\) | M1dep | 2.1 |
| \(\Rightarrow b = 2\) | A1 | 3.2a |
| [10] |
Notes
A1: (2nd) or \(-\overrightarrow{AP}\). Could be seen as part of distance calculation.
M1: (3rd) or \(\mathbf{d} \times \overrightarrow{AP}\)
M1dep: equating the two distances
Alternative method 1
| Scheme | Marks |
|---|---|
| \(\text{Distance from point to plane} = \dfrac{|2 \times 4 + 1 \times 1 + 0 \times 2|}{\sqrt{2^2 + 1^2 + 2^2}}\) | M1* |
| \(= 3\) units | A1 |
| Line is \(\mathbf{r} = 3\mathbf{i} + \mathbf{j} - 5\mathbf{k} + \lambda(2\mathbf{i} + b\mathbf{j} + 3\mathbf{k})\) | M1 |
| \((2\lambda - 1)\mathbf{i} + b\lambda\mathbf{j} + (3\lambda - 5)\mathbf{k}\) | A1 |
| \(\begin{pmatrix} 2\lambda - 1 \\ b\lambda \\ 3\lambda - 5 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ b \\ 3 \end{pmatrix} = 0\) | M1* |
| \(\lambda = \dfrac{17}{13 + b^2}\) | A1 |
| dist from P to line \(\sqrt{\left(2\left(\frac{17}{13 + b^2}\right) - 1\right)^2 + \left(b\left(\frac{17}{13 + b^2}\right)\right)^2 + \left(3\left(\frac{17}{13 + b^2}\right) - 5\right)^2}\) | M1 |
| \(= \dfrac{\sqrt{26b^4 + 387b^2 + 637}}{13 + b^2}\) | A1 |
| So \(\frac{26b^4 + 387b^2 + 637}{(13 + b^2)^2} = 9\) \(\Rightarrow 17b^4 + 153b^2 - 884 = 0\) | M1dep |
| \(\Rightarrow b = 2\) | A1 |
A1: (2nd) vector from P to a point on the line
M1: (3rd) substituting \(\lambda\) into \(\left|\overrightarrow{AP}\right|\)
A1: (4th) simplified
M1dep: equating the two distances
Alternative method 2
| Scheme | Marks |
|---|---|
| \(\text{Distance from point to plane} = \dfrac{|2 \times 4 + 1 \times 1 + 0 \times 2|}{\sqrt{2^2 + 1^2 + 2^2}}\) | M1* |
| \(= 3\) units | A1 |
| Line is \(\mathbf{r} = 3\mathbf{i} + \mathbf{j} - 5\mathbf{k} + \lambda(2\mathbf{i} + b\mathbf{j} + 3\mathbf{k})\) | M1 |
| \((2\lambda - 1)\mathbf{i} + b\lambda\mathbf{j} + (3\lambda - 5)\mathbf{k}\) | A1 |
| \(\sqrt{(2\lambda - 1)^2 + (b\lambda)^2 + (3\lambda - 5)^2}\) | M1* |
| \(\sqrt{(13 + b^2)\lambda^2 - 34\lambda + 26}\) | A1 |
| \(\sqrt{(13 + b^2)\lambda^2 - 34\lambda + 26} = 3\) | M1dep |
| \((13 + b^2)\lambda^2 - 34\lambda + 17 = 0\) | A1 |
| So \((-34)^2 - 4(13 + b^2)(17) = 0\) | M1 |
| \(\Rightarrow b = 2\) | A1 |
A1: (2nd) vector from P to a point on the line
M1*: (2nd) finding magnitude of this vector
A1: (3rd) expanding and simplifying
M1dep: setting distances equal
A1: (4th) correct quadratic equation \(= 0\).
M1: (last) setting discriminant equal to 0
Alternative method 3
| Scheme | Marks |
|---|---|
| \(\text{Distance from point to plane} = \dfrac{|2 \times 4 + 1 \times 1 + 0 \times 2|}{\sqrt{2^2 + 1^2 + 2^2}}\) | M1* |
| \(= 3\) units | A1 |
| Line is \(\mathbf{r} = 3\mathbf{i} + \mathbf{j} - 5\mathbf{k} + \lambda(2\mathbf{i} + b\mathbf{j} + 3\mathbf{k})\) | M1 |
| \(\overrightarrow{AP} = \mathbf{i} + 5\mathbf{k}\) | A1 |
| \(\begin{pmatrix} 1 \\ 0 \\ 5 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ b \\ 3 \end{pmatrix} = \left|\begin{pmatrix} 1 \\ 0 \\ 5 \end{pmatrix}\right|\left|\begin{pmatrix} 2 \\ b \\ 3 \end{pmatrix}\right|\cos\theta\) | M1 |
| \(1 \times 2 + 5 \times 3 = \sqrt{1^2 + 5^2}\sqrt{2^2 + b^2 + 3^2}\cos\theta\) | M1 |
| \(\cos\theta = \dfrac{17}{\sqrt{26}\sqrt{13 + b^2}}\) | A1 |
| \(\sqrt{26}\cos\theta = \dfrac{17}{\sqrt{13 + b^2}}\) | M1* |
| \(26 - \left(\dfrac{17}{\sqrt{13 + b^2}}\right)^2 = 3^2\) | M1dep |
| \(\Rightarrow b = 2\) | A1 |
A1: (2nd) or \(-\overrightarrow{AP}\). Could be seen as part of distance calculation.
M1: (3rd) scalar product formula with \(\overrightarrow{AP} \cdot \mathbf{d}\) to find a value for \(\cos\theta\)
M1: (4th) evaluating scalar product and magnitudes
A1: (3rd) expression for \(\cos\theta\)
M1*: (2nd) expression for the distance from \((3, 1, -5)\) to the foot of the perpendicular to the line
M1dep: using their value correctly with Pythagoras oe to lead to a value for \(b\)