AS June 2022 Paper 1 Q8
8 The line segment \(AB\) is a diameter of a sphere, \(S\). The point \(C\) is any point on the surface of \(S\).
You are now given that \(A\) is the point \((11, 12, -14)\) and \(B\) is the point \((9, 13, 6)\).
| Scheme | Marks | AO |
|---|---|---|
| (If \(C\) is not at \(A\) or \(B\)) whatever plane \(A\), \(B\) and \(C\) lie in, \(AB\) is the diameter of a circle and \(C\) is on the perimeter so angle \(ACB\) is a right angle because it is an angle in a semi-circle. | B1 | 3.1a |
| So \(\overrightarrow{AC}.\overrightarrow{BC} = 0\) since \(\overrightarrow{AC}\) and \(\overrightarrow{BC}\) are perpendicular and the dot product of perpendicular vectors is zero (because \(\cos 90^\circ = 0\)) | B1 | 2.4 |
| In the case where \(C = A\) or \(C = B\) then \(\overrightarrow{AC} = \mathbf{0}\) or \(\overrightarrow{BC} = \mathbf{0}\) and so \(\overrightarrow{AC}.\overrightarrow{BC} = 0\) since the zero vector dotted with any vector is zero. | B1 | 2.1 |
| [3] |
Notes
B1: (1st) Angle \(ACB\) is “an angle in a semi-circle” so \(90^\circ\) oe.
Could choose to prove the circle theorem.
B1: (2nd) A reason must be given. Just “So \(\overrightarrow{AC}.\overrightarrow{BC} = 0\)” is insufficient.
“The dot product of perpendicular vectors is zero.”
Implication must be the correct way around.
B1: (3rd) Must mention both cases but second case can just be covered by “Similarly for \(B\)” or “eg consider \(A\)” oe.
Alternative method
| Scheme | Marks |
|---|---|
| Let the centre of the sphere be at \(O\). Let the coordinates of \(A = (\alpha, \beta, \gamma)\) where \(\alpha^2 + \beta^2 + \gamma^2 = r^2\) Since AB is a diameter, B has coordinates \((-\alpha, -\beta, -\gamma)\) Let C have coordinates \((x, y, z)\), where \(x^2 + y^2 + z^2 = r^2\) \(\overrightarrow{AC} = \begin{pmatrix} x - \alpha \\ y - \beta \\ z - \gamma \end{pmatrix}\) \(\overrightarrow{BC} = \begin{pmatrix} x + \alpha \\ y + \beta \\ z + \gamma \end{pmatrix}\) | B1 |
| Then \(\overrightarrow{AC} \cdot \overrightarrow{BC} = (x - \alpha)(x + \alpha) + (y - \beta)(y + \beta) + (z - \gamma)(z + \gamma)\) | B1 |
| \(\overrightarrow{AC} \cdot \overrightarrow{BC} = (x^2 + y^2 + z^2) - (\alpha^2 + \beta^2 + \gamma^2)\) \(\overrightarrow{AC} \cdot \overrightarrow{BC} = r^2 - r^2 = 0\) | B1 |
B1: (1st) Centre at \(O\) and attempting \(\overrightarrow{AC}\) or \(\overrightarrow{BC}\)
B1: (2nd) \(\overrightarrow{AC}\) and \(\overrightarrow{BC}\) both correct and dot product formed
B1: (3rd) Correctly showing that dot product is 0.
Alternative method 2
| Scheme | Marks |
|---|---|
| Let the centre of the sphere be at \(O\). Let the position vectors of \(A\), \(B\), \(C\) be \(\mathbf{a}, \mathbf{b}, \mathbf{c}\). We have \(|\mathbf{a}| = |\mathbf{b}| = |\mathbf{c}|\) \((= r)\) Then: \(\overrightarrow{AC} \cdot \overrightarrow{BC} = (\mathbf{c} - \mathbf{a}) \cdot (\mathbf{c} - \mathbf{b})\) | B1 |
| \(\overrightarrow{AC} \cdot \overrightarrow{BC} = |\mathbf{c}|^2 - \mathbf{a} \cdot \mathbf{c} - \mathbf{b} \cdot \mathbf{c} + \mathbf{a} \cdot \mathbf{b}\) \(= |\mathbf{c}|^2 - (\mathbf{a} + \mathbf{b}) \cdot \mathbf{c} + \mathbf{a} \cdot \mathbf{b}\) | B1 |
| But \(\mathbf{a} = -\mathbf{b}\) (as A,B ends of diameter) \(\overrightarrow{AC} \cdot \overrightarrow{BC} = |\mathbf{c}|^2 - (\mathbf{a} - \mathbf{a}) \cdot \mathbf{c} - \mathbf{a} \cdot \mathbf{a}\) \(= |\mathbf{c}|^2 - |\mathbf{a}|^2\) \(= r^2 - r^2 = 0\) | B1 |
B1: (1st) Centre at \(O\) and attempting \(\overrightarrow{AC}\) or \(\overrightarrow{BC}\)
B1: (2nd) \(\overrightarrow{AC}\) and \(\overrightarrow{BC}\) both correct and dot product formed
B1: (3rd) Stating and using \(\mathbf{a} = -\mathbf{b}\) to show that dot product is 0.
| Scheme | Marks | AO |
|---|---|---|
| \(\overrightarrow{AC} = \begin{pmatrix} 2p - 11 \\ p - 12 \\ 1 - -14 \end{pmatrix}\) or \(\overrightarrow{BC} = \begin{pmatrix} 2p - 9 \\ p - 13 \\ 1 - 6 \end{pmatrix}\) | M1 | 3.1a |
| \(\overrightarrow{AC} = \begin{pmatrix} 2p - 11 \\ p - 12 \\ 15 \end{pmatrix}\) and \(\overrightarrow{BC} = \begin{pmatrix} 2p - 9 \\ p - 13 \\ -5 \end{pmatrix}\) | A1 | 1.1 |
| \(\overrightarrow{AC}.\overrightarrow{BC} = 0 \Rightarrow \begin{pmatrix} 2p - 11 \\ p - 12 \\ 15 \end{pmatrix}.\begin{pmatrix} 2p - 9 \\ p - 13 \\ -5 \end{pmatrix} =\) \((2p - 11)(2p - 9) + (p - 12)(p - 13) + 15(-5) = 0\) | M1 | 3.1a |
| \(5p^2 - 65p + 180 = 0\) or \(p^2 - 13p + 36 = 0\) | A1 | 1.1 |
| \(p = 4\) or \(p = 9\) | A1 | 1.1 |
| So the possible locations are \((8, 4, 1)\) or \((18, 9, 1)\) | A1FT | 1.1 |
| [6] |
Notes
M1: (1st) Attempt to subtract (in either order)
A1: (1st) Both completely correct (could be \(\overrightarrow{CA} = \begin{pmatrix} 11 - 2p \\ 12 - p \\ -15 \end{pmatrix}\) etc)
M1: (2nd) Attempt to dot their \(\overrightarrow{AC}\) and \(\overrightarrow{BC}\) Dot product must result in a scalar quantity.
“=0” not necessary for M1 here
A1: (2nd) Correct three term quadratic
(must have “= 0” appearing somewhere Might be near start).
A1FT: Solution must be given as coordinates
For FT must have solved a quadratic coming from attempt at dot product
Alternative method
| Scheme | Marks |
|---|---|
| \((|AB|^2 = 2^2 + 1^2 + 20^2 = 405\) \(|AC|^2 = (11 - 2p)^2 + (12 - p)^2 + (-14 - 1)^2\) \(|BC|^2 = (9 - 2p)^2 + (13 - p)^2 + (6 - 1)^2\) | M1 A1 |
| \(|AB|^2 = |AC|^2 + |BC|^2\) \(10p^2 - 130p + 765 = 405\) | M1 |
| \(p^2 - 13p + 36 = 0\) | A1 |
| \(p = 4\) or \(p = 9\) | A1 |
| So the possible locations are \((8, 4, 1)\) or \((18, 9, 1)\) | A1 FT |
M1: (1st) Attempting to find \(|AC|^2\) or \(|BC|^2\)
A1: (1st) Correct expressions for \(|AC|^2\) and \(|BC|^2\)
M1: (2nd) Using Pythagoras
A1: (2nd) Correct 3 term quadratic
A1 FT: Solution must be given as coordinates
For FT must have solved a quadratic coming from attempt at dot product
Alternative method 2
| Scheme | Marks |
|---|---|
| (Centre of sphere is at \((10, 12.5, -4)\)) \(r^2 = (11 - 10)^2 + (12 - 12.5)^2 + (-14 + 4)^2\) \(r^2 = \dfrac{405}{4}\) | M1 |
| Distance OC is: \(|OC|^2 = (2p - 10)^2 + (p - 12.5)^2 + (1 + 4)^2\) \(= \dfrac{405}{4}\) | M1 A1 |
| \(p^2 - 13p + 36 = 0\) | A1 |
| \(p = 4\) or \(p = 9\) | A1 |
| So the possible locations are \((8, 4, 1)\) or \((18, 9, 1)\) | A1 FT |
M1: (1st) Attempting to find radius^2 (could be via diameter)
M1: (2nd) Attempting to find \(|OC|^2\)
A1: (1st) Correct expressions for \(|OC|^2\)
A1: (2nd) Correct 3 term quadratic
A1 FT: Solution must be given as coordinates
For FT must have solved a quadratic coming from attempt at dot product