C4 June 2006 Q5
5. The point \(A\), with coordinates \((0, a, b)\) lies on the line \(l_1\), which has equation
\[\mathbf{r} = 6\mathbf{i} + 19\mathbf{j} - \mathbf{k} + \lambda(\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}).\]
The point \(P\) lies on \(l_1\) and is such that \(OP\) is perpendicular to \(l_1\), where \(O\) is the origin.
Given that \(B\) has coordinates \((5, 15, 1)\),
| Scheme | Marks |
|---|---|
| Equating \(\mathbf{i}\); \(0 = 6 + \lambda \Rightarrow \lambda = -6\) | B1 ⇒ d |
| Using \(\lambda = -6\) and equating \(\mathbf{j}\); \(a = 19 + 4(-6) = -5\) | M1 ⇒ d |
| equating \(\mathbf{k}\); \(b = -1 - 2(-6) = 11\) | A1 |
| (3) |
Notes
B1 \(\lambda = -6\). Can be implied
M1 For inserting their stated \(\lambda\) into either a correct \(\mathbf{j}\) or \(\mathbf{k}\) component. Can be implied.
A1 \(a = -5\) and \(b = 11\)
With no working…
… only one of \(a\) or \(b\) stated correctly gains the first 2 marks.
… both \(a\) and \(b\) stated correctly gains 3 marks.
| Scheme | Marks |
|---|---|
| \(\overrightarrow{OP} = (6 + \lambda)\mathbf{i} + (19 + 4\lambda)\mathbf{j} + (-1 - 2\lambda)\mathbf{k}\) direction vector or \(l_1 = \mathbf{d} = \mathbf{i} + 4\mathbf{j} - 2\mathbf{k}\) | |
| \(\overrightarrow{OP} \perp l_1 \Rightarrow \underline{\overrightarrow{OP} \bullet \mathbf{d} = 0}\) ie. \(\begin{pmatrix}6 + \lambda\\19 + 4\lambda\\-1 - 2\lambda\end{pmatrix} \bullet \begin{pmatrix}1\\4\\-2\end{pmatrix} = 0 \quad \left(\text{or } \underline{x + 4y - 2z = 0}\right)\) | M1 |
| \(\therefore 6 + \lambda + 4(19 + 4\lambda) - 2(-1 - 2\lambda) = 0\) | A1 oe |
| \(6 + \lambda + 76 + 16\lambda + 2 + 4\lambda = 0\) | dM1 |
| \(21\lambda + 84 = 0 \Rightarrow \lambda = -4\) | A1 |
| \(\overrightarrow{OP} = (6 - 4)\mathbf{i} + (19 + 4(-4))\mathbf{j} + (-1 - 2(-4))\mathbf{k}\) | M1 |
| \(\overrightarrow{OP} = 2\mathbf{i} + 3\mathbf{j} + 7\mathbf{k}\) | A1 |
| (6) |
Notes
M1 Allow this statement for M1 if \(\overrightarrow{OP}\) and \(\mathbf{d}\) are defined as above. Allow either of these two underlined statements
A1 oe Correct equation
dM1 Attempt to solve the equation in \(\lambda\)
A1 \(\lambda = -4\)
M1 Substitutes their \(\lambda\) into an expression for \(\overrightarrow{OP}\)
A1 \(2\mathbf{i} + 3\mathbf{j} + 7\mathbf{k}\) or \(P(2, 3, 7)\)
Note: A similar method may be used by using \(\overrightarrow{OP} = (0 + \lambda)\mathbf{i} + (-5 + 4\lambda)\mathbf{j} + (11 - 2\lambda)\mathbf{k}\) and \(\mathbf{d} = \mathbf{i} + 4\mathbf{j} - 2\mathbf{k}\)
\(\overrightarrow{OP} \bullet \mathbf{d} = 0\) yields \(6 + \lambda + 4(-5 + 4\lambda) - 2(11 - 2\lambda) = 0\)
This simplifies to \(21\lambda - 42 = 0 \Rightarrow \lambda = 2\).
\(\overrightarrow{OP} = (0 + 2)\mathbf{i} + (-5 + 4(2))\mathbf{j} + (11 - 2(2))\mathbf{k}\)
\(\overrightarrow{OP} = 2\mathbf{i} + 3\mathbf{j} + 7\mathbf{k}\)
Aliter (b) Way 2
| \(\overrightarrow{OP} = (6 + \lambda)\mathbf{i} + (19 + 4\lambda)\mathbf{j} + (-1 - 2\lambda)\mathbf{k}\) \(\overrightarrow{AP} = (6 + \lambda - 0)\mathbf{i} + (19 + 4\lambda + 5)\mathbf{j} + (-1 - 2\lambda - 11)\mathbf{k}\) direction vector or \(l_1 = \mathbf{d} = \mathbf{i} + 4\mathbf{j} - 2\mathbf{k}\) | |
| \(\overrightarrow{AP} \perp \overrightarrow{OP} \Rightarrow \underline{\overrightarrow{AP} \bullet \overrightarrow{OP} = 0}\) ie. \(\underline{\begin{pmatrix}6 + \lambda\\24 + 4\lambda\\-12 - 2\lambda\end{pmatrix} \bullet \begin{pmatrix}6 + \lambda\\19 + 4\lambda\\-1 - 2\lambda\end{pmatrix} = 0}\) | M1 |
| \(\therefore (6 + \lambda)(6 + \lambda) + (24 + 4\lambda)(19 + 4\lambda) + (-12 - 2\lambda)(-1 - 2\lambda) = 0\) | A1 oe |
| \(36 + 12\lambda + \lambda^2 + 456 + 96\lambda + 76\lambda + 16\lambda^2 + 12 + 24\lambda + 2\lambda + 4\lambda^2 = 0\) \(21\lambda^2 + 210\lambda + 504 = 0\) | dM1 |
| \(\lambda^2 + 10\lambda + 24 = 0 \Rightarrow (\lambda = -6)\quad \underline{\lambda = -4}\) | A1 |
| \(\overrightarrow{OP} = (6 - 4)\mathbf{i} + (19 + 4(-4))\mathbf{j} + (-1 - 2(-4))\mathbf{k}\) | M1 |
| \(\overrightarrow{OP} = 2\mathbf{i} + 3\mathbf{j} + 7\mathbf{k}\) | A1 |
| [6] |
M1 Allow this statement for M1 if \(\overrightarrow{AP}\) and \(\overrightarrow{OP}\) are defined as above. underlined statement
A1 oe Correct equation
dM1 Attempt to solve the equation in \(\lambda\)
A1 \(\lambda = -4\)
M1 Substitutes their \(\lambda\) into an expression for \(\overrightarrow{OP}\)
A1 \(2\mathbf{i} + 3\mathbf{j} + 7\mathbf{k}\) or \(P(2, 3, 7)\)
Note: A similar method to way 2 may be used by using \(\overrightarrow{OP} = (5 + \lambda)\mathbf{i} + (15 + 4\lambda)\mathbf{j} + (1 - 2\lambda)\mathbf{k}\) and \(\overrightarrow{AP} = (5 + \lambda - 0)\mathbf{i} + (15 + 4\lambda + 5)\mathbf{j} + (1 - 2\lambda - 11)\mathbf{k}\)
\(\overrightarrow{AP} \bullet \overrightarrow{OP} = 0\) yields \((5 + \lambda)(5 + \lambda) + (20 + 4\lambda)(15 + 4\lambda) + (-10 - 2\lambda)(1 - 2\lambda) = 0\)
This simplifies to \(21\lambda^2 + 168\lambda + 315 = 0\). \(\lambda^2 + 8\lambda + 15 = 0 \Rightarrow (\lambda = -5)\quad \underline{\lambda = -3}\)
\(\overrightarrow{OP} = (5 - 3)\mathbf{i} + (15 + 4(-3))\mathbf{j} + (1 - 2(-3))\mathbf{k}\)
\(\overrightarrow{OP} = 2\mathbf{i} + 3\mathbf{j} + 7\mathbf{k}\)
| Scheme | Marks |
|---|---|
| \(\overrightarrow{OP} = 2\mathbf{i} + 3\mathbf{j} + 7\mathbf{k}\) \(\overrightarrow{OA} = 0\mathbf{i} - 5\mathbf{j} + 11\mathbf{k}\) and \(\overrightarrow{OB} = 5\mathbf{i} + 15\mathbf{j} + \mathbf{k}\) | |
| \(\overrightarrow{AP} = \pm(2\mathbf{i} + 8\mathbf{j} - 4\mathbf{k}),\ \overrightarrow{PB} = \pm(3\mathbf{i} + 12\mathbf{j} - 6\mathbf{k})\) \(\overrightarrow{AB} = \pm(5\mathbf{i} + 20\mathbf{j} - 10\mathbf{k})\) | M1; A1ft ± |
| As \(\overrightarrow{AP} = \frac{2}{3}(3\mathbf{i} + 12\mathbf{j} - 6\mathbf{k}) = \frac{2}{3}\overrightarrow{PB}\) or \(\overrightarrow{AB} = \frac{5}{2}(2\mathbf{i} + 8\mathbf{j} - 4\mathbf{k}) = \frac{5}{2}\overrightarrow{AP}\) or \(\overrightarrow{AB} = \frac{5}{3}(3\mathbf{i} + 12\mathbf{j} - 6\mathbf{k}) = \frac{5}{3}\overrightarrow{PB}\) or \(\overrightarrow{PB} = \frac{3}{2}(2\mathbf{i} + 8\mathbf{j} - 4\mathbf{k}) = \frac{3}{2}\overrightarrow{AP}\) or \(\overrightarrow{AP} = \frac{2}{5}(5\mathbf{i} + 20\mathbf{j} - 10\mathbf{k}) = \frac{2}{5}\overrightarrow{AB}\) or \(\overrightarrow{PB} = \frac{3}{5}(5\mathbf{i} + 20\mathbf{j} - 10\mathbf{k}) = \frac{3}{5}\overrightarrow{AB}\) etc… | |
| alternatively candidates could say for example that \(\overrightarrow{AP} = 2(\mathbf{i} + 4\mathbf{j} - 2\mathbf{k})\quad \overrightarrow{PB} = 3(\mathbf{i} + 4\mathbf{j} - 2\mathbf{k})\) then the points \(A\), \(P\) and \(B\) are collinear. | A1 |
| \(\therefore \overrightarrow{AP} : \overrightarrow{PB} = 2 : 3\) | B1 oe |
| (4) | |
| (13 marks) |
Notes
M1; A1ft Subtracting vectors to find any two of \(\overrightarrow{AP}\), \(\overrightarrow{PB}\) or \(\overrightarrow{AB}\); and both are correctly ft using candidate’s \(\overrightarrow{OA}\) and \(\overrightarrow{OP}\) found in parts (a) and (b) respectively.
\(\overrightarrow{AP} = \frac{2}{3}\overrightarrow{PB}\) or \(\overrightarrow{AB} = \frac{5}{2}\overrightarrow{AP}\) or \(\overrightarrow{AB} = \frac{5}{3}\overrightarrow{PB}\) or \(\overrightarrow{PB} = \frac{3}{2}\overrightarrow{AP}\) or \(\overrightarrow{AP} = \frac{2}{5}\overrightarrow{AB}\) or \(\overrightarrow{PB} = \frac{3}{5}\overrightarrow{AB}\)
A1 \(A\), \(P\) and \(B\) are collinear. Completely correct proof.
B1 oe \(2 : 3\) or \(1 : \frac{3}{2}\) or \(\sqrt{84} : \sqrt{189}\) aef, allow SC \(\frac{2}{3}\)
Aliter (c) Way 2
| At \(B\); \(\underline{5 = 6 + \lambda}\), \(\underline{15 = 19 + 4\lambda}\) or \(\underline{1 = -1 - 2\lambda}\) or at \(B\); \(\lambda = -1\) | M1 |
| gives \(\lambda = -1\) for all three equations. or when \(\lambda = -1\), this gives \(\mathbf{r} = 5\mathbf{i} + 15\mathbf{j} + \mathbf{k}\) | A1 |
| Hence \(B\) lies on \(l_1\). As stated in the question both \(A\) and \(P\) lie on \(l_1\). \(\therefore\) \(A\), \(P\) and \(B\) are collinear. | A1 |
| \(\therefore \overrightarrow{AP} : \overrightarrow{PB} = 2 : 3\) | B1 oe |
| [4] |
M1 Writing down any of the three underlined equations.
A1 \(\lambda = -1\) for all three equations or \(\lambda = -1\) gives \(\mathbf{r} = 5\mathbf{i} + 15\mathbf{j} + \mathbf{k}\)
A1 Must state \(B\) lies on \(l_1\) \(\Rightarrow\) \(A\), \(P\) and \(B\) are collinear
B1 oe \(2 : 3\) or aef
Beware of candidates who will try to fudge that one vector is multiple of another for the final A mark in part (c).