C4 January 2007 Q7
7. The point \(A\) has position vector \(\mathbf{a} = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k}\) and the point \(B\) has position vector \(\mathbf{b} = \mathbf{i} + \mathbf{j} - 4\mathbf{k}\), relative to an origin \(O\).
\[\mathbf{c} = \mathbf{a} + \mathbf{b}.\]
(1)The diagonals of the rectangle, \(AB\) and \(OC\), meet at the point \(D\).
| Scheme | Marks |
|---|---|
| \(\mathbf{a} = \overrightarrow{OA} = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k} \Rightarrow |\overrightarrow{OA}| = 3\) \(\mathbf{b} = \overrightarrow{OB} = \mathbf{i} + \mathbf{j} - 4\mathbf{k} \Rightarrow |\overrightarrow{OB}| = \sqrt{18}\) \(\overrightarrow{BC} = \pm(2\mathbf{i} + 2\mathbf{j} + \mathbf{k}) \Rightarrow |\overrightarrow{BC}| = 3\) \(\overrightarrow{AC} = \pm(\mathbf{i} + \mathbf{j} - 4\mathbf{k}) \Rightarrow |\overrightarrow{AC}| = \sqrt{18}\) | |
| \(\mathbf{c} = \overrightarrow{OC} = \underline{3\mathbf{i} + 3\mathbf{j} - 3\mathbf{k}}\) | B1 cao |
| (1) |
| Scheme | Marks |
|---|---|
| \(\overrightarrow{OA} \bullet \overrightarrow{OB} = \begin{pmatrix}2\\2\\1\end{pmatrix} \bullet \begin{pmatrix}1\\1\\-4\end{pmatrix} = \underline{2 + 2 - 4} = 0\) or… \(\overrightarrow{BO} \bullet \overrightarrow{BC} = \begin{pmatrix}-1\\-1\\4\end{pmatrix} \bullet \begin{pmatrix}2\\2\\1\end{pmatrix} = \underline{-2 - 2 + 4} = 0\) or… \(\overrightarrow{AC} \bullet \overrightarrow{BC} = \begin{pmatrix}1\\1\\-4\end{pmatrix} \bullet \begin{pmatrix}2\\2\\1\end{pmatrix} = \underline{2 + 2 - 4} = 0\) or… \(\overrightarrow{AO} \bullet \overrightarrow{AC} = \begin{pmatrix}-2\\-2\\-1\end{pmatrix} \bullet \begin{pmatrix}1\\1\\-4\end{pmatrix} = \underline{-2 - 2 + 4} = 0\) | M1 A1 |
| and therefore \(OA\) is perpendicular to \(OB\) and hence \(OACB\) is a rectangle. | A1 cso |
| Area \(= 3 \times \sqrt{18} = 3\sqrt{18} = 9\sqrt{2}\) | M1 M1 A1 |
| (6) |
Notes
M1 An attempt to take the dot product between either \(\overrightarrow{OA}\) and \(\overrightarrow{OB}\), \(\overrightarrow{OA}\) and \(\overrightarrow{AC}\), \(\overrightarrow{AC}\) and \(\overrightarrow{BC}\) or \(\overrightarrow{OB}\) and \(\overrightarrow{BC}\)
A1 Showing the result is equal to zero.
A1 cso perpendicular and \(OACB\) is a rectangle
M1 Using distance formula to find either the correct height or width.
M1 Multiplying the rectangle’s height by its width.
A1 exact value of \(3\sqrt{18}\), \(9\sqrt{2}\), \(\sqrt{162}\) or aef
Aliter 7. (b) (i) Way 2
| \(\mathbf{c} = \overrightarrow{OC} = \pm(3\mathbf{i} + 3\mathbf{j} - 3\mathbf{k})\) \(\overrightarrow{AB} = \pm(-\mathbf{i} - \mathbf{j} - 5\mathbf{k})\) | |
| \(|\overrightarrow{OC}| = \sqrt{(3)^2 + (3)^2 + (-3)^2} = \sqrt{(1)^2 + (1)^2 + (-5)^2} = |\overrightarrow{AB}|\) | M1 |
| As \(|\overrightarrow{OC}| = |\overrightarrow{AB}| = \sqrt{27}\) | A1 |
| then the diagonals are equal, and \(OACB\) is a rectangle. | A1 cso |
| [3] |
M1 A complete method of proving that the diagonals are equal.
A1 Correct result.
A1 cso diagonals are equal and \(OACB\) is a rectangle
Aliter 7. (b) (i) Way 3
| \(\mathbf{a} = \overrightarrow{OA} = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k} \Rightarrow |\overrightarrow{OA}| = 3\) \(\mathbf{b} = \overrightarrow{OB} = \mathbf{i} + \mathbf{j} - 4\mathbf{k} \Rightarrow |\overrightarrow{OB}| = \sqrt{18}\) \(\overrightarrow{BC} = \pm(2\mathbf{i} + 2\mathbf{j} + \mathbf{k}) \Rightarrow |\overrightarrow{BC}| = 3\) \(\overrightarrow{AC} = \pm(\mathbf{i} + \mathbf{j} - 4\mathbf{k}) \Rightarrow |\overrightarrow{AC}| = \sqrt{18}\) \(\mathbf{c} = \overrightarrow{OC} = \pm(3\mathbf{i} + 3\mathbf{j} - 3\mathbf{k}) \Rightarrow |\overrightarrow{OC}| = \sqrt{27}\) \(\overrightarrow{AB} = \pm(-\mathbf{i} - \mathbf{j} - 5\mathbf{k}) \Rightarrow |\overrightarrow{AB}| = \sqrt{27}\) | |
| \((OA)^2 + (AC)^2 = (OC)^2\) or \((BC)^2 + (OB)^2 = (OC)^2\) or \((OA)^2 + (OB)^2 = (AB)^2\) or \((BC)^2 + (AC)^2 = (AB)^2\) or equivalent | |
| \(\Rightarrow \underline{(3)^2 + (\sqrt{18})^2 = \left(\sqrt{27}\right)^2}\) | M1 A1 |
| and therefore \(OA\) is perpendicular to \(OB\) or \(AC\) is perpendicular to \(BC\) and hence \(OACB\) is a rectangle. | A1 cso |
| [3] |
M1 A complete method of proving that Pythagoras holds using their values.
A1 Correct result
A1 cso perpendicular and \(OACB\) is a rectangle
| Scheme | Marks |
|---|---|
| \(\overrightarrow{OD} = \mathbf{d} = \frac{1}{2}(3\mathbf{i} + 3\mathbf{j} - 3\mathbf{k})\) | B1 |
| (1) |
Way 1: using dot product formula
| Scheme | Marks |
|---|---|
| \(\overrightarrow{DA} = \pm\left(\tfrac{1}{2}\mathbf{i} + \tfrac{1}{2}\mathbf{j} + \tfrac{5}{2}\mathbf{k}\right)\) & \(\overrightarrow{DC} = \pm\left(\tfrac{3}{2}\mathbf{i} + \tfrac{3}{2}\mathbf{j} - \tfrac{3}{2}\mathbf{k}\right)\) or \(\overrightarrow{BA} = \pm(\mathbf{i} + \mathbf{j} + 5\mathbf{k})\) & \(\overrightarrow{OC} = \pm(3\mathbf{i} + 3\mathbf{j} - 3\mathbf{k})\) | M1 A1 |
| \(\cos D = (\pm)\dfrac{\begin{pmatrix}0.5\\0.5\\2.5\end{pmatrix} \bullet \begin{pmatrix}1.5\\1.5\\-1.5\end{pmatrix}}{\frac{\sqrt{27}}{2}.\frac{\sqrt{27}}{2}} = (\pm)\underline{\dfrac{\frac{3}{4} + \frac{3}{4} - \frac{15}{4}}{\frac{27}{4}}} = (\pm)\underline{\dfrac{1}{3}}\) | dM1 A1ft |
| \(D = \cos^{-1}\left(-\dfrac{1}{3}\right)\) | ddM1ft |
| \(D = 109.47122\ldots^\circ\) | A1 |
| (6) | |
| (14 marks) |
Notes
M1 Identifies a set of two relevant vectors
A1 Correct vectors \(\pm\)
dM1 Applies dot product formula on multiples of these vectors.
A1ft Correct ft. application of dot product formula
ddM1ft Attempts to find the correct angle \(D\) rather than \(180^\circ - D\).
A1 \(109.5^\circ\) or awrt \(109^\circ\) or \(1.91^c\)
CHECK (the printed mark scheme shows no marks against the dot product line in Way 1; dM1 A1ft are added there as in Ways 2 and 3, so that the part totals 6.)
Aliter (d) Way 2: using dot product formula and direction vectors
| \(\mathrm{d}\overrightarrow{BA} = \pm(\mathbf{i} + \mathbf{j} + 5\mathbf{k})\) & \(\mathrm{d}\overrightarrow{OC} = \pm(\mathbf{i} + \mathbf{j} - \mathbf{k})\) | M1 A1 |
| \(\cos D = (\pm)\dfrac{\begin{pmatrix}1\\1\\-1\end{pmatrix} \bullet \begin{pmatrix}1\\1\\5\end{pmatrix}}{\sqrt{3}.\sqrt{27}} = (\pm)\underline{\dfrac{1 + 1 - 5}{\sqrt{3}.\sqrt{27}}} = (\pm)\underline{\dfrac{1}{3}}\) | dM1 A1ft |
| \(D = \cos^{-1}\left(-\dfrac{1}{3}\right)\) | ddM1ft |
| \(D = 109.47122\ldots^\circ\) | A1 |
| [6] |
M1 Identifies a set of two direction vectors
A1 Correct vectors \(\pm\)
dM1 Applies dot product formula on multiples of these vectors.
A1ft Correct ft. application of dot product formula.
ddM1ft Attempts to find the correct angle \(D\) rather than \(180^\circ - D\).
A1 \(109.5^\circ\) or awrt \(109^\circ\) or \(1.91^c\)
Aliter (d) Way 3: using dot product formula and similar triangles
| \(\mathrm{d}\overrightarrow{OA} = (2\mathbf{i} + 2\mathbf{j} + \mathbf{k})\) & \(\mathrm{d}\overrightarrow{OC} = (\mathbf{i} + \mathbf{j} - \mathbf{k})\) | M1 A1 |
| \(\cos\left(\tfrac{1}{2}D\right) = \dfrac{\begin{pmatrix}2\\2\\1\end{pmatrix} \bullet \begin{pmatrix}1\\1\\-1\end{pmatrix}}{\sqrt{9}.\sqrt{3}} = \underline{\dfrac{2 + 2 - 1}{\sqrt{9}.\sqrt{3}}} = \underline{\dfrac{1}{\sqrt{3}}}\) | dM1 A1ft |
| \(D = 2\cos^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\) | ddM1ft |
| \(D = 109.47122\ldots^\circ\) | A1 |
| [6] |
M1 Identifies a set of two direction vectors
A1 Correct vectors
dM1 Applies dot product formula on multiples of these vectors.
A1ft Correct ft. application of dot product formula.
ddM1ft Attempts to find the correct angle \(D\) by doubling their angle for \(\frac{1}{2}D\).
A1 \(109.5^\circ\) or awrt \(109^\circ\) or \(1.91^c\)
Aliter (d) Way 4: using cosine rule
| \(\overrightarrow{DA} = \tfrac{1}{2}\mathbf{i} + \tfrac{1}{2}\mathbf{j} + \tfrac{5}{2}\mathbf{k}\), \(\overrightarrow{DC} = \tfrac{3}{2}\mathbf{i} + \tfrac{3}{2}\mathbf{j} - \tfrac{3}{2}\mathbf{k}\), \(\overrightarrow{AC} = \mathbf{i} + \mathbf{j} - 4\mathbf{k}\) | |
| \(|\overrightarrow{DA}| = \dfrac{\sqrt{27}}{2},\ |\overrightarrow{DC}| = \dfrac{\sqrt{27}}{2},\ |\overrightarrow{AC}| = \sqrt{18}\) | M1 A1 |
| \(\cos D = \dfrac{\left(\frac{\sqrt{27}}{2}\right)^2 + \left(\frac{\sqrt{27}}{2}\right)^2 - \left(\sqrt{18}\right)^2}{2\left(\frac{\sqrt{27}}{2}\right)\left(\frac{\sqrt{27}}{2}\right)} = \underline{-\dfrac{1}{3}}\) | dM1 A1ft |
| \(D = \cos^{-1}\left(-\dfrac{1}{3}\right)\) | ddM1ft |
| \(D = 109.47122\ldots^\circ\) | A1 |
| [6] |
M1 Attempts to find all the lengths of all three edges of \(\Delta ADC\)
A1 All Correct
dM1 Using the cosine rule formula with correct ‘subtraction’.
A1ft Correct ft application of the cosine rule formula
ddM1ft Attempts to find the correct angle \(D\) rather than \(180^\circ - D\).
A1 \(109.5^\circ\) or awrt \(109^\circ\) or \(1.91^c\)
Aliter (d) Way 5: using trigonometry on a right angled triangle
| \(\overrightarrow{DA} = \tfrac{1}{2}\mathbf{i} + \tfrac{1}{2}\mathbf{j} + \tfrac{5}{2}\mathbf{k}\) \(\overrightarrow{OA} = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k}\) \(\overrightarrow{AC} = \mathbf{i} + \mathbf{j} - 4\mathbf{k}\) Let \(X\) be the midpoint of \(AC\) | |
| \(|\overrightarrow{DA}| = \dfrac{\sqrt{27}}{2},\ |\overrightarrow{DX}| = \tfrac{1}{2}|\overrightarrow{OA}| = \dfrac{3}{2},\ |\overrightarrow{AX}| = \tfrac{1}{2}|\overrightarrow{AC}| = \tfrac{1}{2}\sqrt{18}\) (hypotenuse), (adjacent), (opposite) | M1 A1 |
| \(\sin\left(\tfrac{1}{2}D\right) = \dfrac{\frac{\sqrt{18}}{2}}{\frac{\sqrt{27}}{2}},\quad \cos\left(\tfrac{1}{2}D\right) = \dfrac{\frac{3}{2}}{\frac{\sqrt{27}}{2}}\) or \(\tan\left(\tfrac{1}{2}D\right) = \dfrac{\frac{\sqrt{18}}{2}}{\frac{3}{2}}\) | dM1 A1ft |
| eg. \(D = 2\tan^{-1}\left(\dfrac{\frac{\sqrt{18}}{2}}{\frac{3}{2}}\right)\) | ddM1ft |
| \(D = 109.47122\ldots^\circ\) | A1 |
| [6] |
M1 Attempts to find two out of the three lengths in \(\Delta ADX\)
A1 Any two correct
dM1 Uses correct sohcahtoa to find \(\frac{1}{2}D\)
A1ft Correct ft application of sohcahtoa
ddM1ft Attempts to find the correct angle \(D\) by doubling their angle for \(\frac{1}{2}D\).
A1 \(109.5^\circ\) or awrt \(109^\circ\) or \(1.91^c\)
Aliter (d) Way 6: using trigonometry on a right angled similar triangle \(OAC\)
| \(\overrightarrow{OC} = 3\mathbf{i} + 3\mathbf{j} - 3\mathbf{k}\) \(\overrightarrow{OA} = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k}\) \(\overrightarrow{AC} = \mathbf{i} + \mathbf{j} - 4\mathbf{k}\) | |
| \(|\overrightarrow{OC}| = \sqrt{27},\ |\overrightarrow{OA}| = 3,\ |\overrightarrow{AC}| = \sqrt{18}\) (hypotenuse), (adjacent), (opposite) | M1 A1 |
| \(\sin\left(\tfrac{1}{2}D\right) = \dfrac{\sqrt{18}}{\sqrt{27}},\quad \cos\left(\tfrac{1}{2}D\right) = \dfrac{3}{\sqrt{27}}\) or \(\tan\left(\tfrac{1}{2}D\right) = \dfrac{\sqrt{18}}{3}\) | dM1 A1ft |
| eg. \(D = 2\tan^{-1}\left(\dfrac{\sqrt{18}}{3}\right)\) | ddM1ft |
| \(D = 109.47122\ldots^\circ\) | A1 |
| [6] |
M1 Attempts to find two out of the three lengths in \(\Delta OAC\)
A1 Any two correct
dM1 Uses correct sohcahtoa to find \(\frac{1}{2}D\)
A1ft Correct ft application of sohcahtoa
ddM1ft Attempts to find the correct angle \(D\) by doubling their angle for \(\frac{1}{2}D\).
A1 \(109.5^\circ\) or awrt \(109^\circ\) or \(1.91^c\)