Standard +0.3 This is a straightforward application of the scalar product formula to find the angle between two vectors. Students need to find the diagonal vectors OB and AC, compute their scalar product and magnitudes, then use the formula cos θ = (a·b)/(|a||b|). While it involves multiple steps and 3D coordinates, it's a standard textbook exercise with no conceptual challenges beyond direct application of the method.
2 The points \(O ( 0,0,0 ) , A ( 2,8,2 ) , B ( 5,5,8 )\) and \(C ( 3 , - 3,6 )\) form a parallelogram \(O A B C\). Use a scalar product to find the acute angle between the diagonals of this parallelogram.
\(\pm((3-2)\mathbf{i}+(-3-8)\mathbf{j}+(6-2)\mathbf{k})\) soi
B1
NB \(\mathbf{i}-11\mathbf{j}+4\mathbf{k}\); or B3 for correct use of Cosine Rule using midpoint of diagonals of the parallelogram: \([\cos\theta]=\frac{34.5+28.5-72}{2\sqrt{34.5}\sqrt{28.5}}\) oe
their \(\pm(\mathbf{i}-11\mathbf{j}+4\mathbf{k})\), \(\pm(5\mathbf{i}+5\mathbf{j}+8\mathbf{k})\); both diagonals used; evaluation not essential
M1
if M0 SC2 for \(84°\) (or \(84.5°\)), or \(52(.3°)\) or \(39°\) or \((38.5°\) or \(43(.2°)\) or \(46(.0°)\) found from scalar product or SC1 for the equivalent obtuse angle
1.4 to 1.43 rad; B2 for \(81.7°\) to \(82°\) unsupported or B3+B2 possible for Cosine Rule
[5]
# Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\pm((3-2)\mathbf{i}+(-3-8)\mathbf{j}+(6-2)\mathbf{k})$ soi | B1 | NB $\mathbf{i}-11\mathbf{j}+4\mathbf{k}$; or B3 for correct use of Cosine Rule using midpoint of diagonals of the parallelogram: $[\cos\theta]=\frac{34.5+28.5-72}{2\sqrt{34.5}\sqrt{28.5}}$ oe |
| their $\pm(\mathbf{i}-11\mathbf{j}+4\mathbf{k})$, $\pm(5\mathbf{i}+5\mathbf{j}+8\mathbf{k})$; both diagonals used; evaluation not essential | M1 | if M0 SC2 for $84°$ (or $84.5°$), or $52(.3°)$ or $39°$ or $(38.5°$ or $43(.2°)$ or $46(.0°)$ found from scalar product or SC1 for the equivalent obtuse angle |
| $\pm(1\times5+(-11)\times5+4\times8)=\sqrt{1^2+11^2+4^2}\times\sqrt{5^2+5^2+8^2}\cos\theta$ oe | A1 | Must be fully correct |
| $\theta=\cos^{-1}\frac{\pm18}{\sqrt{138}\times\sqrt{114}}$ | A1 | |
| $81.7°$ to $82°$ | A1 | 1.4 to 1.43 rad; B2 for $81.7°$ to $82°$ unsupported or B3+B2 possible for Cosine Rule |
| **[5]** | | |
2 The points $O ( 0,0,0 ) , A ( 2,8,2 ) , B ( 5,5,8 )$ and $C ( 3 , - 3,6 )$ form a parallelogram $O A B C$. Use a scalar product to find the acute angle between the diagonals of this parallelogram.
\hfill \mbox{\textit{OCR C4 2014 Q2 [5]}}