Easy -1.2 This is a straightforward application of the collinearity test requiring students to show that vectors AB and AC are scalar multiples of each other. It involves basic vector subtraction and checking proportionality—a routine procedure with no problem-solving insight needed, making it easier than average.
## Question 3:
$\overrightarrow{AB} = \begin{pmatrix}-1\\7\\8\end{pmatrix}$ | M1 | AO 3.1a | Attempt to find vector between any two of the points
$\overrightarrow{AC} = \begin{pmatrix}2\\-14\\-16\end{pmatrix}$ | A1 | AO 1.1 | Correct pair of vectors with common point; $\overrightarrow{BC} = \begin{pmatrix}3\\-21\\-24\end{pmatrix}$
AB is parallel to AC | B1 | AO 1.1 |
Common point A so collinear | E1 | AO 2.1 |
**[4 marks]**
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3 Show that points $\mathrm { A } ( 1,4,9 ) , \mathrm { B } ( 0,11,17 )$ and $\mathrm { C } ( 3 , - 10 , - 7 )$ are collinear.
\hfill \mbox{\textit{OCR MEI Paper 3 Q3 [4]}}