Easy -1.2 This is a straightforward application of the distance formula in 3D vectors requiring only vector subtraction and magnitude calculation. It's a routine 'show that' question with no problem-solving element—students simply compute |b-a| = √[(−4)² + 2² + 9²] = √101, making it easier than average.
3 The points \(A\) and \(B\) have position vectors \(\mathbf { a } = \left( \begin{array} { r } 3 \\ 2 \\ - 1 \end{array} \right)\) and \(\mathbf { b } = \left( \begin{array} { r } - 1 \\ 4 \\ 8 \end{array} \right)\) respectively.
Show that the exact value of the distance \(A B\) is \(\sqrt { \mathbf { 1 0 1 } }\).
M1: Attempt to find magnitude of their displacement vector. A1: cao AG. Do not allow final mark when there are missing brackets e.g. \(-4^2 + 2^2 + 9^2\)
## Question 3:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix}-4\\2\\9\end{pmatrix}$ | B1 | AG. Soi. Allow for $\overrightarrow{BA}$ even if wrongly labelled |
| distance $= \sqrt{(-4)^2 + 2^2 + 9^2} = \sqrt{101}$ | M1, A1 [3] | M1: Attempt to find magnitude of their displacement vector. A1: cao AG. Do not allow final mark when there are missing brackets e.g. $-4^2 + 2^2 + 9^2$ |
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3 The points $A$ and $B$ have position vectors $\mathbf { a } = \left( \begin{array} { r } 3 \\ 2 \\ - 1 \end{array} \right)$ and $\mathbf { b } = \left( \begin{array} { r } - 1 \\ 4 \\ 8 \end{array} \right)$ respectively.\\
Show that the exact value of the distance $A B$ is $\sqrt { \mathbf { 1 0 1 } }$.
\hfill \mbox{\textit{OCR MEI Paper 1 2020 Q3 [3]}}