Easy -1.8 This is a straightforward recall question requiring only the computation of a dot product and verification it equals zero. It's a single-step calculation with no problem-solving element, making it significantly easier than average even for Further Maths students.
5 Show that the vectors \(\left[ \begin{array} { c } 1 \\ - 3 \\ 5 \end{array} \right]\) and \(\left[ \begin{array} { l } 7 \\ 4 \\ 1 \end{array} \right]\) are perpendicular.
Calculates scalar product of two vectors. Accept one miscopy of an element, or one incorrect product if elements are not shown.
\(= 7-12+5=0\), \(\therefore\) the two vectors are perpendicular
R1 (AO2.1)
Must include at least one calculation line and \(=0\) linked with perpendicular conclusion. Ignore incorrect vector magnitudes.
## Question 5:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{bmatrix}1\\-3\\5\end{bmatrix} \cdot \begin{bmatrix}7\\4\\1\end{bmatrix} = 1\times7+(-3)\times4+5\times1$ | M1 (AO1.1a) | Calculates scalar product of two vectors. Accept one miscopy of an element, or one incorrect product if elements are not shown. |
| $= 7-12+5=0$, $\therefore$ the two vectors are perpendicular | R1 (AO2.1) | Must include at least one calculation line and $=0$ linked with perpendicular conclusion. Ignore incorrect vector magnitudes. |
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5 Show that the vectors $\left[ \begin{array} { c } 1 \\ - 3 \\ 5 \end{array} \right]$ and $\left[ \begin{array} { l } 7 \\ 4 \\ 1 \end{array} \right]$ are perpendicular.\\
\hfill \mbox{\textit{AQA Further AS Paper 1 2021 Q5 [2]}}