Standard +0.3 This is a standard Further Maths vectors question requiring knowledge that the angle between a line and plane relates to the angle between the line's direction vector and the plane's normal. The calculation involves one dot product, magnitudes, and an inverse trig function—straightforward application of a known formula with minimal problem-solving required.
A plane has equation \(\mathbf{r} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 7\)
A line has equation \(\mathbf{r} = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\)
Calculate the acute angle between the line and the plane.
Give your answer to the nearest \(0.1°\)
[3 marks]
Question 5:
5 | Finds scalar (or vector)
product of the correct
vectors
35o
PI by seeing AWRT | AO1.1a | M1 | 1 1
0 �1 =2
1 1
Moduli of vectors are 2 and 3
Let α be angle between normal & line
2
cosα=
6
Angle between plane & line
=90−α=54.7o
Divides their scalar product
(or magnitude of vector
product) of their vectors by
product of their magnitudes
35o
PI by seeing AWRT | AO1.1a | M1
Deduces the correct angle,
correct to at least 1dp | AO2.2a | A1
Total | 3
Q | Marking Instructions | AO | Marks | Typical Solution
A plane has equation $\mathbf{r} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 7$
A line has equation $\mathbf{r} = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$
Calculate the acute angle between the line and the plane.
Give your answer to the nearest $0.1°$
[3 marks]
\hfill \mbox{\textit{AQA Further Paper 1 2019 Q5 [3]}}