A2 June 2023 Paper 1 Q2

Notes

B1: soi

M1: use of scalar product formula with their normal seen

A1: or 1.2… rads or \(\arccos\frac{5}{14}\)

M1: or \(\frac{\pi}{2} - \text{their } \theta\)

A1: or 0.365 rads

Alternative method 1

Scheme	Marks
Normal to plane is \(-\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}\)	B1
Angle between \(3\mathbf{i} + 2\mathbf{j} + \mathbf{k}\) and the plane is \(\theta\) \(\sin\theta = \dfrac{3 \times (-1) + 2 \times 3 + 1 \times 2}{\sqrt{3^2 + 2^2 + 1^2}\sqrt{(-1)^2 + 3^2 + 2^2}} = \dfrac{5}{\sqrt{14}\sqrt{14}}\)	M2
\(\theta = 20.9^\circ\)	A2

B1: soi

A2: or 0.365 rads

Alternative method 2

Scheme	Marks
Normal to plane is \(-\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}\)	B1
Angle between \(3\mathbf{i} + 2\mathbf{j} + \mathbf{k}\) and \(-\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}\) is \(\theta\) \(\sin\theta = \dfrac{\left|\begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} \times \begin{pmatrix} -1 \\ 3 \\ 2 \end{pmatrix}\right|}{\sqrt{3^2 + 2^2 + 1^2}\sqrt{(-1)^2 + 3^2 + 2^2}} = \dfrac{\left|\begin{pmatrix} 1 \\ -7 \\ 11 \end{pmatrix}\right|}{\sqrt{14}\sqrt{14}}\)	M1
\(\Rightarrow \theta = 69.07\ldots\)	A1
\(90 - \text{their } \theta\)	M1
so angle with plane is \(20.9^\circ\)	A1

B1: soi

M1: complete method with vector product seen

A1: or 1.2… rads or \(\arcsin\frac{\sqrt{171}}{14}\)

M1: or \(\frac{\pi}{2} - \text{their } \theta\)

A1: or 0.365 rads