A2 June 2023 Paper 2 Q11
11 The line \(l_1\) passes through the points \(A(6, 2, 7)\) and \(B(4, -3, 7)\)
(a) Find a Cartesian equation of \(l_1\) [2 marks]
(b) The line \(l_2\) has vector equation \(\mathbf{r} = \begin{bmatrix} 8 \\ 9 \\ c \end{bmatrix} + \mu\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}\) where \(c\) is a constant.
(i) Explain how you know that the lines \(l_1\) and \(l_2\) are not perpendicular. [2 marks]
(ii) The lines \(l_1\) and \(l_2\) both lie in the same plane.
Find the value of \(c\) [5 marks]
| Scheme | Marks | AO |
|---|---|---|
| Obtains a direction vector of \(l_1\) PI | M1 | 1.1a |
| Obtains a correct Cartesian equation of \(l_1\) | A1 | 1.1b |
| (2) |
Typical solution
\[\mathbf{r} = \begin{bmatrix} 6 \\ 2 \\ 7 \end{bmatrix} + \lambda\begin{bmatrix} 2 \\ 5 \\ 0 \end{bmatrix}\]\[x = 6 + 2\lambda,\ y = 2 + 5\lambda,\ z = 7\]\[\frac{x - 6}{2} = \frac{y - 2}{5},\ z = 7\]| Scheme | Marks | AO |
|---|---|---|
| (i) Obtains correct scalar product of their direction vector of \(l_1\) and the direction vector of \(l_2\) | B1 | 1.1b |
| (i) Explains that the lines are not perpendicular because this scalar product is non-zero. | E1 | 2.4 |
| (2) | ||
| (ii) Obtains a vector perpendicular to both lines Or Selects a method to obtain the point of intersection of the two lines. | M1 | 3.1a |
| (ii) Uses scalar product of their normal vector and the position vector of a point on \(l_1\) or \(l_2\) to obtain constant term in equation of plane. PI Or Forms two simultaneous equations in \(\lambda\) and \(\mu\) only. | M1 | 1.1a |
| (ii) Obtains correct equation of plane. Or Obtains correct simultaneous equations. | A1 | 1.1b |
| (ii) Forms and solves equation in \(c\) using their equation of the plane or the solutions to their simultaneous equations. | M1 | 1.1a |
| (ii) Obtains correct value of \(c\) | A1 | 1.1b |
| (5) | ||
| (9 marks) |
Typical solution
(i)
Scalar product of direction vectors
\[= 2 \times 1 + 5 \times 1 + 0 = 7\]The scalar product is non-zero, so the lines are not perpendicular.
(ii)
Normal to plane
\[\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 2 \\ 2 & 5 & 0 \end{vmatrix} = \begin{bmatrix} -10 \\ 4 \\ 3 \end{bmatrix}\]Equation of plane is
\[\mathbf{r} \bullet \begin{bmatrix} -10 \\ 4 \\ 3 \end{bmatrix} = d\]\[d = \begin{bmatrix} -10 \\ 4 \\ 3 \end{bmatrix} \bullet \begin{bmatrix} 6 \\ 2 \\ 7 \end{bmatrix} = -31\]\[\begin{bmatrix} 8 \\ 9 \\ c \end{bmatrix} \bullet \begin{bmatrix} -10 \\ 4 \\ 3 \end{bmatrix} = -31\]\[-80 + 36 + 3c = -31\]\[c = \frac{13}{3}\]