A2 June 2025 Paper 1 Q15
15 Three planes have equations
\[\begin{alignedat}{4} x &\; + \;& 2y &\; - \;& z &\; = \;& 9& \\ x &\; - \;& 3y &\; + \;& 3z &\; = \;& t& \\ 3x &\; + \;& y &\; + \;& z &\; = \;& 4t& \end{alignedat}\]where \(t\) is a constant.
The planes meet along a line of intersection.
(a) Find the value of \(t\) [4 marks]
(b) Find a vector equation of the line of intersection.
Fully justify your answer. [5 marks]
| Scheme | Marks | AO |
|---|---|---|
| Selects an appropriate method Eg eliminates one variable to form at least one equation | M1 | 3.1a |
| Obtains two correct equations in the same two variables and \(t\) | A1 | 1.1b |
| Forms and solves an equation in \(t\) | M1 | 1.1a |
| Obtains 6 | A1 | 1.1b |
| (4) |
Typical solution
\[\begin{alignedat}{4} x &\; + \;& 2y &\; - \;& z &\; = 9 && \quad (1) \\ x &\; - \;& 3y &\; + \;& 3z &\; = t && \quad (2) \\ 3x &\; + \;& y &\; + \;& z &\; = 4t && \quad (3) \end{alignedat}\]\[\begin{aligned} -5y + 4z &= t - 9 && \quad (2) - (1) \\ -5y + 4z &= 4t - 27 && \quad (3) - 3(1) \end{aligned}\]\[t - 9 = 4t - 27\]\[t = 6\]| Scheme | Marks | AO |
|---|---|---|
| Selects an appropriate method eg equates one variable to zero and forms simultaneous equations in two variables. or Sets one variable equal to a multiple of a parametric variable | M1 | 3.1a |
| Obtains one correct point on the line. or Correctly obtains a second variable in terms of the parametric variable | A1 | 1.1b |
| Uses a correct method to obtain a direction vector eg Uses two points on their line to form a direction vector or Obtains the cross product of two normal vectors. or substitutes to find the third variable in terms of the parametric variable | M1 | 1.1a |
| Obtains a correct direction vector. or Correctly obtains the third variable in terms of the parametric variable | A1 | 1.1b |
| Completes a reasoned argument to obtain a correct vector equation. | R1 | 2.2a |
| (5) | ||
| (9 marks) |
Typical solution
\[\begin{alignedat}{3} x &\; + \;& 2y &\; - \;& z &= 9 \\ x &\; - \;& 3y &\; + \;& 3z &= 6 \end{alignedat}\]Let \(x = 0\):
\[\begin{aligned} 2y - z &= 9 \\ -3y + 3z &= 6 \\ -y + z &= 2 \end{aligned}\]\[y = 11,\ z = 13\]Let \(y = 0\):
\[\begin{aligned} x - z &= 9 \\ x + 3z &= 6 \end{aligned}\]\[z = -\frac{3}{4},\ x = \frac{33}{4}\]Two points on the line:
\[A(0, 11, 13),\ B\left(\frac{33}{4}, 0, -\frac{3}{4}\right)\]\[\overrightarrow{AB} = \begin{bmatrix} \frac{33}{4} \\[4pt] -11 \\[4pt] -\frac{55}{4} \end{bmatrix} \quad \text{Direction vector} = \begin{bmatrix} 3 \\ -4 \\ -5 \end{bmatrix}\]\[\mathbf{r} = \begin{bmatrix} 0 \\ 11 \\ 13 \end{bmatrix} + \lambda\begin{bmatrix} 3 \\ -4 \\ -5 \end{bmatrix}\]