A2 June 2020 Paper 2 Q15
15 The points \(A(7, 2, 8)\), \(B(7, -4, 0)\) and \(C(3, 3.2, 9.6)\) all lie in the plane \(\Pi\).
Find an equation of the line \(L_2\) [4 marks]
The point \(D\) is the centre of circle \(G\).
Find the coordinates of \(D\). [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Obtains two vectors in the plane \(\Pi\) | B1 | 1.1a |
| Selects a method to find a vector normal to the plane \(\Pi\) either by taking the vector product of their two vectors in \(\Pi\) or by taking the scalar product of a general vector with their two vectors in \(\Pi\) | M1 | 3.1a |
| Obtains a correct Cartesian equation of the plane \(\Pi\) | A1 | 1.1b |
Typical solution
\[\overrightarrow{AB} = \begin{bmatrix} 0 \\ -6 \\ -8 \end{bmatrix} \qquad \overrightarrow{BC} = \begin{bmatrix} -4 \\ 7.2 \\ 9.6 \end{bmatrix}\]\[\text{Normal vector } \mathbf{n} = \begin{vmatrix} \mathbf{i} & 0 & -5 \\ \mathbf{j} & 3 & 9 \\ \mathbf{k} & 4 & 12 \end{vmatrix} = \begin{bmatrix} 0 \\ -20 \\ 15 \end{bmatrix} \text{ or } \begin{bmatrix} 0 \\ -4 \\ 3 \end{bmatrix}\]Equation of plane \(\Pi\):
\[-4y + 3z = 16\]| Scheme | Marks | AO |
|---|---|---|
| Selects a method to show that \(L_1\) lies in \(\Pi\) either by substituting a general point on \(L_1\) into their equation of \(\Pi\), or by substituting a point on \(L_1\) in their equation of \(\Pi\) and using a scalar product or a vector product to show that the direction vector of \(L_1\) is perpendicular to their normal to \(\Pi\) | M1 | 3.1a |
| Completes a rigorous argument to show that \(L_1\) lies in \(\Pi\) | R1 | 2.1 |
Typical solution
For any point on \(L_1\),
\[y = -0.4 + 3\mu\]and
\[z = 4.8 + 4\mu\]\[\therefore -4y + 3z = -4(-0.4 + 3\mu) + 3(4.8 + 4\mu)\]\[= 1.6 - 12\mu + 14.4 + 12\mu = 16\]So the point lies in the plane \(\Pi\) and therefore \(L_1\) lies in the plane \(\Pi\).
| Scheme | Marks | AO |
|---|---|---|
| Selects a method to show that every point on \(L_1\) is equidistant from \(B\) and \(C\) by finding the general vector from a point on \(L_1\) to \(B\) or \(C\) or by finding the midpoint \((5, -0.4, 4.8)\) of \(BC\) | M1 | 3.1a |
| Finds the vector \(\overrightarrow{BC}\) or forms an expression for the length of a general vector from a point on \(L_1\) to \(B\) or \(C\) | M1 | 1.1a |
| Takes the scalar product of \(\overrightarrow{BC}\) and the direction vector of \(L_1\) or forms an equation for the distances from a point on \(L_1\) to \(B\) and \(C\) being equal. | M1 | 1.1a |
| Completes a reasoned argument to show that every point on \(L_1\) is equidistant from \(B\) and \(C\) | R1 | 2.1 |
Typical solution
Midpoint of \(BC\) is \((5, -0.4, 4.8)\), which lies on \(L_1\)
Consider the direction vectors of \(L_1\) and \(BC\):
\[\begin{bmatrix} 15 \\ 3 \\ 4 \end{bmatrix} \cdot \begin{bmatrix} -5 \\ 9 \\ 12 \end{bmatrix} = -75 + 27 + 48 = 0\]\(\therefore\) \(L_1\) is perpendicular to \(BC\)
Since the midpoint of \(BC\) also lies on \(L_1\) then \(L_1\) is the perpendicular bisector of \(BC\) and hence every point on \(L_1\) is equidistant from \(B\) and \(C\)
| Scheme | Marks | AO |
|---|---|---|
| Deduces that the direction vector of \(L_2\) is perpendicular to \(AB\) or to their normal to \(\Pi\) | M1 | 2.2a |
| Deduces that \(L_2\) passes through the midpoint of \(AB\) | M1 | 2.2a |
| Selects a method to find the direction vector of \(L_2\) by taking the vector product of \(\overrightarrow{AB}\) and their normal to \(\Pi\) | M1 | 3.1a |
| Completes a reasoned mathematical argument to obtain a correct equation of \(L_2\) | R1 | 2.1 |
Typical solution
\(L_2\) is the perpendicular bisector of \(AB\) in the plane \(\Pi\).
Midpoint of \(AB\) is \(M_{AB}\ (7, -1, 4)\)
Direction vector \(\mathbf{s}\) for \(L_2\) is perpendicular to both \(AB\) and \(\mathbf{n}\).
\[\begin{vmatrix} \mathbf{i} & 0 & 0 \\ \mathbf{j} & 3 & -4 \\ \mathbf{k} & 4 & 3 \end{vmatrix} = \begin{bmatrix} 25 \\ 0 \\ 0 \end{bmatrix} \quad \text{so let } \mathbf{s} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}\]\[L_2 : \ \mathbf{r} = \begin{bmatrix} 7 \\ -1 \\ 4 \end{bmatrix} + \lambda\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}\]| Scheme | Marks | AO |
|---|---|---|
| Deduces that \(D\) is at the intersection of \(L_1\) and \(L_2\) | B1 | 2.2a |
| Equates equations of \(L_1\) and their \(L_2\) | M1 | 1.1a |
| Obtains the correct coordinates of \(D\) FT their equation of \(L_2\) Condone position vector of \(D\) | A1 | 1.1b |
| (16 marks) |
Typical solution
\(D\) is the point of intersection of \(L_1\) and \(L_2\).
\[\begin{aligned} 5 + 15\mu &= \lambda + 7 \\ -0.4 + 3\mu &= -1 \\ 4.8 + 4\mu &= 4 \end{aligned}\]\[\mu = -0.2\]\[\mathbf{r} = \begin{bmatrix} 2 \\ -1 \\ 4 \end{bmatrix}\]\[D(2, -1, 4)\]Notes
(corrected from the printed mark scheme: the second equation is printed as \(0.4 + 3\mu = -1\); the \(y\)-component of \(L_1\) is \(-0.4 + 3\mu\))