A2 June 2024 Paper 1 Q12
12 The line \(L_1\) has equation
\[\mathbf{r} = \begin{bmatrix}4 \\ 2 \\ 1\end{bmatrix} + \lambda\begin{bmatrix}1 \\ 3 \\ -1\end{bmatrix}\]The transformation T is represented by the matrix
\[\begin{bmatrix} 2 & 1 & 0 \\ 3 & 4 & 6 \\ -5 & 2 & -3 \end{bmatrix}\]The transformation T transforms the line \(L_1\) to the line \(L_2\)
(a) Show that the angle between \(L_1\) and \(L_2\) is 0.701 radians, correct to three decimal places. [4 marks]
(b) Find the shortest distance between \(L_1\) and \(L_2\)
Give your answer in an exact form. [6 marks]
| Scheme | Marks | AO |
|---|---|---|
| Forms the product of the matrix and the direction/position vector of \(L_1\) | M1 | 2.2a |
| Obtains the correct direction vector of \(L_2\) possibly embedded | A1 | 1.1b |
| Use the scalar product of the direction vectors of \(L_1\) and their \(L_2\) to obtain their correct 28. | M1 | 1.1a |
| Completes a reasoned argument to obtain 0.701 Must see \(\cos\theta = \dfrac{28}{\sqrt{11}\sqrt{122}}\) or \(\cos\theta = 0.7643\ldots\) or \(\theta = 0.7007\ldots\) | R1 | 2.1 |
| (4) |
Typical solution
\[\begin{bmatrix} 2 & 1 & 0 \\ 3 & 4 & 6 \\ -5 & 2 & -3 \end{bmatrix}\begin{bmatrix}1 \\ 3 \\ -1\end{bmatrix} = \begin{bmatrix}5 \\ 9 \\ 4\end{bmatrix} = \text{direction vector of } L_2\]\[\begin{bmatrix}1 \\ 3 \\ -1\end{bmatrix} \cdot \begin{bmatrix}5 \\ 9 \\ 4\end{bmatrix} = 28\]\[\cos\theta = \frac{28}{\sqrt{11}\sqrt{122}}\]\[\theta = 0.701 \text{ (3dp)}\]| Scheme | Marks | AO |
|---|---|---|
| Forms the product of the matrix and a point on \(L_1\) | M1 | 2.2a |
| Obtains a correct point on \(\begin{bmatrix}10 \\ 26 \\ -19\end{bmatrix} + \mu\begin{bmatrix}5 \\ 9 \\ 4\end{bmatrix}\) | A1 | 1.1b |
| Obtains a vector connecting a point on each line. or Obtains the equations of two correct parallel planes PI | B1 | 1.1b |
| Obtains a vector perpendicular to both lines. or Calculates the scalar product of the general vector between the lines with both direction vectors to obtain a pair of simultaneous equations. | M1 | 3.1a |
| Uses a correct method to obtain the shortest distance between the lines. Condone a negative distance or Obtains solutions to their pair of simultaneous equations and obtains the vector between the two closest points | M1 | 3.1a |
| Obtains \(\dfrac{5\sqrt{62}}{31}\) OE | A1 | 1.1b |
| (6) | ||
| (10 marks) |
Typical solution
\[\begin{bmatrix} 2 & 1 & 0 \\ 3 & 4 & 6 \\ -5 & 2 & -3 \end{bmatrix}\begin{bmatrix}4 \\ 2 \\ 1\end{bmatrix} = \begin{bmatrix}10 \\ 26 \\ -19\end{bmatrix}\]Vector connecting \(L_1\) to \(L_2\)
\[\overrightarrow{AB} = \begin{bmatrix}10 \\ 26 \\ -19\end{bmatrix} - \begin{bmatrix}4 \\ 2 \\ 1\end{bmatrix} = \begin{bmatrix}6 \\ 24 \\ -20\end{bmatrix}\]Vector perpendicular to both lines
\[\mathbf{n} = \begin{bmatrix}5 \\ 9 \\ 4\end{bmatrix} \times \begin{bmatrix}1 \\ 3 \\ -1\end{bmatrix} = \begin{bmatrix}-21 \\ 9 \\ 6\end{bmatrix}\]\[\mathbf{n} \cdot \overrightarrow{AB} = -30\]\[\text{Distance} = \left|\frac{\mathbf{n} \cdot \overrightarrow{AB}}{|\mathbf{n}|}\right| = \frac{10}{\sqrt{62}} = \frac{5\sqrt{62}}{31}\]Alternative typical solution 1
\[\begin{bmatrix} 2 & 1 & 0 \\ 3 & 4 & 6 \\ -5 & 2 & -3 \end{bmatrix}\begin{bmatrix}4 \\ 2 \\ 1\end{bmatrix} = \begin{bmatrix}10 \\ 26 \\ -19\end{bmatrix}\]Equation of \(L_2\): \(\mathbf{r} = \begin{bmatrix}10 \\ 26 \\ -19\end{bmatrix} + \mu\begin{bmatrix}5 \\ 9 \\ 4\end{bmatrix}\)
General vector
\[\overrightarrow{AB} = \begin{bmatrix}10 + 5\mu \\ 26 + 9\mu \\ -19 + 4\mu\end{bmatrix} - \begin{bmatrix}4 + \lambda \\ 2 + 3\lambda \\ 1 - \lambda\end{bmatrix} = \begin{bmatrix}6 + 5\mu - \lambda \\ 24 + 9\mu - 3\lambda \\ -20 + 4\mu + \lambda\end{bmatrix}\]Scalar product
\[0 = \begin{bmatrix}1 \\ 3 \\ -1\end{bmatrix} \cdot \begin{bmatrix}6 + 5\mu - \lambda \\ 24 + 9\mu - 3\lambda \\ -20 + 4\mu + \lambda\end{bmatrix} = 98 + 28\mu - 11\lambda\]\[11\lambda - 28\mu = 98\]\[0 = \begin{bmatrix}5 \\ 9 \\ 4\end{bmatrix} \cdot \begin{bmatrix}6 + 5\mu - \lambda \\ 24 + 9\mu - 3\lambda \\ -20 + 4\mu + \lambda\end{bmatrix} = 166 + 122\mu - 28\lambda\]\[28\lambda - 122\mu = 166\]\[\lambda = \frac{406}{31},\quad \mu = \frac{51}{31}\]\[\overrightarrow{AB} = \begin{bmatrix}35/31 \\ -15/31 \\ -10/31\end{bmatrix}\]\[\text{Distance} = \frac{5\sqrt{62}}{31}\]Alternative typical solution 2
\[\begin{bmatrix} 2 & 1 & 0 \\ 3 & 4 & 6 \\ -5 & 2 & -3 \end{bmatrix}\begin{bmatrix}4 \\ 2 \\ 1\end{bmatrix} = \begin{bmatrix}10 \\ 26 \\ -19\end{bmatrix}\]\[\begin{bmatrix}5 \\ 9 \\ 4\end{bmatrix} \times \begin{bmatrix}1 \\ 3 \\ -1\end{bmatrix} = \begin{bmatrix}-21 \\ 9 \\ 6\end{bmatrix}\]Parallel planes
\[\Pi_1:\ 21x - 9y - 6z = 21 \times 4 - 9 \times 2 - 6 \times 1 = 60\]\[\Pi_2:\ 21x - 9y - 6z = 21 \times 10 - 9 \times 26 - 6 \times -19 = 90\]\[\text{distance} = \frac{90 - 60}{\sqrt{21^2 + 9^2 + 6^2}} = \frac{5\sqrt{62}}{31}\]