A-Level Mathematics May/June 2024 Q8(b): The line l₂ has equation r = −2i+j+4k+µ(3i+j – 2k). Find the coordinates of the point o…

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  Let the line from part (a) be $l_1$. Assuming the equation for $l_1$ is $\mathbf{r} = \begin{pmatrix} 5 \\ 2 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$. The equation for $l_2$ is given as $\mathbf{r} = \begin{pmatrix} -2 \\ 1 \\ 4 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}$.

  At the point of intersection, the position vectors are equal: $\begin{pmatrix} 5 \\ 2 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} -2 \\ 1 \\ 4 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}$

  Equating the components gives a system of three simultaneous equations:

  1. i component: $5 + \lambda = -2 + 3\mu \implies \lambda - 3\mu = -7$
  2. j component: $2 + \lambda = 1 + \mu \implies \lambda - \mu = -1$
  3. k component: $-\lambda = 4 - 2\mu \implies 2\mu - \lambda = 4$

  We can solve any two of these equations. Let's use (1) and (2). From (2), we can express $\lambda$ in terms of $\mu$: $\lambda = \mu - 1$.

  Substitute this into (1): $(\mu - 1) - 3\mu = -7$ $-2\mu - 1 = -7$ $-2\mu = -6$ $\mu = 3$

  Now find $\lambda$ using $\lambda = \mu - 1$: $\lambda = 3 - 1 = 2$

  To verify, we can check these values in equation (3): $2(3) - 2 = 6 - 2 = 4$. This is consistent.

  To find the coordinates of the point of intersection, substitute $\lambda = 2$ into the equation for $l_1$ (or $\mu = 3$ into $l_2$). Using $l_1$: $\mathbf{r} = \begin{pmatrix} 5 \\ 2 \\ 0 \end{pmatrix} + 2 \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 5+2 \\ 2+2 \\ 0-2 \end{pmatrix} = \begin{pmatrix} 7 \\ 4 \\ -2 \end{pmatrix}$

  The coordinates of the point of intersection are $(7, 4, -2)$.

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  How the marks are awarded

  • B1FT — Awarded for equating the components of the two line equations to form at least two simultaneous equations, such as $5 + \lambda = -2 + 3\mu$ and $2 + \lambda = 1 + \mu$. The use of different parameters, $\lambda$ and $\mu$, is essential. This is a Follow Through (FT) mark from part (a).
  • M1 — Awarded for the correct method to solve the two simultaneous equations for either $\lambda$ or $\mu$. In the model answer, this is shown by rearranging one equation to make $\lambda$ the subject and substituting it into the other.
  • A1 — Awarded for correctly finding either $\lambda = 2$ or $\mu = 3$. This demonstrates accurate algebraic manipulation.
  • A1 — Awarded for substituting the value of $\lambda$ or $\mu$ back into the corresponding line equation to obtain the correct coordinates of the intersection point, $(7, 4, -2)$. The answer can be given as a position vector or as coordinates.

  Common mistakes

  • Using the same parameter (e.g., $\lambda$) for both line equations, which leads to an invalid system of equations and no solution.
  • Making an algebraic error when solving the simultaneous equations, for example, a sign error when rearranging terms, leading to incorrect values for $\lambda$ and $\mu$.
  • Successfully finding the correct values for $\lambda$ and $\mu$ but then stopping and not substituting them back into a line equation to find the coordinates of the point.
  • Making a calculation error when substituting the parameter value back into the line equation to find the final coordinates, e.g., $5 + 2(1) = 7$ but $0 + 2(-1) = 2$.

  Examiner tip: When finding the intersection of two lines in 3D, always set up and solve a system of simultaneous equations from the vector components, and remember to use different parameters for each line.

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