A-Level Mathematics May/June 2024 Q9(b): The lines l₁ and l₂ also intersect. (b) Find the position vector of the point of inters…

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  Let the lines $l_1$ and $l_2$ be given by the vector equations: $l_1: \mathbf{r} = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -1 \\ a \end{pmatrix}$ $l_2: \mathbf{r} = \begin{pmatrix} -1 \\ -1 \\ -1 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ -2 \\ -2 \end{pmatrix}$

  At the point of intersection, the position vectors are equal for some values of $\lambda$ and $\mu$. A general point on $l_1$ is $\begin{pmatrix} 1 + 2\lambda \\ -2 - \lambda \\ 3 + a\lambda \end{pmatrix}$. A general point on $l_2$ is $\begin{pmatrix} -1 + 3\mu \\ -1 - 2\mu \\ -1 - 2\mu \end{pmatrix}$.

  Equating the components: (i): $1 + 2\lambda = -1 + 3\mu \implies 2\lambda - 3\mu = -2 \quad \cdots(1)$ (j): $-2 - \lambda = -1 - 2\mu \implies -\lambda + 2\mu = 1 \quad \cdots(2)$ (k): $3 + a\lambda = -1 - 2\mu$

  We solve equations (1) and (2) simultaneously. From (2), we can make $\lambda$ the subject: $\lambda = 2\mu - 1$

  Substitute this into (1): $2(2\mu - 1) - 3\mu = -2$ $4\mu - 2 - 3\mu = -2$ $\mu = 0$

  Substitute $\mu = 0$ back to find $\lambda$: $\lambda = 2(0) - 1 = -1$

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

  The position vector of the point of intersection is $-\mathbf{i} - \mathbf{j} - \mathbf{k}$.

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  How the marks are awarded

  • B1 — Correctly expressing the general point of at least one line in component form, such as $(-1 + 3\mu, -1 - 2\mu, -1 - 2\mu)$.
  • M1 — Equating at least two pairs of corresponding components (e.g., the i and j components) to form a system of simultaneous equations in $\lambda$ and $\mu$ and attempting to solve them.
  • A1 — Correctly solving the simultaneous equations to obtain either $\lambda = -1$ or $\mu = 0$.
  • A1 — Substituting the value of $\lambda$ or $\mu$ back into the corresponding line equation to obtain the correct position vector of intersection, $-\mathbf{i} - \mathbf{j} - \mathbf{k}$.

  Common mistakes

  • Making an arithmetic error when solving the simultaneous equations, for example, $2(2\mu - 1)$ becomes $4\mu - 1$, leading to incorrect values for the parameters.
  • Finding the correct values for $\lambda$ and $\mu$ but then stopping, forgetting to substitute one of them back into a line equation to find the position vector.
  • Substituting the value of $\lambda$ into the equation for $l_2$ (or $\mu$ into $l_1$). The parameter for one line must be substituted back into the equation for that same line.
  • Writing the final answer as the values of the parameters (e.g., '$\lambda = -1, \mu = 0$') instead of the position vector that was asked for.

  Examiner tip: To find the intersection of two lines, always set their general parametric points equal to each other and solve the resulting system of simultaneous equations.

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