A-Level Mathematics May/June 2024 Q10(b): Given instead that the lines intersect, find the value of a and the position vector of…

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  Let the two lines be $l_1$ and $l_2$.

  $l_1: \mathbf{r} = \begin{pmatrix} 1 \ 1 \ 2a \end{pmatrix} + \lambda \begin{pmatrix} 3 \ 4 \ a \end{pmatrix}$ $l_2: \mathbf{r} = \begin{pmatrix} -3 \ -1 \ 4 \end{pmatrix} + \mu \begin{pmatrix} -1 \ 2 \ 2 \end{pmatrix}$

  For the lines to intersect, there must be values of $\lambda$ and $\mu$ for which the position vectors are equal.

  The general point on $l_1$ is $\begin{pmatrix} 1+3\lambda \ 1+4\lambda \ 2a+a\lambda \end{pmatrix}$. The general point on $l_2$ is $\begin{pmatrix} -3-\mu \ -1+2\mu \ 4+2\mu \end{pmatrix}$.

  Equating the components: (1) $\mathbf{i}: 1+3\lambda = -3-\mu$ (2) $\mathbf{j}: 1+4\lambda = -1+2\mu$ (3) $\mathbf{k}: 2a+a\lambda = 4+2\mu$

  First, solve equations (1) and (2) simultaneously to find $\lambda$ and $\mu$. From (1): $3\lambda + \mu = -4$ From (2): $4\lambda - 2\mu = -2 \implies 2\lambda - \mu = -1$

  Adding the rearranged equations: $(3\lambda + \mu) + (2\lambda - \mu) = -4 + (-1)$ $5\lambda = -5$ $\lambda = -1$

  Substitute $\lambda = -1$ into $3\lambda + \mu = -4$: $3(-1) + \mu = -4$ $-3 + \mu = -4$ $\mu = -1$

  Now, substitute $\lambda = -1$ and $\mu = -1$ into equation (3) to find $a$. $2a + a(-1) = 4 + 2(-1)$ $2a - a = 4 - 2$ $a = 2$

  Finally, find the position vector of the point of intersection by substituting $\lambda = -1$ into the equation for $l_1$ (or $\mu = -1$ into $l_2$). Using $l_1$ with $\lambda = -1$ and $a=2$: Position vector $= \begin{pmatrix} 1 \ 1 \ 2(2) \end{pmatrix} + (-1) \begin{pmatrix} 3 \ 4 \ 2 \end{pmatrix} = \begin{pmatrix} 1 \ 1 \ 4 \end{pmatrix} - \begin{pmatrix} 3 \ 4 \ 2 \end{pmatrix} = \begin{pmatrix} -2 \ -3 \ 2 \end{pmatrix}$

  The position vector of the point of intersection is $-2\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}$.

  Final Answer: $a=2$, position vector is $-2\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}$.

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  How the marks are awarded

  • B1 — Writing the general point of at least one line in component form, such as (1+3λ, 1+4λ, 2a+aλ) or (-3-μ, -1+2μ, 4+2μ).
  • M1 — Equating the i and j components and attempting to solve them simultaneously for λ and μ. This is shown by the rearrangement and subsequent addition of the first two equations.
  • A1 — Correctly obtaining either λ = -1 or μ = -1 from the simultaneous equations.
  • A1 — Substituting the found values of λ and μ into the third (k-component) equation to correctly find that a = 2.
  • A1 — Substituting the value of λ or μ back into the corresponding line equation to find the correct position vector of intersection, -2i - 3j + 2k.

  Common mistakes

  • Making a sign error when rearranging the component equations, leading to incorrect values for λ and μ.
  • Trying to solve all three equations with three unknowns (λ, μ, a) at once, which is more complex and error-prone than the two-stage method.
  • Stopping after finding the value of 'a' and forgetting to calculate the position vector of the intersection point.
  • Finding the correct coordinates of the intersection point, e.g., (-2, -3, 2), but then writing the final vector in an incorrect format like (-2i, -3j, 2k).

  Examiner tip: When solving for intersecting lines with an unknown constant, first solve the simultaneous equations from the components that do not contain the constant.

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