A-Level Mathematics October/November 2024 Q9(c): The line m has vector equation r = 5i−j+2k+μ(ai-j+3k). The acute angle between the dire…

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  The direction vector of line $l$ is $\mathbf{d}_l = 2\mathbf{i} + \mathbf{j} + 4\mathbf{k}$. The direction vector of line $m$ is $\mathbf{d}_m = a\mathbf{i} - \mathbf{j} + 3\mathbf{k}$.

  The angle $\theta$ between the lines is given by the formula $\cos\theta = \frac{|\mathbf{d}_l \cdot \mathbf{d}_m|}{|\mathbf{d}_l| |\mathbf{d}_m|}$.

  First, calculate the scalar product: $\mathbf{d}_l \cdot \mathbf{d}_m = (2)(a) + (1)(-1) + (4)(3) = 2a - 1 + 12 = 2a + 11$.

  Next, calculate the magnitudes of the direction vectors: $|\mathbf{d}_l| = \sqrt{2^2 + 1^2 + 4^2} = \sqrt{4 + 1 + 16} = \sqrt{21}$. $|\mathbf{d}_m| = \sqrt{a^2 + (-1)^2 + 3^2} = \sqrt{a^2 + 1 + 9} = \sqrt{a^2 + 10}$.

  Substitute these into the angle formula: $\cos\theta = \frac{|2a + 11|}{\sqrt{21} \sqrt{a^2 + 10}}$.

  We are given that $\cos\theta = \frac{1}{\sqrt{6}}$. $\frac{|2a + 11|}{\sqrt{21} \sqrt{a^2 + 10}} = \frac{1}{\sqrt{6}}$.

  This implies: $\frac{2a + 11}{\sqrt{21(a^2 + 10)}} = \pm\frac{1}{\sqrt{6}}$.

  Now, square both sides: $\frac{(2a + 11)^2}{21(a^2 + 10)} = \frac{1}{6}$.

  Cross-multiply to solve for $a$: $6(2a + 11)^2 = 21(a^2 + 10)$. Divide by 3: $2(2a + 11)^2 = 7(a^2 + 10)$.

  Expand the brackets: $2(4a^2 + 44a + 121) = 7a^2 + 70$. $8a^2 + 88a + 242 = 7a^2 + 70$.

  Rearrange into a 3-term quadratic equation: $a^2 + 88a + 172 = 0$.

  Factorise the quadratic: $(a + 2)(a + 86) = 0$.

  The possible values of $a$ are $a = -2$ and $a = -86$.

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  How the marks are awarded

  • M1 — Awarded for correctly calculating the scalar product of the two direction vectors, (2i+j+4k) . (ai-j+3k), resulting in the expression 2a+11.
  • M1* — Awarded for finding the magnitudes of both direction vectors and using the full scalar product formula, setting it up with the scalar product in the numerator and the product of magnitudes in the denominator.
  • A1 — Awarded for obtaining the correct equation (2a+11)/(√21√(10+a²)) = ±1/√6. The ± is essential as the formula for the acute angle between lines uses the modulus of the scalar product.
  • DM1 — Awarded for the dependent method of squaring both sides, expanding, and correctly simplifying to form a 3-term quadratic equation (a² + 88a + 172 = 0), and attempting to solve it.
  • A1 — Awarded for obtaining the two correct final values, a = -2 and a = -86.

  Common mistakes

  • Forgetting that the formula for an acute angle between lines involves the modulus of the scalar product, i.e., cosθ = |a.b|/(|a||b|). This leads to ignoring the ± and only finding one of the two possible values for a.
  • Making an arithmetic error when calculating the scalar product, for example, 2a - 1 + 12 = 2a - 13, which invalidates all subsequent steps.
  • Incorrectly squaring an expression, such as (2a + 11)² becoming 4a² + 121, which prevents the formation of the correct quadratic equation.
  • Using a position vector (e.g., 5i-j+2k) in the scalar product formula instead of the direction vector (ai-j+3k), which is a fundamental misunderstanding of the topic.

  Examiner tip: When finding an unknown in a vector equation involving an angle, remember that the absolute value in the acute angle formula cosθ = |a.b|/(|a||b|) often generates a ± case, leading to a quadratic equation and multiple solutions.

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