A-Level Mathematics May/June 2024 Q10(a): The equations of two straight lines are r=i+j+2ak+λ(3i+4j+ak) and r=-3i-j+4k+μ(−i+2j+2k…

1 answer

• ▲0
  accepted ✓

  Let the direction vectors of the two lines be $\mathbf{d}_1$ and $\mathbf{d}_2$. From the equations of the lines, we have: $\mathbf{d}_1 = 3\mathbf{i} + 4\mathbf{j} + a\mathbf{k} = \begin{pmatrix} 3 \\ 4 \\ a \end{pmatrix}$ $\mathbf{d}_2 = -\mathbf{i} + 2\mathbf{j} + 2\mathbf{k} = \begin{pmatrix} -1 \\ 2 \\ 2 \end{pmatrix}$

  The angle $\theta$ between the direction vectors is related by the scalar product formula: $\cos \theta = \frac{\mathbf{d}_1 \cdot \mathbf{d}_2}{|\mathbf{d}_1| |\mathbf{d}_2|}$

  First, calculate the scalar product: $\mathbf{d}_1 \cdot \mathbf{d}_2 = (3)(-1) + (4)(2) + (a)(2) = -3 + 8 + 2a = 5 + 2a$

  Next, calculate the magnitudes of the direction vectors: $|\mathbf{d}_1| = \sqrt{3^2 + 4^2 + a^2} = \sqrt{9 + 16 + a^2} = \sqrt{25 + a^2}$ $|\mathbf{d}_2| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$

  We are given that the acute angle between the lines is $\frac{1}{4}\pi$. This means the angle $\theta$ between their direction vectors is either $\frac{1}{4}\pi$ or $\pi - \frac{1}{4}\pi = \frac{3}{4}\pi$. Therefore, $\cos \theta = \cos(\frac{1}{4}\pi) = \frac{\sqrt{2}}{2}$ or $\cos \theta = \cos(\frac{3}{4}\pi) = -\frac{\sqrt{2}}{2}$. This can be written as $\cos \theta = \pm \frac{\sqrt{2}}{2}$.

  Substitute these into the scalar product formula: $\frac{5 + 2a}{3\sqrt{25 + a^2}} = \pm \frac{\sqrt{2}}{2}$

  To solve for $a$, we square both sides of the equation: $(\frac{5 + 2a}{3\sqrt{25 + a^2}})^2 = (\pm \frac{\sqrt{2}}{2})^2$ $\frac{(5 + 2a)^2}{9(25 + a^2)} = \frac{2}{4} = \frac{1}{2}$

  Now, cross-multiply to eliminate the fractions: $2(5 + 2a)^2 = 9(25 + a^2)$ $2(25 + 20a + 4a^2) = 225 + 9a^2$ $50 + 40a + 8a^2 = 225 + 9a^2$

  Rearrange into a standard quadratic form ($Ax^2+Bx+C=0$): $0 = (9a^2 - 8a^2) - 40a + (225 - 50)$ $a^2 - 40a + 175 = 0$

  Factorise the quadratic equation: $(a - 5)(a - 35) = 0$

  This gives two possible values for $a$. The possible values are $a = 5$ and $a = 35$.

  ---

  How the marks are awarded

  • M1 — Correctly calculating the scalar product of the two direction vectors: $(3)(-1) + (4)(2) + (a)(2)$, which simplifies to $5 + 2a$.
  • M1 — Using the full scalar product formula by finding the moduli of both direction vectors, dividing the scalar product by their product, and equating the result to $\pm \cos(\frac{1}{4}\pi)$ or $\pm \frac{\sqrt{2}}{2}$.
  • A1 — Stating a correct equation in any form, such as $\frac{5 + 2a}{3\sqrt{25 + a^2}} = \pm \frac{\sqrt{2}}{2}$.
  • DM1 — Correctly squaring both sides of the equation and rearranging it into a three-term quadratic equation, $a^2 - 40a + 175 = 0$. This mark is dependent on the previous two method marks.
  • A1 — Obtaining the first correct value, $a = 5$, from solving the quadratic equation.
  • A1 — Obtaining the second correct value, $a = 35$, from solving the quadratic equation.

  Common mistakes

  • Only considering the positive case for the angle, i.e., equating to $+\frac{\sqrt{2}}{2}$ only. This leads to a linear equation in $a$ and misses the second solution.
  • Making an algebraic error when squaring the equation, such as incorrectly expanding $(5+2a)^2$ as $25+4a^2$ or failing to square the '3' in the denominator.
  • Using the position vectors (e.g., $\mathbf{i}+\mathbf{j}+2a\mathbf{k}$) instead of the direction vectors (e.g., $3\mathbf{i}+4\mathbf{j}+a\mathbf{k}$) in the scalar product formula.
  • Making an error in calculating the magnitude of a vector, for example $\sqrt{25+a^2} = 5+a$, which is incorrect.

  Examiner tip: For problems involving angles between lines, always start by identifying the correct direction vectors and then apply the scalar product formula, remembering that the acute angle between lines means the angle between their direction vectors could be acute or obtuse.

  ---

  AI-generated model answer, grounded in the official Cambridge mark scheme and reviewed by the MarkScheme team. Mark your own answer to this question →

  @markscheme_answers

Your answer

Sign in to answer this question.