A-Level Mathematics May/June 2025 Q9(b): Using a scalar product, find the exact value of cos BAC.

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  To find the angle BAC, we need the vectors $\vec{AB}$ and $\vec{AC}$.

  From the information given in the question, the relevant direction vectors are: $\vec{AB} = \begin{pmatrix} -3 \\ 5 \\ 1 \end{pmatrix}$ and $\vec{AC} = \begin{pmatrix} 1 \\ 7 \\ 3 \end{pmatrix}$.

  The formula for the angle between two vectors using the scalar product is: $\cos(\angle BAC) = \frac{\vec{AB} \cdot \vec{AC}}{|\vec{AB}| |\vec{AC}|}$

  First, find the scalar product (dot product) of $\vec{AB}$ and $\vec{AC}$: $\vec{AB} \cdot \vec{AC} = (-3)(1) + (5)(7) + (1)(3)$ $= -3 + 35 + 3$ $= 35$

  Next, find the magnitude of each vector: $|\vec{AC}| = \sqrt{1^2 + 7^2 + 3^2} = \sqrt{1 + 49 + 9} = \sqrt{59}$ $|\vec{AB}| = \sqrt{(-3)^2 + 5^2 + 1^2} = \sqrt{9 + 25 + 1} = \sqrt{35}$

  Now, substitute these values back into the cosine formula: $\cos(\angle BAC) = \frac{35}{\sqrt{35} \sqrt{59}}$

  To simplify, we can write the denominator as a single square root: $\cos(\angle BAC) = \frac{35}{\sqrt{35 \times 59}} = \frac{35}{\sqrt{2065}}$

  Alternatively, we can simplify by cancelling $\sqrt{35}$: $\cos(\angle BAC) = \frac{\sqrt{35} \times \sqrt{35}}{\sqrt{35} \sqrt{59}} = \frac{\sqrt{35}}{\sqrt{59}}$

  The question asks for the exact value, so the final answer is: $\cos(\angle BAC) = \frac{35}{\sqrt{2065}}$

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  How the marks are awarded

  • B1 — Correctly stating or finding a direction vector for the line AC, which is (1, 7, 3) or its negative equivalent.
  • M1 — Correctly applying the scalar product formula to the two vectors originating from vertex A, AB and AC, and obtaining the value 35.
  • M1 — Using the correct overall process for the cosine rule: dividing the calculated scalar product (35) by the product of the magnitudes of the two vectors.
  • A1 — Obtaining the final, correct exact value for cos BAC. The answer 35/√2065 is a valid exact form as required.

  Common mistakes

  • Using the wrong vectors, for example BA and AC, which do not originate from the same point, leading to an incorrect angle (the supplement).
  • Making an arithmetic error in the scalar product calculation, particularly with negative numbers.
  • Incorrectly calculating the magnitude of a vector, for instance by forgetting to square the components before adding them.
  • Finding the angle θ itself in degrees or radians, instead of providing the exact value of cos(θ) as the question explicitly requests.

  Examiner tip: Always ensure the two vectors used in the scalar product formula for an angle are directed away from the vertex where the angle is being measured (e.g., for angle A, use vectors AB and AC).

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