A-Level Mathematics October/November 2024 Q9(c): Using a scalar product, calculate angle ABC.

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  To find the angle ABC, we need to use the scalar product formula for the vectors originating from the point B, which are $\overrightarrow{BA}$ and $\overrightarrow{BC}$.

  From the previous parts of the question, we have the position vectors: $\vec{A} = 5\mathbf{i} + 2\mathbf{j} - \mathbf{k}$ $\vec{B} = 3\mathbf{i} + 5\mathbf{j} + 3\mathbf{k}$ $\vec{C} = 0\mathbf{i} - \mathbf{j} + 4\mathbf{k}$

  First, find the direction vectors $\overrightarrow{BA}$ and $\overrightarrow{BC}$: $\overrightarrow{BA} = \vec{A} - \vec{B} = (5\mathbf{i} + 2\mathbf{j} - \mathbf{k}) - (3\mathbf{i} + 5\mathbf{j} + 3\mathbf{k}) = 2\mathbf{i} - 3\mathbf{j} - 4\mathbf{k}$ $\overrightarrow{BC} = \vec{C} - \vec{B} = (0\mathbf{i} - \mathbf{j} + 4\mathbf{k}) - (3\mathbf{i} + 5\mathbf{j} + 3\mathbf{k}) = -3\mathbf{i} - 6\mathbf{j} + \mathbf{k}$

  The scalar product formula is $\overrightarrow{BA} \cdot \overrightarrow{BC} = |\overrightarrow{BA}| |\overrightarrow{BC}| \cos(\theta)$, where $\theta$ is the angle ABC.

  Calculate the scalar product: $\overrightarrow{BA} \cdot \overrightarrow{BC} = (2)(-3) + (-3)(-6) + (-4)(1)$ $= -6 + 18 - 4$ $= 8$

  Calculate the magnitudes of the vectors: $|\overrightarrow{BA}| = \sqrt{2^2 + (-3)^2 + (-4)^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$ $|\overrightarrow{BC}| = \sqrt{(-3)^2 + (-6)^2 + 1^2} = \sqrt{9 + 36 + 1} = \sqrt{46}$

  Now, substitute these values into the formula to find $\cos(\theta)$: $\cos(\theta) = \frac{\overrightarrow{BA} \cdot \overrightarrow{BC}}{|\overrightarrow{BA}| |\overrightarrow{BC}|} = \frac{8}{\sqrt{29} \sqrt{46}}$

  Finally, calculate the angle $\theta$: $\theta = \cos^{-1}\left(\frac{8}{\sqrt{29 \times 46}}\right) = \cos^{-1}\left(\frac{8}{\sqrt{1334}}\right)$ $\theta = 77.347...^\circ$

  Angle ABC = $77.3^\circ$ (to 1 decimal place).

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  How the marks are awarded

  • B1FT — Correctly finding the two vectors originating from the angle's vertex, B. This is awarded for stating BA = 2i - 3j - 4k and BC = -3i - 6j + k (or their negatives AB and CB).
  • M1 — Correctly applying the method to calculate the scalar (dot) product of the two vectors found. The calculation (2)(-3) + (-3)(-6) + (-4)(1) = 8 earns this mark.
  • M1 — Using the full scalar product formula correctly. This involves finding the magnitudes of both vectors, dividing the scalar product by the product of the magnitudes, and setting up the inverse cosine: cos⁻¹(8 / (√29 * √46)).
  • A1 — Obtaining the final correct answer of 77.3° or 1.35 radians. The answer must be correctly rounded to at least 3 significant figures.

  Common mistakes

  • Using inconsistent vector directions, for example finding the angle between AB and BC. This calculates the supplementary angle (180° - θ) because one vector points towards B and one points away.
  • Making a sign error when calculating the scalar product, such as (-3)(-6) = -18 instead of +18.
  • Forgetting to square negative components when calculating a vector's magnitude, for example √((-3)²) becomes -3 instead of 3 or √(-9).
  • Using the wrong vectors, for example AB and AC, which would find the angle at vertex A (angle BAC) instead of B.

  Examiner tip: When finding the angle at a vertex, such as angle ABC at vertex B, ensure both vectors used in the scalar product either point away from the vertex (BA, BC) or both point towards it (AB, CB).

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