IBDP Maths AHL 3.12 Concept of a vector AA HL Paper 2- Exam Style Questions- New Syllabus

(a)(i)
Applying the scalar product formula: \( \mathbf{a} \cdot \mathbf{c} = (-5)(p) + (7)(-6) = -5p – 42 \).
\( \mathbf{a} \cdot \mathbf{c} = \boxed{-5p – 42} \)

(a)(ii)
Applying the scalar product formula: \( \mathbf{b} \cdot \mathbf{c} = (-8)(p) + (9)(-6) = -8p – 54 \).
\( \mathbf{b} \cdot \mathbf{c} = \boxed{-8p – 54} \)

(b)
The condition is that the angle \( \theta \) between \( \mathbf{a} \) and \( \mathbf{c} \) equals the angle between \( \mathbf{b} \) and \( \mathbf{c} \).

Using the formula \( \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}||\mathbf{v}|} \), we set up the equality: \( \frac{-5p – 42}{\sqrt{(-5)^2 + 7^2} \cdot \sqrt{p^2 + (-6)^2}} = \frac{-8p – 54}{\sqrt{(-8)^2 + 9^2} \cdot \sqrt{p^2 + (-6)^2}} \)

Since the denominators \( \sqrt{p^2 + 36} \) are common and non-zero, they cancel, leaving: \( \frac{-5p – 42}{\sqrt{74}} = \frac{-8p – 54}{\sqrt{145}} \)

Cross-multiplying and solving for \( p \): \( \sqrt{145}(-5p – 42) = \sqrt{74}(-8p – 54) \) \( -5p\sqrt{145} – 42\sqrt{145} = -8p\sqrt{74} – 54\sqrt{74} \) \( 8p\sqrt{74} – 5p\sqrt{145} = 42\sqrt{145} – 54\sqrt{74} \) \( p(8\sqrt{74} – 5\sqrt{145}) = 42\sqrt{145} – 54\sqrt{74} \) \( p = \frac{42\sqrt{145} – 54\sqrt{74}}{8\sqrt{74} – 5\sqrt{145}} \)

Numerical evaluation yields \( p \approx 4.78727… \).
\( p = \boxed{4.79} \) (to three significant figures)