Q.(a) A bridge has a parabolic arch that is 10 m high in the centre and 30 m wide at the bottom. Find the height of the arch 6 m from the centre, on either sides. OR

(a) Models the arch as a downward-opening parabola fitted through the base points, then evaluates its height at $6$ m from the centre; (b) uses the dot product of two unit position vectors, computed two ways, to derive the cosine-difference formula. Both alternatives answered below.

(a) Height of the parabolic arch $6$ m from the centre

1. Set up coordinates. Take the origin at the base of the arch directly below the vertex, with $x$ measured horizontally and $y$ measured vertically upward. The vertex (highest point) is at $(0,10)$, and the arch is $30$ m wide at the bottom, so it meets the ground at $(\pm15,0)$.

2. General downward parabola with vertex $(0,10)$. $x^2=-4a(y-10)$ for some $a>0$; write it as $x^2=k(10-y)$ with $k=4a>0$.

3. Fit using the base point $(15,0)$.

$15^2=k(10-0)\ \Rightarrow\ 225=10k\ \Rightarrow\ k=22.5$

So the arch is $x^2=22.5(10-y)$.

4. Evaluate at $x=6$ (i.e. $6$ m from the centre).

$36=22.5(10-y)\ \Rightarrow\ 10-y=\dfrac{36}{22.5}=1.6\ \Rightarrow\ y=8.4$

5. By symmetry the same height, $8.4$ m, occurs on either side of the centre.

(b) Vector proof of $\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$

1. Unit vectors. Let $\hat a=\cos\alpha\,\hat i+\sin\alpha\,\hat j$ and $\hat b=\cos\beta\,\hat i+\sin\beta\,\hat j$ — unit vectors making angles $\alpha,\beta$ with the $x$-axis, so $|\hat a|=|\hat b|=1$.

2. Dot product by components. …