Find the angle between the two vectors [1, -1, 1] and [2, 1, 4].

$\text{radians} = \text{degrees} \times \frac{\pi}{180^\circ}$.

(ii) $25^\circ 30'$

$30' = \frac{30}{60}^\circ = 0.5^\circ$.
So, $25^\circ 30' = 25.5^\circ$.

$\text{radians} = \text{degrees} \times \frac{\pi}{180^\circ}$.

b) Prove that $\cos(30^\circ-x)=\frac{\sqrt{3}\cos x+\sin x}{2}$

$\cos(A-B) = \cos A \cos B + \sin A \sin B$.
Let $A = 30^\circ$ and $B = x$.

This is equal to the Right Hand Side (RHS).

Therefore, the identity is proven.

c) A constant force of $\vec{F}=6\vec{i}-2\vec{j}+\vec{k}$ newtons displaces an object from $A=\vec{i}+\vec{j}+2\vec{k}$ to $B=3\vec{i}-\vec{j}+3\vec{k}$ (in metres). Find the work done in newton metres.

QUESTION TWO

a) Find the equation of the graph of the type $y=ax^2+bx+c$ which crosses the x-axis at the points $(1,0)$ and $(4,0)$ and also passes through the point $(3,-4)$.

The equation can be written in factored form as $y = a(x-1)(x-4)$.

Substitute $x=3$ and $y=-4$ into the equation.

b) The equation $kx^2-2x-7=0$ has two real roots. Find the value of the constant $k$.

In the given equation $kx^2-2x-7=0$:
$A = k$
$B = -2$
$C = -7$

c) For the given function $f(x)=x^2-6x+8$, find:

(i) The x-intercept

(ii) The y-intercept

(iii) The line of symmetry

In $f(x)=x^2-6x+8$, we have $a=1$ and $b=-6$.

(iv) Hence sketch the graph

• x-intercepts: $(2,0)$ and $(4,0)$
• y-intercept: $(0,8)$
• Line of symmetry: $x=3$
• Vertex: The x-coordinate of the vertex is on the line of symmetry, $x=3$. Substitute $x=3$ into $f(x)$ to find the y-coordinate. $$f(3) = (3)^2 - 6(3) + 8$$ $$f(3) = 9 - 18 + 8$$ $$f(3) = -1$$ So, the vertex is $(3,-1)$. Since $a=1 > 0$, the parabola opens upwards.

(Self-correction: I cannot draw a graph here, but I can describe it.)
The graph is a parabola opening upwards, passing through $(2,0)$, $(4,0)$, and $(0,8)$. Its lowest point (vertex) is at $(3,-1)$, and it is symmetric about the vertical line $x=3$.

QUESTION THREE

a) A is a point with coordinates $(1,0,3)$ and B is a point with coordinates $(3,-2,6)$. Find, using column vectors,

(i) $\vec{AB}$

(ii) The position vector of the point C, which is the mid-point of AB.

If $A=(x_1, y_1, z_1)$ and $B=(x_2, y_2, z_2)$, the midpoint $C$ is $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)$.

b) Vectors $\vec{OA}$ and $\vec{OB}$ are given by $\vec{OA}=\begin{pmatrix}3\\-1\\2\end{pmatrix}$ and $\vec{OB}=\begin{pmatrix}6\\-4\\-5\end{pmatrix}$. Calculate the scalar product $\vec{OA} \cdot \vec{OB}$.

c) Find the angle between the two vectors $\begin{pmatrix}1\\-1\\1\end{pmatrix}$ and $\begin{pmatrix}2\\1\\4\end{pmatrix}$.

Calculate the dot product $\vec{u} \cdot \vec{v}$.

$\cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}$.