Use the Question 11 Writing Booklet (a) Solve the quadratic equation $z^{2} - 3z + 4 = 0$, where $z$ is a complex number - HSC - SSCE Mathematics Extension 2 - Question 11 - 2023 - Paper 1

To find the angle $heta$ between vectors $q$ and $b$, we can use the formula:

$ext{cos}( heta) = \frac{|q \cdot b|}{|q||b|}$

First, we need to compute the dot product $q \cdot b$: $q \cdot b = (1)(-1) + (2)(4) + (-3)(2) = -1 + 8 - 6 = 1$

Next, we find the magnitudes of the vectors: $|q| = \sqrt{(1)^2+(2)^2+(-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$ $|b| = \sqrt{(-1)^2+(4)^2+(2)^2} = \sqrt{1 + 16 + 4} = \sqrt{21}$

Thus, $ext{cos}( heta) = \frac{1}{\sqrt{14} \cdot \sqrt{21}}$

To find the angle: $heta = \text{cos}^{-1} \left( \frac{1}{\sqrt{14} \cdot \sqrt{21}} \right)$ Upon calculating, you can get: $heta \approx 87^{\circ}$.

To find a vector equation for the line through points $A$ and $B$, we first find the direction vector:

$ext{AB} = B - A = \begin{pmatrix} 0 - (-3) \ 2 - 1 \ 3 - 5 \end{pmatrix} = \begin{pmatrix} 3 \ 1 \ -2 \end{pmatrix}$

The vector equation of a line can be expressed as: $R(t) = A + t \cdot (B - A)$ where $t$ is a parameter. Thus, the vector equation is: $R(t) = \begin{pmatrix} -3 \ 1 \ 5 \end{pmatrix} + t \begin{pmatrix} 3 \ 1 \ -2 \end{pmatrix}, \quad t \in \mathbb{R}$.

To show that $CD$ is a parallelogram, we must demonstrate that opposite sides are equal and parallel. We know that in parallelogram $ABEF$, we have:

$AB = EF \text{ (opposite sides are equal)}$ $AB = DC \text{ (opposite sides are equal)}$

Since $AB$ and $DC$ are both equal to $AB$, we can conclude: $CD = AB \text{ (equal)}$ Thus, both pairs of opposite sides $AB$ and $CD$, as well as $AE$ and $BF$, are equal, verifying that $CD$ is also a parallelogram.

The equation of motion is given as: $\frac{dx}{dt} = -9(x - 4)$ This describes simple harmonic motion.

To find the period ($T$) we use: $T = 2 \pi \sqrt{\frac{m}{k}}$ where $m = 3$ and $k = 9$. This gives: $T = 2 \pi \sqrt{\frac{3}{9}} = 2 \pi \frac{1}{3} = \frac{2\pi}{3}$

The central point of motion is where the velocity is zero, which is at $x = 4$. Therefore, the central point is: $x_{c} = 4$.

To solve the integral, we can use the method of partial fractions:

Assume: $\frac{5x - 3}{(x + 1)(x - 3)} = \frac{A}{x + 1} + \frac{B}{x - 3}$

Multiplying through by $(x + 1)(x - 3)$ gives: $5x - 3 = A(x - 3) + B(x + 1)$

Next, we expand and collect like terms to find $A$ and $B$. Solving the resulting equations, we can determine that:

1. Substituting appropriate values gives: $B = 2$
2. Solving the equations, we find: $A = 3$

Now the integral becomes: $\int(\frac{3}{x + 1} + \frac{2}{x - 3}) ext{d}x$

Integrating term by term leads to: $= 3 \ln|x + 1| + 2 \ln|x - 3| + C$

Thus evaluating from 0 to 1 gives you the result.