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 the vectors $extbf{q}$ and $extbf{b}$, we use the formula:

$extbf{q} \cdot \textbf{b} = |\textbf{q}| |\textbf{b}| \cos \theta$

First, calculate the dot product:

$\textbf{q} \cdot \textbf{b} = (1)(-1) + (2)(4) + (-3)(2) = -1 + 8 - 6 = 1$

Next, we find the magnitudes:

$|\textbf{q}| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$

$|\textbf{b}| = \sqrt{(-1)^2 + (4)^2 + (2)^2} = \sqrt{1 + 16 + 4} = \sqrt{21}$

Thus,

$1 = \sqrt{14} \sqrt{21} \cos \theta$

Solving for $heta$:

$\cos \theta = \frac{1}{\sqrt{14} \sqrt{21}}$

Calculating: $\theta = \cos^{-1}(\frac{1}{\sqrt{14 \cdot 21}}) \approx 87^{\circ}$

To find the vector equation of the line, we first find the direction vector $extbf{AB}$:

$\textbf{AB} = \textbf{B} - \textbf{A} = (0, 2, 3) - (-3, 1, 5) = (3, 1, -2)$

Using point $A(-3, 1, 5)$, the vector equation can be expressed as:

$\textbf{r}(t) = \textbf{A} + t \textbf{AB}$

where $t \in \mathbb{R}$. Therefore:

$\textbf{r}(t) = (-3, 1, 5) + t(3, 1, -2)$

In parallelogram $ABCD$, we have:

$\textbf{AB} = \textbf{DC}$

For parallelogram $ABEF$:

$\textbf{AB} = \textbf{EF}$

From the properties of parallelograms, if two sides are equal to each other and are parallel, then:

$\textbf{CD} = \textbf{FE}$

Thus, $CD$ and $FE$ are also parallel with equal length, confirming that $CDEF$ is a parallelogram.

The equation describes simple harmonic motion. To analyze it, we first rewrite the equation to identify the properties:

The standard form for simple harmonic motion is:

$\frac{dx}{dt} = -k(x - A)$

Comparing gives:

• Here, $k = 9$ and $A = 4$.

To find the period $T$, we use:

$T = \frac{2\pi}{\sqrt{k}} = \frac{2\pi}{\sqrt{9}} = \frac{2\pi}{3}$

The central point of motion is at $A = 4$. Therefore:

• Period: $\frac{2\pi}{3}$
• Central point: $x = 4$.

To evaluate the integral, we first perform partial fraction decomposition:

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

Multiplying through by $(x + 1)(x - 3)$, we equate:

$5x - 3 = A(x - 3) + B(x + 1)$

Setting up the system of equations to solve for $A$ and $B$ gives:

• We find:

$A = 2, \quad B = 3$

Thus,

$\int \frac{5x - 3}{(x + 1)(x - 3)} dx = \int \left( \frac{2}{x + 1} + \frac{3}{x - 3} \right) dx$

The antiderivatives are:

$2 \ln|x + 1| + 3 \ln|x - 3| + C$

Evaluating from 0 to 1 will yield:

$\left[ 2 \ln|1 + 1| + 3 \ln|1 - 3| \right] - \left[ 2 \ln|0 + 1| + 3 \ln|0 - 3| \right]$