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 between the vectors $q = i + 2j - 3k$ and $b = -i + 4j + 2k$, we use the formula:

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

First, we compute the dot product:

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

Next, we find the magnitudes:

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

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

Now, substituting these values in:

$\cos(\theta) = \frac{1}{\sqrt{14} \cdot \sqrt{21}} = \frac{1}{\sqrt{294}}$

Calculating the angle:

$\theta = \cos^{-1}(\frac{1}{\sqrt{294}}) \approx 87^{\circ}$

Let vector $A = \begin{pmatrix} -3 \\ 1 \\ 5 \end{pmatrix}$ and vector $B = \begin{pmatrix} 0 \\ 2 \\ 3 \end{pmatrix}$. The direction vector is:

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

The vector equation of the line is given by:

$\mathbf{r} = A + t \overrightarrow{AB}$

Therefore:

$\mathbf{r} = \begin{pmatrix} -3 \\ 1 \\ 5 \end{pmatrix} + t \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}, \quad t \in \mathbb{R}$

To show that quadrilateral CD is a parallelogram, we need to confirm that opposite sides are equal and parallel. Since ABCD and ABEF are given as parallelograms, we know that:

1. $AB = DC$ (opposite sides) and they are equal in length.
2. $AD = BC$ which we already have as part of our parallelogram property.
3. Therefore, by the properties of parallelograms, since one pair of opposite sides in quadrilateral CD are equal in length and parallel, it implies that CD is a parallelogram too.

The differential equation is given by:

$\frac{dx}{dt} = -9(x - 4)$

To find the period of the motion, we recognize that this describes simple harmonic motion where:

1. The coefficient of $x$ is constant, $n = 3$.
2. The central point (or equilibrium position) is at $x = 4$.

The period ($T$) is given by:

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

Thus, the period is approximately:

$T = \frac{2\pi}{3}$, and the central point of motion is at: $x = 4$.

To evaluate the integral:

$\int_{0}^{2} \frac{5x - 3}{(x + 1)(x - 3)} dx$,

we will use partial fraction decomposition:

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

Equating coefficients:

1. $5 = -3A + B$,
2. $-3 = 4A - 3B$.

Solving gives:

1. For $A$: $A = 2$
2. For $B$: $B = 5$

Now substituting back:

$\int \left( \frac{2}{x + 1} + \frac{-3}{x - 3} \right) dx \rightarrow 2 \ln |x + 1| - 3 \ln |x - 3| + C$

Evaluating from 0 to 2:

This leads to a final solution, which can be computed to find $\left[ 2\ln(3) - 3\ln(1) \right] - \left[ 2\ln(1) - 3\ln(3) \right]$