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

The angle between the vectors $q = i + 2j - 3k$ and $b = -i + 4j + 2k$ can be found using the dot product formula:

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

First, we calculate the dot product:

$q \cdot b = (1)(-1) + (2)(4) + (-3)(2) = -1 + 8 - 6 = 1.$
Next, we find the magnitudes:

$|q| = \sqrt{(1^2 + 2^2 + (-3)^2)} = \sqrt{1 + 4 + 9} = \sqrt{14},$
$|b| = \sqrt{((-1)^2 + 4^2 + 2^2)} = \sqrt{1 + 16 + 4} = \sqrt{21}.$
Then substituting into the formula gives us:

$\cos\theta = \frac{1}{\sqrt{14} \cdot \sqrt{21}} \approx 0.174.$

Now, calculate the angle: $\theta = \cos^{-1}(0.174) \approx 86.6^\circ \text{ (to the nearest degree)}.$

To find a vector equation of the line through points A and B, we can represent the points as vectors:

$A = \begin{pmatrix}-3 \\ 1 \\ 5 \end{pmatrix}, ext{ and } B = \begin{pmatrix}0 \\ 2 \\ 3 \end{pmatrix}.$
We find the direction vector $\vec{AB}$:

$\vec{AB} = B - A = \begin{pmatrix} 0 \\ 2 \\ 3 \end{pmatrix} - \begin{pmatrix} -3 \\ 1 \\ 5 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}.$
Thus, the vector equation of the line is:

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

To show that quadrilateral $CD$ is a parallelogram, consider the previous result:

$AB = DC ext{ (opposite sides of a parallelogram are equal in length and parallel)}.$
Additionally, for parallelogram $ABEF$, we have:

$AB = EF ext{ (similarly, opposite sides are equal)}.$
Thus, since both pairs of opposite sides $CD$ and $AB$ are equal, we conclude that $CD$ is also a parallelogram.

The equation of motion is given by:

$x'' = -9(x - 4).$
This can be rearranged into standard form for simple harmonic motion: $x'' + 9x = 36.$
Identifying parameters, we have: $n^2 = 9 \Rightarrow n = 3 \, \text{ and } \, c = 4.$
Thus, the period $T$ can be found using:

$T = \frac{2\pi}{n} = \frac{2\pi}{3}.$
The central point of motion is: $x = 4.$

To evaluate the integral:

$\int_0^1 \frac{5x - 3}{(x + 1)(x - 3)} \, dx,$
we first perform partial fraction decomposition:

$\frac{5x - 3}{(x + 1)(x - 3)} = \frac{A}{x + 1} + \frac{B}{x - 3}.$
Multiplying through by the denominator:

$5x - 3 = A(x - 3) + B(x + 1).$
By equating coefficients and solving, we can find the values of $A$ and $B$. Continuing with the evaluated integral gives:

$\int_0^1 \frac{5x - 3}{(x + 1)(x - 3)} \, dx = -\ln(3) - 3\ln(1) = -\ln(3) + 0 = -\ln(3).$