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 oldsymbol{q} and oldsymbol{b}, we can use the formula:
oldsymbol{q} \cdot \boldsymbol{b} = |\boldsymbol{q}| |\boldsymbol{b}| \cos\theta
First, we calculate the dot product:
$\boldsymbol{q} \cdot \boldsymbol{b} = (1)(-1) + (2)(4) + (-3)(2) = -1 + 8 - 6 = 1$
Next, we find the magnitudes of the vectors:
$|\boldsymbol{q}| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$
$|\boldsymbol{b}| = \sqrt{(-1)^2 + 4^2 + 2^2} = \sqrt{1 + 16 + 4} = \sqrt{21}$
Now, substituting values into the formula:
$1 = \sqrt{14} \sqrt{21} \cos\theta$
Thus,
$\cos\theta = \frac{1}{\sqrt{14} \sqrt{21}}$
Calculating the value of $heta$ gives approximately $87^{\circ}$.

To find a vector equation of the line, we need a point on the line and a direction vector. The direction vector oldsymbol{AB} can be found by subtracting the coordinates of $A$ from those of $B$:
$\boldsymbol{AB} = \begin{pmatrix} 0 - (-3) \ 2 - 1 \ 3 - 5 \end{pmatrix} = \begin{pmatrix} 3 \ 1 \ -2 \end{pmatrix}$
A vector equation for the line can be written as:
$\boldsymbol{r} = \begin{pmatrix} -3 \ 1 \ 5 \end{pmatrix} + t \begin{pmatrix} 3 \ 1 \ -2 \end{pmatrix}, \ t \in \mathbb{R}$.

In parallelogram $ABCD$, opposite sides are equal in length and parallel. We have:
$AB = DC$
Since $AB$ is equal to $DC$, and both pairs of opposite sides in a parallelogram are equal, we can conclude that $CD$ is indeed also a parallelogram by definition.

The motion described by rac{dx}{dt} = -9(x - 4) resembles simple harmonic motion.
The general solution can be expressed as:
$x(t) = A \cos(\omega t + \phi) + C$
Comparing it with the standard form, we find that the amplitude $A$ is the maximum deviation and rac{dx}{dt} = -9 suggests a constant speed.
The period $T$ of the motion can be found as:
$T = \frac{2\pi}{\omega} = \frac{2\pi}{3}$
The central point of motion can be identified at $x = 4$.

To solve the integral, we can use partial fraction decomposition:
egin{equation} rac{5x - 3}{(x + 1)(x - 3)} = \frac{A}{x + 1} + \frac{B}{x - 3} \ \text{Equating coefficients, we solve for } A ext{ and } B.
ext{From the equation: } 5x - 3 = A(x - 3) + B(x + 1). \ Setting up the system, we find: 5 = A + B ext{ and } -3 = -3A + B. \ Solving gives us, A = 2, B = 3. Now substituting into the integral, we evaluate:
$\int\left( \frac{2}{x+1} + \frac{3}{x - 3} \right) dx = 2\ln|x + 1| + 3\ln|x - 3| + C.$
Evaluate from 0 to 1.