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, we use the formula:

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

1. Calculate the dot product: $q \cdot b = (1)(-1) + (2)(4) + (-3)(2) = -1 + 8 - 6 = 1$

2. Compute the magnitudes: $|q| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$ $|b| = \sqrt{(-1)^2 + 4^2 + 2^2} = \sqrt{1 + 16 + 4} = \sqrt{21}$

3. Substitute into the angle equation: $\cos \theta = \frac{1}{\sqrt{14} \cdot \sqrt{21}}$ $\theta = \cos^{-1} \left(\frac{1}{\sqrt{294}}\right) \approx 86.87°$

Thus, the angle is approximately 87°.

To find the vector equation of the line, we first determine the direction vector:

1. The direction vector is given by: $\overline{AB} = B - A = (0 - (-3), 2 - 1, 3 - 5) = (3, 1, -2)$

2. The vector equation of the line can be written as: $\mathbf{r} = \mathbf{A} + t \cdot (3, 1, -2)$ where (\mathbf{A} = (-3, 1, 5)) and (t \in \mathbb{R}.$$

To show that quadrilateral $CD$ is a parallelogram:

1. By definition, for quadrilaterals, opposite sides must be equal:

   • Since $AB = DC$ and $AD = BC$, we see that both pairs of opposite sides are equal and parallel.

2. Thus, $CD$ is a parallelogram due to the property that opposite sides of $ABCD$ are equal in a parallelogram.

To analyze the simple harmonic motion equation, we rearrange to:\n $\dot{x} = -9(x - 4)$\n

1. The equation can be rewritten as:$\dot{x} = -k(x - A)$ where (A = 4) (the central point) and (k = 9).

2. The period (T) is calculated using: $T = 2\pi \sqrt{\frac{m}{k}}$\n Assuming mass ( m = 1 ), then we have: $T = 2\pi \sqrt{\frac{1}{9}} = \frac{2\pi}{3}$ Hence, the period is ( T = \frac{2\pi}{3} ) and the central point of motion is ( x = 4 ).

To solve the integral:

1. We can use partial fractions to decompose: $\frac{5x - 3}{(x + 1)(x - 3)} = \frac{A}{x + 1} + \frac{B}{x - 3}$ Solving for A and B: $5x - 3 = A(x - 3) + B(x + 1)$ By equating coefficients, set up the system:
   • For (x^1) terms: (A + B = 5)
   • For constant terms: (-3A + B = -3)\n Solving gives A and B:
   • Replace back in integral, evaluate each part: $\int_0^1 \frac{A}{x + 1}dx + \int_0^1 \frac{B}{x - 3}dx$ leading to the final evaluation of the integral.