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 cosine formula:

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

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

3. Use the dot product and magnitudes to find the angle: $\cos(\theta) = \frac{\textbf{q} \cdot \textbf{b}}{|\textbf{q}| |\textbf{b}|} = \frac{1}{\sqrt{14} \sqrt{21}}$

4. Calculate: $\theta = \cos^{-1}(\frac{1}{\sqrt{14 \times 21}}) \approx 86.656$

5. Rounding to the nearest degree, the angle is approximately 87 degrees.

To find the vector equation of the line, we first determine the direction vector from A to B:

1. Calculate the direction vector: \textbf{AB} = \textbf{B} - \textbf{A} = \begin{pmatrix} 0 \ -3 \ 2 \ end{pmatrix}

2. The parametric equation for the line can be expressed as: $\textbf{r}(t) = \textbf{A} + t\textbf{AB} = \begin{pmatrix} -3 \ 1 \ 5 \end{pmatrix} + t\begin{pmatrix} 3 \ 1 \ -2 \end{pmatrix}, \ t \in \mathbb{R}$

Hence, the vector equation is: $\textbf{r}(t) = \begin{pmatrix} -3 \ 1 \ 5 \end{pmatrix} + t\begin{pmatrix} 3 \ 1 \ -2 \end{pmatrix}$.

To show that $\textbf{CD}$ is a parallelogram:

1. Recall that in a parallelogram, opposite sides are equal in length and parallel.

   • From the parallelogram property, we know: $\textbf{AB} = \textbf{DC}$ and $\textbf{AD} = \textbf{BC}$

2. Analyze both sets of opposite sides:

   • Given that $\textbf{AB} = \textbf{DC}$ (as identified in parallelogram ABCD), it implies that sides CD and AB are equal in length and direction.
   • Furthermore, since $\textbf{AD} = \textbf{BC}$, the condition for CD as a parallelogram is satisfied.

Thus, we conclude that $\textbf{CD}$ is also a parallelogram.

The equation governing the simple harmonic motion is given as:

1. Start with: $\frac{dx}{dt} = -9(x - 4)$

2. Rearranging gives: $\frac{dx}{x - 4} = -9 dt$

3. Integrating both sides leads to: $\int \frac{1}{x - 4} dx = -9 \int dt$

4. The solutions yield:

   • The central point of motion is x = 4.
   • The standard period T in simple harmonic motion can be determined as: $T = \frac{2\pi}{n}$ where n = 3 thus, $T = \frac{2\pi}{3}$.

To solve the integral,

1. First, perform partial fraction decomposition: $\frac{5x - 3}{(x + 1)(x - 3)} = \frac{A}{x + 1} + \frac{B}{x - 3}$

2. By matching coefficients:

   • Equate the numerators, $5x - 3 = A(x - 3) + B(x + 1)$

3. Solving gives:

   • From values found, determine constants A and B.

4. Finally, compute: \int \left( A rac{1}{x + 1} + B rac{1}{x - 3} \right)dx yielding the solution of that limit integral.