Use the Question 12 Writing Booklet (a) The vector $ extbf{a}$ is $$egin{pmatrix} 1 \ 2 \ 3 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\end{pmatrix}$$ and the vector $ extbf{b}$ is $$egin{pmatrix} 2 \ 0 \ -4 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \end{pmatrix}$$ (i) Find \( \frac{\textbf{a} \cdot \textbf{b}}{\textbf{b} \cdot \textbf{b}} \textbf{b} \) - HSC - SSCE Mathematics Extension 2 - Question 12 - 2024 - Paper 1

To show this, calculate:

$\textbf{a} - \frac{\textbf{a} \cdot \textbf{b}}{\textbf{b} \cdot \textbf{b}} \textbf{b} = \begin{pmatrix} 1 \ 2 \ 3 \end{pmatrix} - \begin{pmatrix} -1 \ 0 \ 2 \end{pmatrix} = \begin{pmatrix} 1 - (-1) \ 2 - 0 \ 3 - 2 \end{pmatrix} = \begin{pmatrix} 2 \ 2 \ 1 \end{pmatrix}.$

Now, calculate the dot product with ( \textbf{b} ):

$\begin{pmatrix} 2 \ 2 \ 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \ 0 \ -4 \end{pmatrix} = 2 \cdot 2 + 2 \cdot 0 + 1 \cdot (-4) = 4 - 4 = 0.$

Since the dot product is zero, this implies that ( \textbf{a} - \frac{\textbf{a} \cdot \textbf{b}}{\textbf{b} \cdot \textbf{b}} \textbf{b} ) is perpendicular to ( \textbf{b} ).

We start with a partial fractions assumption:

$\frac{3x^2 + 2x + 1}{(x - 1)(x^2 + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1}$

Multiply through by ( (x-1)(x^2 + 1) ) to clear the denominators:

$3x^2 + 2x + 1 = A(x^2 + 1) + (Bx + C)(x - 1).$

Expanding both sides, we get:

$3x^2 + 2x + 1 = Ax^2 + A + Bx^2 - Bx + Cx - C.$

Combining like terms yields:

$3x^2 + 2x + 1 = (A + B)x^2 + (C - B)x + (A - C).$

Matching coefficients gives us:

• For ( x^2 ): ( A + B = 3 )
• For ( x ): ( C - B = 2 )
• Constant: ( A - C = 1 )

From the second equation: ( C = B + 2 ).

Substituting into the first, we have:

( A + (B + 2) = 3 \Rightarrow A + B = 1 \Rightarrow A = 1 - B ).

Substituting ( A ) in the third equation gives:

( (1 - B) - (B + 2) = 1 \Rightarrow 1 - B - B - 2 = 1 \Rightarrow 1 - 2B - 2 = 1 \Rightarrow -2B = 2 \Rightarrow B = -1 ).

Then, ( C = -1 + 2 = 1 ).

And finally, substituting ( B ) into the first gives:

( A + (-1) = 3 \Rightarrow A = 4. )

Now we have ( A = 4, B = -1, C = 1 ). Thus:

$\int \frac{4}{x-1} + \frac{-x + 1}{x^2 + 1} dx.$

Integrating term by term gives:

$4 \ln |x - 1| - \frac{1}{2} \ln(x^2 + 1) + \tan^{-1}(x) + C.$

From the equation (|z|^2 = z + 8 + 12i), we know:

$|z|^2 = |a + bi|^2 = a^2 + b^2.$

Thus, the left-hand side equals:

$|z|^2 = z + 8 + 12i = (a + bi) + 8 + 12i = (a + 8) + (b + 12)i.$

This means the imaginary part must be zero, so:

$b + 12 = 0 \Rightarrow b = -12.$

This confirms that ( b = -12 ).

Substituting ( b = -12 ) into the equation:

(|z|^2 = a^2 + (-12)^2 = a^2 + 144),

and from the condition, we can set:

\sqrt{a^2 + 144} = a + 8 - 12i,$$$$ { ext{since }}\sqrt{12^2 + 8^2} = |8 + 12i|.

To solve for ( z ), we square both sides:

$a^2 + 144 = (a + 8)^2 + 144 \Rightarrow a^2 + 144 = a^2 + 16a + 64 + 144 \Rightarrow 0 = 16a + 64 \Rightarrow a = -4.$

Thus, we have:

$z = a + bi = -4 - 12i.$

Therefore, ( z = -4 - 12i ).

Consider the equation:

$(n + 1)^4 - 79n^4 = 2.$

Expanding ( (n + 1)^4 ):

$(n+1)^4 = n^4 + 4n^3 + 6n^2 + 4n + 1.$

Substituting gives:

$n^4 + 4n^3 + 6n^2 + 4n + 1 - 79n^4 = 2.$

Rearranging effectively results in:

Notice that this polynomial is dominated by the \( -78n^4 \) term for large \( n \), indicating that: * If \( n \) is large, it will be negative. * If \( n \) is small, checking integer values will also yield negative values. Since the left side cannot equal 2 for any integer \( n \), we conclude that there is no integer \( n \) such that \( (n + 1)^4 - 79n^4 = 2 \).

The vector equation of a line through points ( A(3, 5, -4) ) and ( B(7, 0, 2) ) can be expressed as:

1. Find the direction vector ( \textbf{d} = \textbf{B} - \textbf{A} = \begin{pmatrix} 7 - 3 \ 0 - 5 \ 2 - (-4) \end{pmatrix} = \begin{pmatrix} 4 \ -5 \ 6 \end{pmatrix} )

Thus, the line ( \ell ) can be represented in vector form as: $\begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 3 \ 5 \ -4 \end{pmatrix} + \lambda \begin{pmatrix} 4 \ -5 \ 6 \end{pmatrix}$, where ( \lambda \in \mathbb{R} ).

To determine whether ( C(10, 5, -2) ) lies on the line ( \ell ), we must check if there exists a value of ( \lambda ) such that:

$\begin{pmatrix} 10 \ 5 \ -2 \end{pmatrix} = \begin{pmatrix} 3 \ 5 \ -4 \end{pmatrix} + \lambda \begin{pmatrix} 4 \ -5 \ 6 \end{pmatrix}$.

This gives us the following equations:

1. ( 10 = 3 + 4\lambda ) \Rightarrow ( \lambda = \frac{7}{4} ),
2. ( 5 = 5 - 5\lambda ) \Rightarrow ( 0 = -5\lambda ) so ( \lambda = 0 ),
3. ( -2 = -4 + 6\lambda ) \Rightarrow ( 6\lambda = 2 ) \Rightarrow ( \lambda = \frac{1}{3} ).

As we can see, there are inconsistent values for ( \lambda ). Thus, point ( C(10, 5, -2) ) does not lie on the line ( \ell ).