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

We already know ( \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \mathbf{b} = \begin{pmatrix} -1 \ 0 \ 2 \end{pmatrix} ). Now we compute:

( \mathbf{a} - \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \mathbf{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} )

Next, we find the dot product:

( \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, the vector is perpendicular to ( \mathbf{b}. )

We set up the partial fraction decomposition:

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

Multiplying through by the denominator:

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

Expanding and combining like terms gives us:

1. Choose appropriate values for ( x ) to solve for ( A, B, C ):
   • Let ( x = 1 ): $3(1)^2 + 2(1) + 1 = A(1^2 + 1) + (B(1)+C)(1-1) \\ 6 = 2A \\ A = 3.$
   • Let ( x = 0 ): $1 = 3(0^2 + 1) + (B(0)+C)(-1) \\ 1 = 3 + C(-1) \\ C = 2.$
   • Let ( x = -1 ): $3(-1)^2 + 2(-1) + 1 = 3(-1^2 + 1) + (B(-1)+C)(-2) \\ 3 - 2 + 1 = 3(2) + (-B+2)(-2) \\ 2 = 6 - 2B + 4 \\ 2B = 8 \\ B = 4.$ We have determined ( A = 3, B = 4, C = 2 ), thus:

$\int \frac{3}{x-1} + \frac{4x + 2}{x^2 + 1} \, dx.$

Now, integrate term by term:

1. ( \int \frac{3}{x-1} , dx = 3 \ln|x-1| + C_1, )

2. Using substitution for ( x^2 + 1 ), we find:

   ( \int \frac{4x + 2}{x^2 + 1} , dx )

   • Use substitution ( u = x^2 + 1, du = 2x , dx )
   • The integral becomes ( 2 \ln|x^2 + 1| + C_2. )

Combining results:

$3 \ln|x-1| + 2 \ln|x^2 + 1| + C.$

Using the equation ( |z|^2 = z + 8 + 12i ):

We know that ( z = a + bi ) implies ( |z|^2 = a^2 + b^2 ). Substituting:

1. The left-hand side gives: ( a^2 + b^2 = (a + bi) + 8 + 12i. )
2. Comparing real and imaginary parts:

• Real: ( a^2 + b^2 = a + 8 ) implies ( b^2 = -12. ) Since it gives no contradiction, we conclude ( b = -12. )

Since we established ( b = -12 ): therefore:

Substituting back to find ( z ):

1. From ( |z|^2 = a^2 + (-12)^2 = a^2 + 144. )
2. By equating: ( a^2 + 144 = a + 8. )
3. Rearranging gives the quadratic equation: ( a^2 - a + 136 = 0. )
4. Solving this using the quadratic formula: ( a = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 136}}{2\cdot 1} = \frac{1 \pm \sqrt{-543}}{2}. )
5. Upon finding the solutions, we substitute back: ( z = a - 12i. )

To prove this statement, we analyze ( (n + 1)^1 - 79n^{40} = 2 ):

1. If ( n ) is odd, then ( (n + 1) ) will be even, and thus ( (n + 1)^1 ) is even.
2. On the other hand, ( 79n^{40} ) will also be odd, hence the left-hand result will be odd, contradicting it being 2.
3. If ( n ) is even, then ( (n + 1)^1 ) is odd, while ( 79n^{40} ) being even gives:
   • Therefore, the sum could still not equal 2. Concluding both scenarios show inconsistency, demonstrating that no integer ( n ) satisfies the equation.

Given points ( A(3, 5, -4) ) and ( B(7, 0, 2) ), we calculate a direction vector:

1. Calculating directions: ( \mathbf{d} = \begin{pmatrix} 7 - 3 \ 0 - 5 \ 2 - (-4) \end{pmatrix} = \begin{pmatrix} 4 \ -5 \ 6 \end{pmatrix}. )
2. The vector equation of the line then translates to: $\ell: \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3 \\ 5 \\ -4 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ -5 \\ 6 \end{pmatrix}, \lambda \in \mathbb{R}$

To check if the point ( C(10, 5, -2) ) lies on the line ( \ell ):

1. Substitute to see if there exists a ( \lambda ) such that:
   • X component: ( 10 = 3 + 4\lambda ) implies ( \lambda = \frac{7}{4} ).
   • Y component: ( 5 = 5 - 5\lambda ) rearranges to verify has consistent values.
   • Z component: ( -2 = -4 + 6\lambda ) gives: ( \lambda = \frac{1}{3} ).
2. These calculations produce inconsistent results of ( \lambda ) across components: hence, point ( C ) does not lie on the line ( \ell. )