The projection of the vector \( \begin{pmatrix} 6 \\ 7 \end{pmatrix} \) onto the line \( y = 2x \) is \( \begin{pmatrix} 4 \\ 8 \end{pmatrix} \) - HSC - SSCE Mathematics Extension 1 - Question 9 - 2020 - Paper 1

To find the projection of the vector ( \mathbf{v} = \begin{pmatrix} 6 \ 7 \end{pmatrix} ) onto the line, we can use the formula: [ \text{proj}{\mathbf{n}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{n}}{\mathbf{n} \cdot \mathbf{n}} \mathbf{n} ] First, calculate the dot products: [ \mathbf{v} \cdot \mathbf{n} = 6 \cdot 1 + 7 \cdot (-2) = 6 - 14 = -8 ] [ \mathbf{n} \cdot \mathbf{n} = 1 \cdot 1 + (-2) \cdot (-2) = 1 + 4 = 5 ] Therefore, [ \text{proj}{\mathbf{n}} \mathbf{v} = \frac{-8}{5} \begin{pmatrix} 1 \ -2 \end{pmatrix} = \begin{pmatrix} -\frac{8}{5} \ \frac{16}{5} \end{pmatrix} ]

To find the reflection point A, we first find the coordinates of the foot of the perpendicular, which is given by: [ \mathbf{p} = \begin{pmatrix} 6 \ 7 \end{pmatrix} + \text{proj}_{\mathbf{n}} \mathbf{v} = \begin{pmatrix} 6 \ 7 \end{pmatrix} + \begin{pmatrix} -\frac{8}{5} \ \frac{16}{5} \end{pmatrix} = \begin{pmatrix} 6 - \frac{8}{5} \ 7 + \frac{16}{5} \end{pmatrix} = \begin{pmatrix} \frac{30}{5} - \frac{8}{5} \ \frac{35}{5} + \frac{16}{5} \end{pmatrix} = \begin{pmatrix} \frac{22}{5} \ \frac{51}{5} \end{pmatrix} ] The coordinates of point A will be the reflection across the line, thus: [ A = \begin{pmatrix} 2 \ 9 \end{pmatrix} ]