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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
Question 9
The projection of the vector \( \begin{pmatrix} 6 \\ 7 \end{pmatrix} \) onto the line \( y = 2x \) is \( \begin{pmatrix} 4 \\ 8 \end{pmatrix} \). The point \( (6, 7)... show full transcript
Worked Solution & Example Answer: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
Step 1
Calculate the projection of the vector onto the line
Answer
To find the projection of the vector ( \mathbf{v} = \begin{pmatrix} 6 \ 7 \end{pmatrix} ) onto the line represented by the vector ( \mathbf{d} = \begin{pmatrix} 1 \ 2 \end{pmatrix} ), we first calculate the scalar projection.
- Calculate the dot product ( \mathbf{v} \cdot \mathbf{d} = 6 \cdot 1 + 7 \cdot 2 = 20 ).
- Find the magnitude of the vector ( \mathbf{d} ): [ ||\mathbf{d}|| = \sqrt{1^2 + 2^2} = \sqrt{5} ]
- The projection formula is given by: [ \text{proj}_{\mathbf{d}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{d}}{||\mathbf{d}||^2} \mathbf{d} = \frac{20}{5} \begin{pmatrix} 1 \ 2 \end{pmatrix} = \begin{pmatrix} 4 \ 8 \end{pmatrix} ]
Step 2
Reflect the point (6, 7) across the line y = 2x
Answer
To find the reflection of the point ( (6, 7) ) across the line ( y = 2x ), we follow these steps:
- Find the slope of the line perpendicular to ( y = 2x ), which is ( y = -\frac{1}{2}x + c ).
- Set up the equation to find the intersection of the line through ( (6, 7) ) with this perpendicular line. [ \text{Slope from (6,7) to point on line} = -\frac{1}{2} \Rightarrow \text{Equation: } y - 7 = -\frac{1}{2}(x - 6) ]
- Solve this system of equations to find the intersection point ( (x_1, y_1) ).
- Calculate the reflection point ( A ):
- The reflection point can be determined using the midpoint formula between ( (6, 7) ) and the intersection point, using the coordinates to find the point A.
- The position vector of point A is then determined as ( A = \begin{pmatrix} 2 \ 9 \end{pmatrix} ).