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) $ is reflected in the line $ y = 2x $ to a point A. What is the position vector of the point A?

SSCE HSC Mathematics Extension 1 Vectors in two dimensions: The projection of the vector \(

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}-$-HSC-SSCE Mathematics Extension 1-Question 9-2020-Paper 1.png$

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

The projection of the vector onto the line

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Answer

To find the projection of the vector ( \mathbf{v} = \begin{pmatrix} 6 \ 7 \end{pmatrix} ) onto the line ( y = 2x ), we first write the direction vector of the line. The slope is 2, hence the direction vector ( \mathbf{d} = \begin{pmatrix} 1 \ 2 \end{pmatrix} ).

The projection formula is given by:

$\text{proj}_{\mathbf{d}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{d}}{\mathbf{d} \cdot \mathbf{d}} \mathbf{d}$

Calculating ( \mathbf{v} \cdot \mathbf{d} ): ( (6)(1) + (7)(2) = 6 + 14 = 20 )

Calculating ( \mathbf{d} \cdot \mathbf{d} ): ( (1)(1) + (2)(2) = 1 + 4 = 5 )

Thus, the projection is:

$\text{proj}_{\mathbf{d}} \mathbf{v} = \frac{20}{5} \begin{pmatrix} 1 \\ 2 \end{pmatrix} = 4 \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 4 \\ 8 \end{pmatrix}$

Step 2

Reflecting the point in the line

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Answer

We need to reflect the point ( (6, 7) ) across the line ( y = 2x ). The formula for reflection across a line ( Ax + By + C = 0 ) is:

$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix} - 2 \frac{Ax + By + C}{A^2 + B^2} \begin{pmatrix} A \\ B \end{pmatrix}$

For the line ( y - 2x = 0 ), we have ( A = -2, B = 1, C = 0 ).

First, we compute: ( Ax + By + C = -2(6) + (7) + 0 = -12 + 7 = -5 )

Next, calculate ( A^2 + B^2 ):
( (-2)^2 + (1)^2 = 4 + 1 = 5 )

Now we can substitute into the reflection formula:

$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 6 \\ 7 \end{pmatrix} - 2 \frac{-5}{5} \begin{pmatrix} -2 \\ 1 \end{pmatrix}$

This results in: $\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 6 \\ 7 \end{pmatrix} + 2 \begin{pmatrix} -2 \\ 1 \end{pmatrix} = \begin{pmatrix} 6 - 4 \\ 7 + 2 \end{pmatrix} = \begin{pmatrix} 2 \\ 9 \end{pmatrix}$

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

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

The projection of the vector onto the line

Reflecting the point in the line

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