Question

For each of the following pairs of vectors x and y, find the vector projection p of x onto y.

[ -1 ] [ 3 ]
x = [ 1 ] and y = [ -1 ]
[ -2 ] [ 4 ]

Answer 1

The vector projection
`\( \text{proj}_y(\mathbf{x}) \)` of vector
`\(\mathbf{x}\)` onto vector
`\(\mathbf{y}\)` is
`\(\begin{bmatrix} -(18)/(13) \\ (6)/(13) \\ -(24)/(13) \end{bmatrix}\)` given vectors
`\(\mathbf{x}\) and \(\mathbf{y}\)`.

The vector projection
`\( \text{proj}_y(\mathbf{x}) \)` of vector
`\(\mathbf{x}\)` onto vector
`\(\mathbf{y}\)` is given by:

`\[ \text{proj}_y(\mathbf{x}) = \frac{\mathbf{x} \cdot \mathbf{y}}{\|\mathbf{y}\|^2} \cdot \mathbf{y} \]`

Given vectors
`\(\mathbf{x}\) and \(\mathbf{y}\)`:

`\[ \mathbf{x} = \begin{bmatrix} -1 \\ 1 \\ -2 \end{bmatrix}, \quad \mathbf{y} = \begin{bmatrix} 3 \\ -1 \\ 4 \end{bmatrix} \]`

1. Calculate the dot product
`\(\mathbf{x} \cdot \mathbf{y}\)`:

`\[ \mathbf{x} \cdot \mathbf{y} = (-1)(3) + (1)(-1) + (-2)(4) \]`

2. Calculate the norm
`(\(\|\mathbf{y}\|\))` of vector
`\(\mathbf{y}\)`:

`\[ \|\mathbf{y}\| = √(3^2 + (-1)^2 + 4^2) \]`

3. Substitute these values into the formula:

`\[ \text{proj}_y(\mathbf{x}) = \frac{\mathbf{x} \cdot \mathbf{y}}{\|\mathbf{y}\|^2} \cdot \mathbf{y} \]`

Perform the calculations to find the vector projection.

`\[ \mathbf{x} = \begin{bmatrix} -1 \\ 1 \\ -2 \end{bmatrix}, \quad \mathbf{y} = \begin{bmatrix} 3 \\ -1 \\ 4 \end{bmatrix} \]`

1. Calculate the dot product
`\(\mathbf{x} \cdot \mathbf{y}\)`:

`\[ \mathbf{x} \cdot \mathbf{y} = (-1)(3) + (1)(-1) + (-2)(4) = -3 - 1 - 8 = -12 \]`

2. Calculate the norm
`(\(\|\mathbf{y}\|\))` of vector
`\(\mathbf{y}\)`:

`\[ \|\mathbf{y}\| = √(3^2 + (-1)^2 + 4^2) = √(9 + 1 + 16) = √(26) \]`

3. Substitute these values into the formula:

`\[ \text{proj}_y(\mathbf{x}) = \frac{\mathbf{x} \cdot \mathbf{y}}{\|\mathbf{y}\|^2} \cdot \mathbf{y} \]\[ \text{proj}_y(\mathbf{x}) = (-12)/(26) \cdot \begin{bmatrix} 3 \\ -1 \\ 4 \end{bmatrix} \]`

4. Perform the calculation:

`\[ \text{proj}_y(\mathbf{x}) = -(6)/(13) \cdot \begin{bmatrix} 3 \\ -1 \\ 4 \end{bmatrix} = \begin{bmatrix} -(18)/(13) \\ (6)/(13) \\ -(24)/(13) \end{bmatrix} \]`

Therefore, the vector projection
`\( \text{proj}_y(\mathbf{x}) \)` of vector
`\(\mathbf{x}\)` onto vector
`\(\mathbf{y}\)` is
`\(\begin{bmatrix} -(18)/(13) \\ (6)/(13) \\ -(24)/(13) \end{bmatrix}\)`.