\[ y = \begin {bmatrix} 5 \\ 3 \end {bmatrix} \]

\[ u = \begin {bmatrix} 4 \\ 9 \end {bmatrix} \]

The question aims to find the **distance** between **vector y** to the line through **u** and the **origin.**

The question is based on the concept of **vector multiplication, dot product,** and **orthogonal projection. Dot product** of two vectors is the multiplication of corresponding terms and then the **summing** of their **output.** The **projection** of a **vector** onto a **plane** is known as the **orthogonal projection** of that **plane.**

## Expert Answer

The **orthogonal projection** of **y** is given by the formula as:

\[ \hat {y} = \dfrac{ y . u }{ u . u } u \]

We need to calculate the **dot products** of the **vectors** in the above formula. The **dot product** of **y** and **u** is given as:

\[ y . u = (5, 3) . (4, 9) \]

\[ y . u = 20 + 27 \]

\[ y . u = 47 \]

The **dot product** of **u** with itself is given as:

\[ u . u = (4, 9) . (4, 9) \]

\[ u .u = 16 + 81 \]

\[ u . u = 97 \]

Substituting the values in the above equation, we get:

\[ \hat {y} = \dfrac{ 47 }{ 97 } u \]

\[ \hat {y} = \dfrac{ 47 }{ 97 } \begin {bmatrix} 4 \\ 9 \end {bmatrix} \]

\[ \hat {y} = \begin {bmatrix} \frac{ 188 }{ 97 } \\ \frac{ 423 }{ 97 } \end {bmatrix} \]

We need to find the **difference** of $\hat {y}$ from y, which is given as:

\[ y\ -\ \hat {y} = \begin {bmatrix} 5 \\ 3 \end {bmatrix}\ -\ \begin {bmatrix} \frac{ 188 }{ 97 } \\ \frac{ 423 }{ 97 } \end {bmatrix} \]

\[ y\ -\ \hat {y} = \begin {bmatrix} \frac{ 297 }{ 97 } \\ \frac{ -132 }{ 97 } \end {bmatrix} \]

Finding the **distance,** we take the **square root** of the **sum** of **squared terms** of the **vector.** The **distance** is given as:

\[ d = \sqrt{ \dfrac{ 88209 }{ 9409 } + \dfrac{ 17424 }{ 9409 }} \]

\[ d = \sqrt{ \dfrac{ 1089 }{ 97 }} \]

\[ d = \dfrac{ 33 }{ \sqrt {97} } \]

\[ d = 3.35 units \]

## Numerical Result

The **distance** from **vector** **y** to the line through **vector u** and the **origin** is calculated to be:

\[ d = 3.35 units \]

## Example

Compute the **distance** from the given **vector y** to the line through the **vector** **u** and the **origin** if the **orthogonal projection** of **y** is given.

\[ y = \begin {bmatrix} 1 \\ 3 \end {bmatrix} \]

\[ \hat {y} = \begin {bmatrix} 22/13 \\ 33/13 \end {bmatrix} \]

\[ u = \begin {bmatrix} 2 \\ 3 \end {bmatrix} \]

The **distance** is calculated using the same **distance formula,** which is given as:

\[ d = 1.61 units \]