Questions: Find the distance from the point Q to the line l. Q=(3,3), l with equation [x, y]=[-1, 3]+t[1, -1]

Solution Steps

To find the distance from a point \( Q \) to a line \( \ell \) in 2D, we can use the formula for the distance from a point to a line given in parametric form. The line \( \ell \) is given in the form: \[ \left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{r} -1 \\ 3 \end{array}\right]+t\left[\begin{array}{r} 1 \\ -1 \end{array}\right] \] We can use the following steps:

1. Identify a point \( P \) on the line \( \ell \) and the direction vector \( \mathbf{d} \).
2. Compute the vector \( \mathbf{PQ} \) from \( P \) to \( Q \).
3. Project \( \mathbf{PQ} \) onto \( \mathbf{d} \) to find the component of \( \mathbf{PQ} \) in the direction of the line.
4. Subtract this projection from \( \mathbf{PQ} \) to get the perpendicular component.
5. The magnitude of this perpendicular component is the distance from \( Q \) to the line \( \ell \).

\begin{pmatrix} -1 \\ 3 \end{pmatrix} + t \begin{pmatrix} 1 \\ -1 \end{pmatrix} \] From this, we identify a point \( P \) on the line as \( P = (-1, 3) \) and the direction vector \( \mathbf{d} = \begin{pmatrix} 1 \\ -1 \end{pmatrix} \).

The vector \( \mathbf{PQ} \) from \( P \) to \( Q \) is: \[ \mathbf{PQ} = Q - P = \begin{pmatrix} 3 \\ 3 \end{pmatrix} - \begin{pmatrix} -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \end{pmatrix} \]

The projection length of \( \mathbf{PQ} \) onto \( \mathbf{d} \) is: \[ \text{proj\_length} = \frac{\mathbf{PQ} \cdot \mathbf{d}}{\|\mathbf{d}\|} = \frac{\begin{pmatrix} 4 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -1 \end{pmatrix}}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2.8284 \] The projection vector is: \[ \text{proj\_vector} = \left( \frac{\text{proj\_length}}{\|\mathbf{d}\|} \right) \mathbf{d} = \left( \frac{2.8284}{\sqrt{2}} \right) \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \end{pmatrix} \]

The perpendicular component of \( \mathbf{PQ} \) is: \[ \text{perpendicular\_vector} = \mathbf{PQ} - \text{proj\_vector} = \begin{pmatrix} 4 \\ 0 \end{pmatrix} - \begin{pmatrix} 2 \\ -2 \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \end{pmatrix} \]

The distance from point \( Q \) to the line \( \ell \) is the magnitude of the perpendicular component: \[ \text{distance} = \|\text{perpendicular\_vector}\| = \sqrt{2^2 + 2^2} = \sqrt{8} = 2.8284 \]

Final Answer