• We will investigate the distance from a point to a line in both \(2\) and \(3\) dimensions, starting with \(\mathbb{R}^2\).

Find the distance from the point \(Q\,(-4,3)\) to the line \(l: 5x+y+8=0\).

Solution

Let the line orthogonal to \(l\) and passing through \(Q\) meet \(l\) at \(P\), as shown.

The distance from \(Q\) to \(l\) is the length of \(QP\), which is equal to \(\lvert \textbf{proj }(\overrightarrow{QP_0} \text{ onto } \vec{n}) \rvert\), where \(P_0\) is any point on \(l\) and \(\vec{n}\) is the normal to \(l\).

Find the distance from the point \(Q\,(-4,3)\) to the line \(l: 5x+y+8=0\).

Solution

By choosing \(P_0\,(0,-8)\), we get \(\overrightarrow{QP_0}=(4,-11)\).

Since a normal to \(l\) is \(\vec{n}=(5,1)\), then

Therefore, the distance from the point \(Q\,(-4,3)\) to the line \(l: 5x+y+8=0\) is \(\dfrac{9\sqrt{26}}{26}\).