Distance from point to line

For a line l_1 and point A:

• l_1:r=\begin{pmatrix}1\\1\\-3\end{pmatrix}+\lambda\begin{pmatrix}2\\-2\\-1\end{pmatrix}
• A=(1,2,-1)
• l_1 has direction vector \begin{pmatrix}2\\-2\\-1\end{pmatrix}
• Let B be the point where AB makes a right angle to l_1
• We need to find B so that \vec{AB}\cdot d=0
• \vec{OB}=\begin{pmatrix}1+2\lambda\\1-2\lambda\\-3-\lambda\end{pmatrix}
• \vec{AB}=\vec{OB}-\vec{OA}
  • =\begin{pmatrix}1+2\lambda\\1-2\lambda\\-3-\lambda\end{pmatrix}-\begin{pmatrix}1,2,-1\end{pmatrix}
  • =\begin{pmatrix}1+2\lambda-1\\1-\lambda-2\\-3-\lambda--1\end{pmatrix}
  • =\begin{pmatrix}2\lambda\\-1-2\lambda\\-2-\lambda\end{pmatrix}
• For l_1 and \vec{AB} to be perpendicular:
  • \begin{pmatrix}2\lambda\\-1-2\lambda\\-2-\lambda\end{pmatrix}\cdot\begin{pmatrix}2\\-2\\-1\end{pmatrix}=0
  • 4\lambda-2(-1-2\lambda)-1(-2-\lambda)
  • $(4\lambda+4\lambda+\lambda)+2+2=0
  • 9\lambda+4=0
  • 9\lambda=-4
  • \lambda=-\frac49
• We’ve found \lambda=-\frac49!
• \vec{AB}=\begin{pmatrix}2\lambda\\-1-2\lambda\\-2-\lambda\end{pmatrix}
  • \vec{AB}=\begin{pmatrix}2(-\frac49)\\-1-2(-\frac49)\\-2-(\frac49)\end{pmatrix}
  • \begin{pmatrix}-\frac89\\-\frac19\\-\frac{14}9\end{pmatrix}
• The perpendicular distance is of length \vec{AB}:
  • \sqrt{(-\frac89)^2+(-\frac19)^2\\+(-\frac{14}9)^2}
  • =\frac{\sqrt{261}}9
  • =\frac{\sqrt{29}}3

Find the distance between the point (2,3,4) and the line l:r=\begin{pmatrix}1\\0\\2\end{pmatrix}+\lambda\begin{pmatrix}3\\-1\\4\end{pmatrix}

• The line l has direction vector \begin{pmatrix}3\\-1\\4\end{pmatrix}
• Let B be the point where AB makes a right angle to l
• We need to find B so that \vec{AB}\cdot d=0
• \vec{OB}=\begin{pmatrix}1+3\lambda\\0-\lambda\\2+4\lambda\end{pmatrix}
• \vec{AB}=\vec{OB}-\vec{OA}
  • =\begin{pmatrix}1+3\lambda\\0-\lambda\\2+4\lambda\end{pmatrix}-\begin{pmatrix}2,3,4\end{pmatrix}
  • =\begin{pmatrix}1+3\lambda-2\\0-\lambda-3\\2+4\lambda-4\end{pmatrix}
  • =\begin{pmatrix}3\lambda-1\\-3-\lambda\\4\lambda-2\end{pmatrix}
• For l and \vec{AB} to be perpendicular:
  • \begin{pmatrix}3\lambda-1\\-3-\lambda\\4\lambda-2\end{pmatrix}\cdot\begin{pmatrix}3\\-1\\4\end{pmatrix}=0
  • 3(3\lambda-1)-1(-3-\lambda)+4(4\lambda-2)=0
  • (9\lambda+\lambda+16\lambda)+(-3+3-8)=0
  • 26\lambda-8=0
  • 26\lambda=8
  • \lambda=\frac{4}{13}
• We’ve found \lambda=\frac{4}{13}!
• \vec{AB}=\begin{pmatrix}3\lambda-1\\-3-\lambda\\4\lambda-2\end{pmatrix}
  • \vec{AB}=\begin{pmatrix}3(\frac{4}{13})-1\\-3-(\frac{4}{13})\\4(\frac{4}{13})-2\end{pmatrix}
  • =\begin{pmatrix}\frac{12}{13}-\frac{13}{13}\\-\frac{39}{13}-\frac{4}{13}\\\frac{16}{13}-\frac{26}{13}\end{pmatrix}
  • =\begin{pmatrix}-\frac{1}{13}\\-\frac{43}{13}\\-\frac{10}{13}\end{pmatrix}
• The perpendicular distance is of length \vec{AB}:
  • \sqrt{(-\frac{1}{13})^2+(-\frac{43}{13})^2+(-\frac{10}{13})^2}
  • =\frac{\sqrt{1+1849+100}}{13}
  • =\frac{\sqrt{1950}}{13}
  • =\frac{5\sqrt{78}}{13}
• So the distance from the point (2,3,4) to the line l is \frac{5\sqrt{78}}{13}.

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