Point on line & Foot of perpendicular

Check if point lines on line

Any point that lies on a line must satisfy the equation of the line.

$$ l: \textbf{r} = \left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) + \lambda \left( \begin{matrix} 2 \\ -3 \\ 1 \end{matrix} \right), \lambda \in \mathbb{R} $$

Checking if the point $(3, 4, 5)$ lies on the line: \begin{align} \text{Let } \left( \begin{matrix} 3 \\ 4 \\ 5 \end{matrix} \right) & = \left( \begin{matrix} 1 + 2 \lambda \\ 1 - 3 \lambda \\ 1 + \lambda \end{matrix} \right) \\ \\ 3 & = 1 + 2\lambda \implies \lambda = 1 \\ 4 & = 1 - 3\lambda \implies \lambda = - 1 \\ 5 & = 1 + \lambda \implies \lambda = 4 \\ \\ \therefore (3, 4, 5) & \text{ does not lie on } l \end{align}

Checking if the point $(-2, 5.5, -0.5)$ lies on the line: \begin{align} \text{Let } \left( \begin{matrix} -2 \\ 5.5 \\ -0.5 \end{matrix} \right) & = \left( \begin{matrix} 1 + 2 \lambda \\ 1 - 3 \lambda \\ 1 + \lambda \end{matrix} \right) \\ \\ -2 & = 1 + 2\lambda \implies \lambda = -1.5 \\ 5.5 & = 1 - 3\lambda \implies \lambda = - 1.5 \\ -0.5 & = 1 + \lambda \implies \lambda = -1.5 \\ \\ \therefore (-2, 5.5, -0.5) & \text{ lies on } l \end{align}

Foot of perpendicular from point to line

$$ l: \textbf{r} = \left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) + \lambda \underbrace{ \left( \begin{matrix} 2 \\ -3 \\ 1 \end{matrix} \right)}_\textbf{d}, \lambda \in \mathbb{R} $$

There are two ways to find the coordinates of the foot of the perpendicular $N$ from the point $P(-1, 2, 3)$ to the line $l$.

\begin{align} l: \textbf{r} & = \left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) + \lambda \left( \begin{matrix} 2 \\ -3 \\ 1 \end{matrix} \right), \lambda \in \mathbb{R} \\ \\ \text{Since } N \text{ lies on } l, \overrightarrow{ON} & = \left( \begin{matrix} 1 + 2\lambda \\ 1 - 3\lambda \\ 1 + \lambda \end{matrix} \right), \text{ for a certain value of } \lambda \\ \\ \overrightarrow{PN} & = \overrightarrow{ON} - \overrightarrow{OP} \\ & = \left( \begin{matrix} 1 + 2\lambda \\ 1 - 3\lambda \\ 1 + \lambda \end{matrix} \right) - \left( \begin{matrix} -1 \\ 2 \\ 3 \end{matrix} \right) \\ & = \left( \begin{matrix} 2 + 2\lambda \\ -1 - 3\lambda \\ -2 + \lambda \end{matrix} \right) \\ \\ \text{Since } \overrightarrow{PN} \perp l, \phantom{0} \overrightarrow{PN} \cdot \left( \begin{matrix} 2 \\ -3 \\ 1 \end{matrix} \right) & = 0 \\ \left( \begin{matrix} 2 + 2\lambda \\ -1 - 3\lambda \\ -2 + \lambda \end{matrix} \right) \cdot \left( \begin{matrix} 2 \\ - 3 \\ 1 \end{matrix} \right) & = 0 \\ 2(2 + 2\lambda) - 3(-1 - 3\lambda) + (-2 + \lambda) &= 0 \\ 4 + 4\lambda + 3 + 9\lambda - 2 + \lambda & = 0 \\ 14 \lambda + 5 & = 0 \\ 14 \lambda & = -5 \\ \lambda & = -{5 \over 14} \\ \\ \overrightarrow{ON} & = \left( \begin{matrix} 1 + 2 \left(-{5 \over 14}\right) \\ 1 - 3\left(-{5 \over 14}\right) \\ 1 + \left(-{5 \over 14}\right) \end{matrix} \right) \\ & = \left( \begin{matrix} {2 \over 7} \\ {29 \over 14} \\ {9 \over 14} \end{matrix} \right) \\ \\ \therefore & \phantom{.} N \left( {2 \over 7}, {29 \over 14}, {9 \over 14} \right) \end{align}

\begin{align} l: \textbf{r} & = \left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) + \lambda \underbrace{ \left( \begin{matrix} 2 \\ -3 \\ 1 \end{matrix} \right)}_\textbf{d}, \lambda \in \mathbb{R} \\ \\ \text{Let } A & \text{ denote the point } (1, 1, 1) \\ \\ \text{Projection vector, } \overrightarrow{AN} & = \left( \overrightarrow{AP} \cdot \hat{ \textbf{d} } \right) \hat{ \textbf{d} } \\ \\ \overrightarrow{AP} & = \overrightarrow{OP} - \overrightarrow{OA} \\ & = \left( \begin{matrix} -1 \\ 2 \\ 3 \end{matrix} \right) - \left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) \\ & = \left( \begin{matrix} -2 \\ 1 \\ 2 \end{matrix} \right) \\ \\ \hat{ \textbf{d} } & = {1 \over | \textbf{d} | } \textbf{d} \\ & = {1 \over \sqrt{2^2 + (-3)^2 + 1^2} } \left( \begin{matrix} 2 \\ -3 \\ 1 \end{matrix} \right) \\ & = {1 \over \sqrt{14} } \left( \begin{matrix} 2 \\ -3 \\ 1 \end{matrix} \right) \\ \\ \overrightarrow{AN} & = \left[ \left( \begin{matrix} -2 \\ 1 \\ 2 \end{matrix} \right) \cdot {1 \over \sqrt{14} } \left( \begin{matrix} 2 \\ -3 \\ 1 \end{matrix} \right) \right] {1 \over \sqrt{14} } \left( \begin{matrix} 2 \\ -3 \\ 1 \end{matrix} \right) \\ & = {1 \over \sqrt{14}} \left[ \left( \begin{matrix} -2 \\ 1 \\ 2 \end{matrix} \right) \cdot \left( \begin{matrix} 2 \\ -3 \\ 1 \end{matrix} \right) \right] {1 \over \sqrt{14} } \left( \begin{matrix} 2 \\ -3 \\ 1 \end{matrix} \right) \\ & = {1 \over \sqrt{14}} [-4 + (-3) +2] {1 \over \sqrt{14} } \left( \begin{matrix} 2 \\ -3 \\ 1 \end{matrix} \right) \\ & = {1 \over 14} (-5) \left( \begin{matrix} 2 \\ -3 \\ 1 \end{matrix} \right) \\ & = -{5 \over 14} \left( \begin{matrix} 2 \\ -3 \\ 1 \end{matrix} \right) \\ \\ \overrightarrow{ON} & = \overrightarrow{OA} + \overrightarrow{AN} \\ & = \left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) - {5 \over 14} \left( \begin{matrix} 2 \\ -3 \\ 1 \end{matrix} \right) \\ & = \left( \begin{matrix} {2 \over 7} \\ {29 \over 14} \\ {9 \over 14} \end{matrix} \right) \\ \\ \therefore & \phantom{.} N \left( {2 \over 7}, {29 \over 14}, {9 \over 14} \right) \end{align}