Answer to Question #137169 in Linear Algebra for Brightness

Question

What is the relationship between the vector (2,1,2) and the line. L defined by the following parametric equations
x=1+2t
y=2+t
z=3-2t

Answer 1

$\textsf{The general vector equation of a straight line is,}\\ (x, y, z) = a + \lambda d \hspace{0.1cm}\textsf{where} \hspace{0.1cm} \lambda\hspace{0.1cm}\textsf{is a parameter and}\\d\hspace{0.1cm} \textsf{is the direction vector.}\\ \textsf{Let the vector equation of the line be defined as}\hspace{0.1cm} L_1.\\ (x, y, z) = (1, 2, 3) + (2, 1, -2)t\\ \textsf{The vector equation of the line perpendicular to}\hspace{0.1cm} L_1\\\textsf{passing through line}\hspace{0.1cm} P(x_0, y_0, z_0)\hspace{0.1cm}\textsf{is given by}\\ ax + by + cz = ax_0 + by_0 + cz_0\hspace{0.1cm}..... (1)\\ \textsf{Define the vector equation of the line}\\\textsf{perpendicular to line passing through}\hspace{0.1cm}P \hspace{0.1cm}\textsf{as}\hspace{0.1cm} L_2. \\ (a, b, c)\hspace{0.1cm}\textsf{in}\hspace{0.1cm}(1) \hspace{0.1cm}\textsf{is the direction vector}\\\textsf{of the line passing through}\hspace{0.1cm}L_1, \hspace{0.1cm}\textsf{thus,}\hspace{0.1cm} (a, b, c) = (2, 1, -2).\\ \therefore 2x + y - 2z = ax_0 + by_0 + cz_0.\\ \textsf{If} \hspace{0.1cm}L_2 \hspace{0.1cm}\textsf{passes through}\hspace{0.1cm} M(2, 1, 2),\\ \begin{aligned} 2x + y - 2z &= (2, 1, -2) \cdot (2, 1, 2)\\&= 4 + 1 - 4 = 1\\&=1 \end{aligned}\\ \textsf{Thus,}\hspace{0.1cm} 2x + y - 2z = 1\hspace{0.1cm} \textsf{passes through the point}\hspace{0.1cm}(2, 1, 2) \\\textsf{and is perpendicular to}\hspace{0.1cm} L_1.\\ \textsf{To find the point where the lines}\hspace{0.1cm} x(t), \\y(t) \hspace{0.1cm}\textsf{and}\hspace{0.1cm} z(t)\hspace{0.1cm} \textsf{intersects the plane,}\\\textsf{substitute the parametric equations}\\\textsf{of the vector equation of the line}\\ 2x + y - 2z = 1\\ 2(1 + 2t) + (2 + t) - 2(3 - 2t) = 1\\ 2 + 4t + 2 + t - 6 +4t = 1\\ -2 + 9t = 1\\ \displaystyle 9t = 3, t = \frac{1}{3}\\ \displaystyle\textsf{Thus, at}\hspace{0.1cm} t = \frac{1}{3}, \hspace{0.1cm} L_1 \hspace{0.1cm}\textsf{intersects}\hspace{0.1cm}L_2 \\ \displaystyle\textsf{The coordinates of a point N on}\hspace{0.1cm} L_1 \hspace{0.1cm} \textsf{at}\hspace{0.1cm}t = \frac{1}{3} \hspace{0.1cm}\textsf{are}\\\textsf{given by}\hspace{0.1cm}\left(\frac{5}{3}, \frac{7}{3},\frac{7}{3}\right).\\ \displaystyle\textsf{Thus, the vector equation of the line passing}\\\textsf{through point}\hspace{0.1cm} M\hspace{0.1cm}\&\hspace{0.1cm} N \\\textsf{is the line passing through}\hspace{0.1cm}(2, 1, 2)\hspace{0.1cm}\textsf{and}\\\left(\frac{5}{3}, \frac{7}{3}, \frac{7}{3}\right) \hspace{0.1cm}\textsf{which is given by}\\ \displaystyle(x, y, z) = (2, 1, 2) + \lambda\left(\left(\frac{5}{3}, \frac{7}{3}, \frac{7}{3}\right) - (2, 1, 2)\right)\\ \displaystyle(x, y, z) = (2, 1, 2) + \lambda\left(\frac{-1}{3}, \frac{4}{3}, \frac{1}{3}\right) \hspace{0.1cm}......(2)\\ \textsf{Thus, equation}\hspace{0.1cm} (2) \hspace{0.1cm}\textsf{shows the relationship}\\\textsf{between} \hspace{0.1cm}(2, 1, 2) \hspace{0.1cm}\textsf{and the vector equation of the line}\hspace{0.1cm}L_1.$