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Parametric equations provide a way to represent a geometric object, like a line, by using parameters. In the case of a line, we use a direction vector and a point on the line. For example, to find the parametric equations of the line through the points \((4, -1, 2)\) and \((3, 1, 4)\), we first determine a direction vector: \((3 – 4, 1 – (-1), 4 – 2) = (-1, 2, 2)\). Using one point, say \((4, -1, 2)\), and the direction vector, we write the parametric equations as follows: \ x = 4 - t \, \ y = -1 + 2t \, \ z = 2 + 2t \. These equations describe a line where \ t \ represents any real number.

The cross product of two vectors gives a third vector that is perpendicular to the plane formed by the original two. This is useful for finding the normal vector to a plane, which is essential in defining the plane's equation. For instance, to find the plane through the points \((0,0,0), (1,2,3), (2,1,1)\), we first find two direction vectors: \textbf{v}_1 = (1-0, 2-0, 3-0) = (1, 2, 3)\ and \textbf{v}_2 = (2-0, 1-0, 1-0) = (2, 1, 1)\. The cross product of \ \textbf{v}_1\ and \ \textbf{v}_2\ is given by: \ \textbf{v}_1 \times \textbf{v}_2 = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ 1 & 2 & 3 \ 2 & 1 & 1 \ \textbf{a_{i},k} = (2-3, 3-2, 1-4) = (-1, 1, -3)\}\. This normal vector is used to write the plane's equation as \'\textbf{-x + y - 3z = 0 ''''.

The distance formula allows us to find the shortest distance between a point and a geometric object like a plane or a line. For a plane, the formula is: \ d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}\. For example, the distance from the point \((1, 1, 1)\) to the plane \3x - 2y + 6z = 12\ is: \ \frac{|3(1) - 2(1) + 6(1) - 12|}{\sqrt{3^2 + (-2)^2 + 6^2}}\ = \frac{|-5|}{\/7}\f5'\

The angle between a plane and a line can be found by using the dot product between the plane's normal vector and the line's direction vector. The formula is: \ cos(\theta) = \frac{(a, b, c) \ \textbf{dot} \ (d, e, f)}{ \sqrt{a^2 + b^2 + c^2} * \sqrt{d^2 + e^2 + f^2}}\. Here, \ theta\ is the angle between the two vectors. For instance, to find the angle between the plane \ 3x - 2y + 6z = 12)\ and the line with parametric equations \ x = 2 + t, y = 1 - 2t, z = -1 + 2t\r>[\(3 - 2, 6 // (1, -2, 2)\ be:: (\ theate), =). t\.{3, }]. This can be d \ \cos becoming\ = arccos(value'' = \)