91Ó°ÊÓ

For each pair of points, we need to determine the direction vector, it is the difference between two points. Define the direction vector \(\bold{d} = \langle d_x, d_y, d_z\rangle\), where \(d_x = x_2 - x_1, d_y = y_2 - y_1, d_z = z_2 - z_1\): (a) \(\bold{d}_a = \langle -5-3, 7-(-2), 1-4\rangle = \langle -8, 9, -3\rangle\) (b) \(\bold{d}_b = \langle -3-2, -6-4, 0-0\rangle = \langle -5, -10, 0\rangle\) (c) \(\bold{d}_c = \langle 3-3, 7-7, -8-2\rangle = \langle 0, 0, -10\rangle\) (d) \(\bold{d}_d = \langle 3-(-2), 9-(-1), 7-5\rangle = \langle 5, 10, 2\rangle\)

Now, we will plug in the direction vectors, as well as one of the points, into the parametric equation of a line in 3D space, which has the form \(\bold{r} = \bold{p} + t\bold{d}\), where \(\bold{r} = \langle x, y, z\rangle\), \(\bold{p}\) is a point on the line, \(t\) is a scalar parameter, and \(\bold{d}\) is the direction vector. (a) \(\bold{r}_a = \langle 3, -2, 4\rangle + t\langle -8, 9, -3\rangle = \langle 3-8t, -2+9t, 4-3t\rangle\) (b) \(\bold{r}_b = \langle 2, 4, 0\rangle + t\langle -5, -10, 0\rangle = \langle 2-5t, 4-10t, 0\rangle\) (c) \(\bold{r}_c = \langle 3, 7, 2\rangle + t\langle 0, 0, -10\rangle = \langle 3, 7, 2-10t\rangle\) (d) \(\bold{r}_d = \langle -2, -1, 5\rangle + t\langle 5, 10, 2\rangle = \langle -2+5t, -1+10t, 5+2t\rangle\) Now, we have the parametric equations for each line: (a) \(x = 3-8t\), \(y = -2+9t\), \(z = 4-3t\) (b) \(x = 2-5t\), \(y = 4-10t\), \(z = 0\) (c) \(x = 3\), \(y = 7\), \(z = 2-10t\) (d) \(x = -2+5t\), \(y = -1+10t\), \(z = 5+2t\)