1. Determine how the following lines interact. a. (x, y, z) = (-2, 1, 3) + t(1, 1, 5) ; (x, y, z) = (-3, 0, 2) + s(-1, 2, -3) b. (x, y, z) = (1, 2, 0) + t(1, 1, -1) ; (x, y, z) = (3, 4, -1) + s(2, 2, -2) c. x = 2 + t, y= -1 + 2t, z= -1 - t ; x= -1 - 2s, y= -1 -1s, z = 1 + s d. (x, y, z) = (1, -1, 2) + t(2, -1, 3) ; x = -3 - 4s, y = 1 + 2s, z = -4 -6s 2. Show that the two lines with equations (x, y, z) = (-1, 3, -4) + t(1, -1, 2) and (x, y, z) = (5, -3, 2) + s(-2, 2, 2) are perpendicular. Determine how the two lines interact. 3. Find the point of intersection of the line (x, y, z) = (1, -2, 1) + t(4, -3, -2) and the plane x - 2y + 3z = -8. 4. Determine how the following lines and planes interact. a. (x, y, z) = (-1, 2, 3) + t(1, 2, 2) ; 2x + 3y - 4z + 9 = 0. b. x = 2 - t, y = 3 - t, z= -1 + t; 3x - y+ 2z -1 = 0. 5. Determine the interaction of the line of intersection of the planes x + y - z = 1 and 3x + y + z = 3 with the line of intersection of the planes 2x - y + 2z = 4 and 2x + 2y + z = 1. 6. Determine the "shortest" distance between the point P(-3, 1,1) and the plane 2x - y + z - 5 = 0 Hint : Of all possible points that you could choose on this plane, find the point on the plane that makes the "shortest" distance between itself and the point P(-3,1,1). Find an equation of a line with P(-3,1,1) that intersects the plane 2x-y+z - 5=0, and then calculate the point of intersection of this line and plane. Finally, modify the 2-space distance formula for use in three-space. Use a diagram to help you visualize this situation