Question

Find an equation for the plane containing the two (parallel) lines v, = (0, 1, -3) + +(2, 5, -1) and v, = (3,-1,0) + t(2, 5, -1).

Answer 1

Solution! you have to find the equation of plane Containing two parallel line. = (0,1-3) + t < 2, 5, -17 V2 = ( 3,-1,0) + f <2,5,-1). col. 3) && 2, 5, -177 (0,1,-3) (3,-1,0) from the given information you know one kector in the plane and two points (co, 1-7), (3,-1,0) with the help of these two points you can find another vector in the plane. e (3,-1,0) (0 - 3) another vector = 23,01,07 To 1-37 19 <3, 1-1, 0-(3) <3, -2,37 19

Now, you known two acolods in the plane. < 2, 5, -17, & <3.-23). find cross booduct of these two vector, to find normal vector (n) to the plane = < 25.- 13 x 23.-2, 73 - مه لرم 5 1 15- 2 -1 2 12 5 1-2 31 13 31 - î(15-9) - 5 ( 6+3) + Ê (-4-15) = 131 - 99 - 19 n = 13% -99-ig î (normal vector to the plane) (xo, Yo, 20) = (0,4l, -3) ( Point in the plane) Equation of Plane aca-20) + bly. Yo) a G (Z-20) = 0 where La, b, c): normal vector to the plane. (sco, Yo, 2o) Point in the plane

equation of Plane. - > 13 CX-o) - g (4-1) 19 (Z -(+3) = 0 150 - gy+9 - 192 - 57=6 132-gy - 197 = 48.