Question

Determine whether the given lines intersect. If so, find the point of intersection. (If not, enter NOT.) x = 6 + t, y = 3 + t, z = -1 + 2t x = 8 + 2s, y = 9+ 4s, z = -3 + S (x, y, z) = eBook

Answer 1

Given lines will intersect if there exist real values for t and s such that 6+t=8+2s, 3+t=9+4s and -1+2t=-3+s

Given two lines are no 6tt, ya 37t, zar1+2+ 2 n=8+25, y=9+45 , Z =-345 ར sine Suppose these lines are intersecting, then oht, SG IR such that 6&t=8+25 ; 3t =9+45 -1+2t =-3t5 zs - tot 2 = 0 ; 45-d +6=0 ; S-2t - 2 = 0 equation we have to solve these three to get values of sandt. take first two equations 25 - t +20 45-df6 substract 3 from get as s = -2 25 +4=0 we get substitute sa-2 in - 846 - 2 ta 3 s-at-2 - Now substitute sa-2, += 2 in - 2 + 4- 2-o een G Therefore solutions of equations ② and ③ sentirties is the => ta-2, S = -2 solution of equartion &

Therefore we are getting real values for sandt for ean This shows that the given lines are intersect. To find the intersect point, substitute s=-2 in equation 2 or substitute tar2 in equation . Substitute ta-2 in @ we get na 6-2 = 4 ; 9=3-2=1, z=-1+21-2) = - 5 Therefore the intersecting point is (ny, z) (4, 1,-5).