Question

uestion 6 Let a,b,c be the last digits of your student ID. Find the symmetric equation of the line that passes through the point (a+b+c, a+b+c,a+b+c) and the point of intersection between the lines x= -a +at.y = -b+bt.z=(a+b+c+1) and x = a +as, y =b+bs.2= -s+a+b+c Attach File Browse My Computer Browse Content Collection

where a=4 b=5 c=8

Answer 1

(6) Here a X z=18€ And x = a +as, y=6+65, Z=-5+2+b+e general point on il., x=4+45, y=5+58, Z=-5+17 of intersection (0o, 18) is. Here Q = 4, b= 5, e=8 at btc - 4 +5+8=17 1 = (17,17,177) ..16+b+e, 9+67e, aubte) Given lines are : Qual, y=-b+bt, 2 =(a+b+e+132 lil, x=-4+4+, y = -5+5+, Now, the coordinates of a first line are (-4+4+, -5+5+, 18t). And the coordinates of a general point on first line are (4+45, 5+55,-5+17) If the lines intersect, then they have a common point. so, for some values of t and so we must have - 4+4t = 4+45, 5+55= -5+54 and 187 =-5+17 7 t-s=2 t-s=2 and 18t+ s =17 18t+s=17 => 18 (2+3)+5=17 5) 36 +185+5=17 195=-19 S = -1 ait- 2-1=1 secting point is (-4+4.1, -5+5.1,18.1) =(0,9,18) : the symmetric equation of the line that passes through the point (17, 17, 17 ) and the poine t = 2+s & the intenso Z-17 7-17 Y-17 -17 cie, 2-17 18-17 x-17 -17 2-17 0-17 0-17