Question

2. Find distance from point S(-2, 3, 4) to the line x = 3 - 2t, y = –2 + 3t, 2 = 5 - 6t Write plane equation passing through point S and par- allel to the given line. Show calculation steps clear and cleanly.

Answer 1

Use :- Answer :- here point sis (-2,3,4) 31= 3-2t, g=-2 + 3t, z=5-6t line can also written as :- A-3 y+2 z-5 - 6 at Direction ratios o of line is (-2,3-6) of Distance between line and print SL-2,3, 4) As clearly Coordinates A tam be written in terms of para arameter 4. A[3-24, 3-24, -2+35,5-877 Now, Direction of 3A 19-21-1-2), -213t-3, 5-67-4 5-2t, -5+3+, 1-67] wil be. [ As, SA and line perpendicular to product of their DRs will be 5-2t) x -2 + (5+3+) x3 +11-6t)x -6 CS. Scan 0+ 4+ - 15+9+ -6 +36 4 =0 CamScanner

49xt = 31 5 t = 31 49 so A- 2+3x 31 5-6431 49 49 3-2x31 49 1147-62 49 147-186- -98 +93 49 77136] فا 9 A 85 49² 1 49 59 49 5 ] Estos. Distance between two points is given by Je, -22) 24 (% Y.)? (2,-2,² १ So, between A 2 S that is t Length of SA | SA) = 105-12] + [3] +63 4] 2 134 49 + t 49 a137 49 z 598 29 494 49 f 59829 49. IS AL 59829 49 cs Scanned w30 CamScanner

Now equation of plane can be obtained as s(-2,3,4) Play, el foc here let p be any arbitrary point on plane. A[85 so. Direction ratios of pe is 490 4999 รา De-L2), Y-3, 2-4) = (*+2,4-3, 2-4) Now as in figure clearly. Ps and As es perpendicular to each other. So produal their direction Ratios will be zerb of DRs of SA (5-2t, 15+3t, kot) by eart by putting to So, DRs 49 **39 1-6831 - 5-2x31 a5+3431 49 59 152 137 49 49 49 9 So, product of [ps x As? To (32+2) *** +(9-3)*152 +(2-4) x 137 CS =) 49 Cams 49

59 (x+2) + 152(4-3) + 7 4 137 (2-4) 590 + 1524 +137 2 42x59-34152-4413770 5981+ 1524 +137 Z -889 = 0 so, this is required equation of plane.