Question

(8 points) Find the points of intersection in R3 of the line L(t) = (3-1, -2+1, 3t) and the unit sphere: x2 + y2 + x2 = 1 (Hint: Use x = 3t - 1, y = -2t + 1 and 2 = 3t in the equation of the sphere, and solve for t.)

Answer 1

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Any Giver line is L(+)= (34-1, -2+ +1,34) A boint (x,y,z) on the line will be (x, y, 2) = (3x), – 22+), 3 t) X-3t-1 y=- 28th 2= 3t Given sphere is x² + y² +2²=) for intersection point, but . and in =) (3x - 1)² + (-22 +13+ (3d) ² = 1 - gt² + 1 - 6t+4+²+1-4 t +9+ ² = 1 (a - b) = a + b²-2abj

- 297²-lot + 2 = 1 => 227 - 10t+1=0 3) t= - (0) IVF-10)2-4(22) (1) ㄴ because at² + b& tC=0 = = -b- Jo u ac da t = lot J5oo - 88 44 3) t- lot vie 44 1 #- 10 123 44 t= 5+ 3

e) - 5+ 3 6 t = 5-3 - 6 put & in a 5+53 ) - 3 - 1 - 15+353-22 - - 7+33 93 2) X- – 7+33 put ③ in y=- 2 ( 5+3 +1 - y = –10-23 +22

12-23 - 2 ( 6-3) y 아B Now 2오 but in 25 3(59) 154 3F - Ⓒ So one point of intersection is (by 강 3, 68 154353 / / ||

Now put in x= 3/5-3 35-9 - 15-33- 22 -7-33 x = -(7+353 but @ in ti Y=-215- 22

- y = -lot 253 +22 y= 1 2+2 33 2 (6+ 3) 22 y = 6+3 | but 2= 3 5 - 3 15-33 - 12 Second boint of intersection is l by 0,0 ) - (7+3v3), 6+3 15-33 22. So boints of intersection are - 7+33 (63) 15+33 and 22 (-74 (-17+353), 6+3 15-33 1 22 TU