HW3 - {X @233; Q Q r”; 3;, ‘ Q 95% €51 my? $34 we? 5...

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Ag. x} i éfgrgwgsé ;/ a: & «a i»; ,«V £44 a“ {v «A; Q 7% i J KEN} SJ 5% is E? QM .9 r: ,9 h” g , 9. 5 5.. as 2 w , . . g .9 a {w € 5*” a a“ m: in? 9 a?» m g, M W £2 y. {a :fi. 5, < win a fly 3% § 935., my é sg z a fi &., 8 _ Q ,N «HQ 2% 3 «my 5?; Z “12;; E $5 3:323.» as 9: g g .8 A is” (Kw, 9;; we» 4», 3a.. ,5 8 a i 5 .3. w e Z a g5 fig :we 5, 2. 2w é : w? u ; me a: _ aw: ,x v an ad, W a? a ,5, s «new A: :w‘r wank,“ as g g Q»? 5 way. {2 , g; ea My} my V}. g E 13.53 You are given that the two planes are parallel which means, in particular, that if we choose a point P on one plane (say plane 1 with til), then P’s distance to the other plane (plane 2 with d2) will be the same distance as that of any other point on plane I to plane 2. In other words, the fact that the planes are parallel reduces this problem to a question of finding the distance between any point P on plane 1 to plane 2. This is then an already solved problem, whose solution is given on pg. 837. For our case, plane 2 is given by am + by + cz + d2 2 0, and the point P on plane 1 we’ll label as P(a:0, ya, 20). Then by eq. 9 on pg. 837, D: i‘1$t)+b90+020+d217 (1) Va? +b2+02 but then again, since P is presumed on plane 1, it must satisfy the plane 1 equation: auto + byg + czo + d1 2 0. (2) From (2) we can solve for (1170 + byg a» czo 2 wdl, and then plugging this back into (1) yields the answer. 13.5c Let’s label these three points by A 2 (1,1,-—3), B 2 (2, 4, *1), and C 2 (*3, 1, *1), To find the equation of a line in 3~space, we need a point on that line, and a direction vector. To find a point on the line, let’s find a point P(x, 3;, 2) which is equidistant from A, B. C. There are several ways of doing this, but here is the most straightforward. We have the three distances: “A e \/(1' --1)‘2 + (y *1)? ~ (2 + 3)2 (3) dz: : \/(l‘ - 2)2 + (y 2 4)2 ~ (2 + 1)2 (4) dc: : x/(LL‘ + 3)2 + (y -— 1)2 — (z + 1):2 (5) and we express the equidistant condition by setting dA 2 dB 2 dc. From (1/; 2 d3, d3 2 rig, and (1A 2 dc you get, respectively, 22: + 6y e- 42' 2 10, (6) 100:. M 69 2 10, (7) 8.1", w 42 2 0. (8) These equations actually give us a bit more than we need. However, one can easily see that the point (0. 5/ 3, 0) satisfies (6)482; and is thus a point on our fine Now we need to find the direction vector, We can do this by forming the vectors AB and AC and taking their cross product. This yields the direction vector v 2 (6, 210, 12). Thus, the equation of the line is r(t) : (0,0) +/:(6,-—10,12). (9) 13.73 We can rewrite the equation of the sphere in standard form by completing the square. This yields {152 + y2 + (3 2 4)? 2 16, showing that this is a sphere of radius 4 and with center (0, 0, ln spherical coordinates, this equation is p 2 8cos Now, if we set :1: O in the equation of the cone, then we see that the resulting equation, 2 2 V’lel, describes two lines in the airplane These two lines form an angle 0 with respect to the z~axis, and one can find this angle by using tan 2 l/tfi. Define A 2arctan (lit/’77) (which ends up being a: 20.70 or 0.361 radians). Now, since the two surfaces (the sphere and cone) intersect, and the solid is the region contained inside the intersection, we now know that the description is: ngggcoséflgcpgA,0§(}€27r, 13.7b-a This surface is similar to the surface above. It looks just like an ice cream cone filled with a spherical scoop of ice cream. l3.7b~f This surface is what we discussed in lab, The level curves look roughly like cylindrical wedges, and the final surface iooks something like the Colosseum ef ancient Rome, 13.7b—g This equation represents a plane, Wmm(mmwmw _ 14.2a) ?.2 5.2 2 3.2 ...
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HW3 - {X @233; Q Q r”; 3;, ‘ Q 95% €51 my? $34 we? 5...

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