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Ex2 :- given P(6,125°, -3), Q(3,-1,4) find the distance from :a- P to the origin b-Q perpendicular to the z-axis c- P to the Q Solution:-

Ex2 :- given P(6,125°, -3), Q(3,-1,4) find the distance from :a- P to the origin b-Q perpendicular to the z-axis c- P to the Q Solution:-

Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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Ex2 :- given P(6,125°,-3), Q(3,-1,4) find the distance from :a- P to the origin b-Q perpendicular to the z-axis c- P to the Q Solution:- • Q(3, -1, 4) (0, 0, 4) P (6, 125 a- P(p, Ø, z) X-pcosø = 6cos125 = -3.44 y=psinø = 6sin125 = 4.915 ale z=-3 V(-3.44)2 + (4.915)? + (3)² = v11.84 + 24.15 + 9 = 6.71 b- p = x2 + y2 = 9+1= 3.16 c- FPQ = (3 +3.44)āx + (-1– 4.915)āy + (4 + 3)āz =6.44āx – 5.915āy + 7āz : rPQ = V41.47 + 34.98 + 49 = 11.2 I did not understand the point (b) why he extracted (p)which is mentioned in the question the point perpendicular to the z axis
Transcribed Image Text:Ex2 :- given P(6,125°,-3), Q(3,-1,4) find the distance from :a- P to the origin b-Q perpendicular to the z-axis c- P to the Q Solution:- • Q(3, -1, 4) (0, 0, 4) P (6, 125 a- P(p, Ø, z) X-pcosø = 6cos125 = -3.44 y=psinø = 6sin125 = 4.915 ale z=-3 V(-3.44)2 + (4.915)? + (3)² = v11.84 + 24.15 + 9 = 6.71 b- p = x2 + y2 = 9+1= 3.16 c- FPQ = (3 +3.44)āx + (-1– 4.915)āy + (4 + 3)āz =6.44āx – 5.915āy + 7āz : rPQ = V41.47 + 34.98 + 49 = 11.2 I did not understand the point (b) why he extracted (p)which is mentioned in the question the point perpendicular to the z axis
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