Find the distance between the two parallel planes 2x-y+z=1 and 2x-y+z=3 for both equations, if we let x and y = 0 then a point that exists on the first plane is (0, 0, 1) and for the second plane you have (0, 0, 3) distance is typically given as: distance= sqrt((x2-x1)^(2)+(y2-y1)^(2)+(z2-z1)^(2)) substituting the points of both planes we get: sqrt((0-0)^(2)+(0-0)^(2)+(3-1)^(2)) = sqrt(2^(2))=2 units, is this the correct way to go about this?
Find the distance between the two parallel planes 2x-y+z=1 and 2x-y+z=3 for both equations, if we let x and y = 0 then a point that exists on the first plane is (0, 0, 1) and for the second plane you have (0, 0, 3) distance is typically given as: distance= sqrt((x2-x1)^(2)+(y2-y1)^(2)+(z2-z1)^(2)) substituting the points of both planes we get: sqrt((0-0)^(2)+(0-0)^(2)+(3-1)^(2)) = sqrt(2^(2))=2 units, is this the correct way to go about this?
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.3: Lines
Problem 31E
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Find the distance between the two parallel planes 2x-y+z=1 and 2x-y+z=3
for both equations, if we let x and y = 0 then a point that exists on the first plane is (0, 0, 1) and for the second plane you have (0, 0, 3) distance is typically given as:
distance= sqrt((x2-x1)^(2)+(y2-y1)^(2)+(z2-z1)^(2))
substituting the points of both planes we get:
sqrt((0-0)^(2)+(0-0)^(2)+(3-1)^(2)) = sqrt(2^(2))=2 units, is this the correct way to go about this?
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