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Qestion 2 a). Use vector projection to explain the distance in part (iii) equation b). find the distance between point (2,4,8) and the plane 2x+y+z=5.

Qestion 2 a). Use vector projection to explain the distance in part (iii) equation b). find the distance between point (2,4,8) and the plane 2x+y+z=5.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Solve part B
P (ii) (1) Distance between į →d= (iii) Topic: P Distance S. two points in IR ³: distane is the length of PS or (x₂-x₁)² + (Y₂-Y₂)² + (7₁-7₁) ² S -57 Distance from a point S to a line Containing P Parallel to V. d = | PS x V | |VI Distance from with normal vector n P= (x₁, y₁, Z₁) S = (x₂, Y₂, 7₂) point S to a plane Containy P egn of plane: 4(x-x)+ Bly-yo) + c(Z - 2₂) = 0 n = < A, B, C distance = from S to plane is I distance IPS.1 E
Transcribed Image Text:P (ii) (1) Distance between į →d= (iii) Topic: P Distance S. two points in IR ³: distane is the length of PS or (x₂-x₁)² + (Y₂-Y₂)² + (7₁-7₁) ² S -57 Distance from a point S to a line Containing P Parallel to V. d = | PS x V | |VI Distance from with normal vector n P= (x₁, y₁, Z₁) S = (x₂, Y₂, 7₂) point S to a plane Containy P egn of plane: 4(x-x)+ Bly-yo) + c(Z - 2₂) = 0 n = < A, B, C distance = from S to plane is I distance IPS.1 E
Pestion 2 a). Use vector projection to explain the distance in part (iii) equation b). find the distance between point (2,4,8) and the plane 2x+y+z=5.
Transcribed Image Text:Pestion 2 a). Use vector projection to explain the distance in part (iii) equation b). find the distance between point (2,4,8) and the plane 2x+y+z=5.
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