Show that there are no points (x, y, z) satisfying 2x - 3y + z - 6 = 0 and lying on the line v = (8, -8, -4) + t(1, 1, 1). If (x, y, z) lies on the line v, then x = y = and z = . Therefore, substituting these values in for x, y, and z, 2x - 3y + z - 6% D However, we are given that 2x - 3y + z - 6 = 0, and since this does not match, there are no points (x, y, z) lying on the line v.

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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Show that there are no points (x, y, z) satisfying 2x
Зу + z - 6
= 0 and lying on the line v =
(8, -8, -4) + t(1, 1, 1).
If (x, y, z) lies on the line v, then x =
and z =
- Therefore, substituting these values in for x, y, and z,
2х — Зу + Z - 6 %3
However, we are given that 2x – 3y + z - 6 = 0, and since this does not
match, there are no points (x, y, z) lying
on the line v.
Transcribed Image Text:Show that there are no points (x, y, z) satisfying 2x Зу + z - 6 = 0 and lying on the line v = (8, -8, -4) + t(1, 1, 1). If (x, y, z) lies on the line v, then x = and z = - Therefore, substituting these values in for x, y, and z, 2х — Зу + Z - 6 %3 However, we are given that 2x – 3y + z - 6 = 0, and since this does not match, there are no points (x, y, z) lying on the line v.
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