1. A line passes through the points M(0, 1, 4) and N(1, 4, 5). Find a vector equation of the line. [x, y, z] = [0, 1, 4] + i[1, 4, 5], t eR [x, y, z] = [1, 3, 1] + :[0, 1, 4), t e R [x, y, z] = [1, 3, 1] + i[1, 4, 5], t eR [x, y, z] = [0, 1, 4] + :[1, 3, 1], t eR а. с. %3D %3D b. d. %3D

Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter1: Vectors
Section1.3: Lines And Planes
Problem 18EQ
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1. A line passes through the points M(0, 1, 4) and N(1, 4, 5). Find a vector equation of the line.
[x, y, z] = [0, 1, 4] + [1, 4, 5], t e R
[x, y, z] = [1, 3, 1] + {[1, 4, 5], t eR
а.
с.
%3D
b.
[x, y, z] = [1, 3, 1] + t[0, 1, 4], t eR
d.
[x, y, z] = [0, 1, 4] + [1, 3, 1], t eR
2. A line has x-intercept 7 and z-intercept 5. Find a vector equation of the line.
[x, y, z] = [0, 0, 5] + ¿[-7, 0, 0], t e R
[x, y, z] = [-7, 0, 5] + [0, 0, 5], t e R
а.
с.
b.
[x, y, z] = [0, 0, 5] + ¿[-7, 0, 5], t e R d.
[x, y, z] = [-7, 0, 5] + [7, 0, 0], t eR
3. Write the scalar equation of the plane with normal vector = [0, –1, 0] and having a y-intercept 3.
y+ 3 = 0
X+Z+3 = 0
a.
с.
b.
y- 3 = 0
d.
Зу - 3 - 0
4. The vector equation of a plane is [x, y, z] = [1, 1, 1] + s[2, 1, 3] + i[-3, 2, 4], s, t E R. Find a scalar
equation of the plane.
2x +y+ 3z + 3= 0
с.
2x + 17y – 7z – 12 = 0
а.
%3D
b.
2х + 17у — 7z+ 12 3D 0
d.
x + y + z + 10 = 0
Transcribed Image Text:1. A line passes through the points M(0, 1, 4) and N(1, 4, 5). Find a vector equation of the line. [x, y, z] = [0, 1, 4] + [1, 4, 5], t e R [x, y, z] = [1, 3, 1] + {[1, 4, 5], t eR а. с. %3D b. [x, y, z] = [1, 3, 1] + t[0, 1, 4], t eR d. [x, y, z] = [0, 1, 4] + [1, 3, 1], t eR 2. A line has x-intercept 7 and z-intercept 5. Find a vector equation of the line. [x, y, z] = [0, 0, 5] + ¿[-7, 0, 0], t e R [x, y, z] = [-7, 0, 5] + [0, 0, 5], t e R а. с. b. [x, y, z] = [0, 0, 5] + ¿[-7, 0, 5], t e R d. [x, y, z] = [-7, 0, 5] + [7, 0, 0], t eR 3. Write the scalar equation of the plane with normal vector = [0, –1, 0] and having a y-intercept 3. y+ 3 = 0 X+Z+3 = 0 a. с. b. y- 3 = 0 d. Зу - 3 - 0 4. The vector equation of a plane is [x, y, z] = [1, 1, 1] + s[2, 1, 3] + i[-3, 2, 4], s, t E R. Find a scalar equation of the plane. 2x +y+ 3z + 3= 0 с. 2x + 17y – 7z – 12 = 0 а. %3D b. 2х + 17у — 7z+ 12 3D 0 d. x + y + z + 10 = 0
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