1. A line passes through the points A (7,-4) & B (-5, -2). Find a vector equation of the line. [x.y] = [7.-4] + t[-5,-2] b) [x.y] = [7.-4] +t[−2,−6] c) [x.y] = [-5.-2] + [-12,2] d) [x.y] = [12.-2] + t[7,4]
1. A line passes through the points A (7,-4) & B (-5, -2). Find a vector equation of the line. [x.y] = [7.-4] + t[-5,-2] b) [x.y] = [7.-4] +t[−2,−6] c) [x.y] = [-5.-2] + [-12,2] d) [x.y] = [12.-2] + t[7,4]
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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![1. A line passes through the points A (7,-4) & B (-5, -2). Find a vector equation of the line.
[x.y] = [7,-4] + t[−5,−2]
b) [x.y] = [7,-4] + t[-2,-6]
c) [x.y] = [-5, -2] + t[-12,2]
d) [x.y] = [12,-2] + t[7,4]
2. A line has slope-3 and x-intercept 5. Find a vector equation of the line.
a) [x.y] = [-3, 1] + t[5,0]
b) [x.y] = [0,5] + t[−3,1]
c) [x.y] = [5,0] + t[-1,3]
d) [x.y] = [-1, -3] + t[0,5]
3. Write the scalar equation of the plane with normal vector n = [3,2,1] and passing
through the point (4, 5, 6).
28 = 0
a 3x + 2y + z
b) 4x + 5y + 6z+ 28 = 0
c) 3x +2y+z+28 = 0
d) 4x + 5y + 6z - 28 = 0
4. A plane passes through the origin and has the direction vectors [-1, -2, -3] and [-1, 3, -2].
Find a scalar equation of the plane.
a) -x - 2y3z = 0
b) 13x + y - 5z = 0
c) -x + 3y2z = 0
d) 5y + z = 0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5e9b6dff-d19a-42af-bc37-34d630a6af1a%2Fcb6bb745-777d-4e51-9deb-8bc0e410ceef%2Fuhnealg_processed.png&w=3840&q=75)
Transcribed Image Text:1. A line passes through the points A (7,-4) & B (-5, -2). Find a vector equation of the line.
[x.y] = [7,-4] + t[−5,−2]
b) [x.y] = [7,-4] + t[-2,-6]
c) [x.y] = [-5, -2] + t[-12,2]
d) [x.y] = [12,-2] + t[7,4]
2. A line has slope-3 and x-intercept 5. Find a vector equation of the line.
a) [x.y] = [-3, 1] + t[5,0]
b) [x.y] = [0,5] + t[−3,1]
c) [x.y] = [5,0] + t[-1,3]
d) [x.y] = [-1, -3] + t[0,5]
3. Write the scalar equation of the plane with normal vector n = [3,2,1] and passing
through the point (4, 5, 6).
28 = 0
a 3x + 2y + z
b) 4x + 5y + 6z+ 28 = 0
c) 3x +2y+z+28 = 0
d) 4x + 5y + 6z - 28 = 0
4. A plane passes through the origin and has the direction vectors [-1, -2, -3] and [-1, 3, -2].
Find a scalar equation of the plane.
a) -x - 2y3z = 0
b) 13x + y - 5z = 0
c) -x + 3y2z = 0
d) 5y + z = 0
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