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2. (a) With reference to a Cartesian coordinate system 0xyz, find the plane that contains the points (2, 1, 3), (3, 0, 2) and (-1, 1, 4) (Hint. One way of doing this is to find two vectors that lie on the plane and use them to obtain a vector perpendicular to the plane.) (b) A straight line perpendicular to the plane in part (a) passes through the point (-2, 5, -2) Give a parametric representation for the straight line. (c) Find the point where the plane in part (a) and the straight line in part (b) intersect.

2. (a) With reference to a Cartesian coordinate system 0xyz, find the plane that contains the points (2, 1, 3), (3, 0, 2) and (-1, 1, 4) (Hint. One way of doing this is to find two vectors that lie on the plane and use them to obtain a vector perpendicular to the plane.) (b) A straight line perpendicular to the plane in part (a) passes through the point (-2, 5, -2) Give a parametric representation for the straight line. (c) Find the point where the plane in part (a) and the straight line in part (b) intersect.

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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hello, i got the workings already, however i am unsure as to why each step is done, is it possible for you to explain them to me step by step with explaination and wiht drawn illustration. thanks

equation is x-2 y+3z=9 (Check that the other two points also satisfy this equation.) (b) A vector parallel to the straight line is i-2j+3k. Thus, a parametric representation of the straight line is X= -2+t y= 5– 2t Z= -2+3t (c) To find the required point of intersection, substitute the above parametric equations (-2+ 1) – 2(5– 2t) + 3(-2+3t) = 9. We obtain t= 27/14. Thus, the co-ordinates of the point of intersection are given by for the line into the equation of the plane, that is, x= -2+ (27/14) = -1/14 y= 5–2(27/14) = 8/7 z=-2+3(27/14) = 53/14
Transcribed Image Text:equation is x-2 y+3z=9 (Check that the other two points also satisfy this equation.) (b) A vector parallel to the straight line is i-2j+3k. Thus, a parametric representation of the straight line is X= -2+t y= 5– 2t Z= -2+3t (c) To find the required point of intersection, substitute the above parametric equations (-2+ 1) – 2(5– 2t) + 3(-2+3t) = 9. We obtain t= 27/14. Thus, the co-ordinates of the point of intersection are given by for the line into the equation of the plane, that is, x= -2+ (27/14) = -1/14 y= 5–2(27/14) = 8/7 z=-2+3(27/14) = 53/14
2. (a) With reference to a Cartesian coordinate system 0xyz, find the plane that contains the points (2, 1, 3), (3, 0, 2) and (-1, 1, 4) (Hint. One way of doing this is to find two vectors that lie on the plane and use them to obtain a vector perpendicular to the plane.) (b) A straight line perpendicular to the plane in part (a) passes through the point (-2, 5, -2) Give a parametric representation for the straight line. (c) Find the point where the plane in part (a) and the straight line in part (b) intersect. (Solution) (a) The two vectors on the plane are: (3– 2) i+ (0–1) j+ (2– 3)k = i-j-k (3+1) i+ (0–1) j+ (2 – 4)k = 4i– j-2k %3D A cross product of the two vectors gives a normal vector to the plane. We have: (i-j-k)×(4i- j- 2k) = i– 2j+3k) Thus, the equation of the plane may be written as X- 2 y+3z= d. Use one point to work out d, e.g. use (2, 1, 3), d = 2-2+9 = 9. Thus, the required
Transcribed Image Text:2. (a) With reference to a Cartesian coordinate system 0xyz, find the plane that contains the points (2, 1, 3), (3, 0, 2) and (-1, 1, 4) (Hint. One way of doing this is to find two vectors that lie on the plane and use them to obtain a vector perpendicular to the plane.) (b) A straight line perpendicular to the plane in part (a) passes through the point (-2, 5, -2) Give a parametric representation for the straight line. (c) Find the point where the plane in part (a) and the straight line in part (b) intersect. (Solution) (a) The two vectors on the plane are: (3– 2) i+ (0–1) j+ (2– 3)k = i-j-k (3+1) i+ (0–1) j+ (2 – 4)k = 4i– j-2k %3D A cross product of the two vectors gives a normal vector to the plane. We have: (i-j-k)×(4i- j- 2k) = i– 2j+3k) Thus, the equation of the plane may be written as X- 2 y+3z= d. Use one point to work out d, e.g. use (2, 1, 3), d = 2-2+9 = 9. Thus, the required
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