Understanding Equations in Space: Lines & Planes

[bobey]

Homework Statement

describe the set of points in space whose coordinate satisfy the given equation or pair of equations:

(i) z=2y (ii) 3x=4y, z=1

if (i) or (ii) represents a line in space, give a unit vector that is parallel to the line. If (i) or (ii) represents a plane, give a unit vector that is normal to the plane.

Homework Equations

The Attempt at a Solution

z=2y is a line parallel to x-axis consisting of all points of the form (0,z,z)

3x=4y, z=1 is a plane perpendicular to the z=1 axis consisting of all points of the form (3x,3x,1)

(i) is a line, thus the unit vector that is parallel to the line is v/|v|= (0i-2j+k)/5

(ii) is a plane, thus a unit vector that is normal to the plane is v/|v|= (3i-4j+k)/26

is my answer is correct? pls help me... tq

 

[Mark44]

Homework Statement

describe the set of points in space whose coordinate satisfy the given equation or pair of equations:

(i) z=2y (ii) 3x=4y, z=1

if (i) or (ii) represents a line in space, give a unit vector that is parallel to the line. If (i) or (ii) represents a plane, give a unit vector that is normal to the plane.

Homework Equations

The Attempt at a Solution

z=2y is a line parallel to x-axis consisting of all points of the form (0,z,z)

No and no. This is not a line, and the form is not (0, z, z).

3x=4y, z=1 is a plane perpendicular to the z=1 axis consisting of all points of the form (3x,3x,1)

Each of the equations 3x = 4y and z = 1 is a plane, so together they do not determine a plane. There is no z = 1 axis.

(i) is a line, thus the unit vector that is parallel to the line is v/|v|= (0i-2j+k)/5

No, it's not a line.

(ii) is a plane, thus a unit vector that is normal to the plane is v/|v|= (3i-4j+k)/26

No, it's not a plane.

is my answer is correct? pls help me... tq

 

[HallsofIvy]

R³ as "three degrees of freedom"- you are free to choose any numbers for x, y, and z you wish. Adding one equation restricts that by "one degree of freedom" reducing from three to two degrees of freedom: You could choose any values for two of x, y, or z, then use the equation to solve for the third. If you have two equations, that reduces from three to one degree of freedom- with two equations, choose whatever value for one of x, y, or z you wish, leaving you with two equations in two unknowns so that you can solve for those values.

The upshot of all that is that a single equation in R³ gives you a surface (and if the equation is linear, a plane) while two equations gives you a curve (if the equations are linear, a line). That is, you have (a) and (b) reversed. Since (a) has one linear equation, it is a plane. Since (b) has two linear equations, it is a line.

If a plane is given by an equation like Ax+ By+ Cz= D, then [itex]A\vec{i}+ B\vec{j}+ C\vec{k}[/tex] is normal to the plane. (a) has equation z= 2y which is the same as (0)x+ (-2)y+ (1)z= 0.

If a line is given by two equations, you could interpret them as equations of planes (so your line is the intersection of the two planes), find vectors normal to them and take the cross product which will be a vector normal to both normal vectors, so in both planes, so in the direction of the line of intersection.

Another way to do that is to use the two equations to write two of the coordinates as linear functions of the third. If, for example, the equations are 3x= 9y, z= 2y, you can write x= 3y, z= 2y so that a vector form of the line is $x\vec{i}+ y\vec{j}+ z\vec{k}= 3y\vec{i}+ y\vec{j}+ 2y\vec{k}= y(3\vec{i}+ \vec{j}+ 2\vec{k})$. There "y" is the parameter and the vector $3\vec{i}+ \vec{j}+ \vec{k}$ is a vector in the direction of the plane.