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sachin

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- Homework Statement
- Can (y+z)x1=(y1+z1)x be an equation of a straight line in 3 dimensional space?

- Relevant Equations
- (x-x1)/a = (y-y1)/b = (z-z1)/c

Can we say,

(y + z ) x1 = (y1 + z1) x is also an equation of a straight line in 3 dimensional space,

where (x1,y1,z1) and (x,y,z) are the coordinates of a given point and a variable point respectively on a 3D line that passes through the origin,

have seen equation of a straight line in 3 dimensional space as,

(x-x1)/a = (y-y1)/b = (z-z1)/c

where (x1,y1,z1) is a known point in the line and a,b,c are the direction ratios (drs) of the line,

I am trying to modify the equation into another form which one will look like the equation of a line in a 2D case like ax + by + c = 0,

My approach,

Let (x1,y1,z1) and (x,y,z) be the coordinates of any given and a variable point respectively on a 3D line that passes through the origin, as direction ratios are just the length of the bases of the triangles formed in the space,

a = x1, b = y1 , c = z1,

now, instead of working on direction ratios measured by the cosine functions, I am taking the tan functions and using tan θ = opposite side/ base, along the y axis,

tan β = y1/x1 = y/x,

along the z axis,

tan γ = z1/x1 = z/x,

where β and γ are the angles made by the line with y and z axes respectively, so,we are getting,

y = (y1/x1) x,

z = (z1/x1) x,

y + z = (y1/x1) x + (z1/x1) x,

y + z = (y1+ z1) x/x1,

(y + z ) x1 = (y1 + z1 ) x,

which resembles the equation of a 2D line passing through the origin ax + by = 0,

my approach is to simply the equation of a 3D line into simple linear equation of three variables,

also in books it is given ax + by + cz + d = 0 is not the equation of a straight line but is the equation of a plane what I am finding is not the case as there a,b,c are not related to the plane but on the properties of a line what is perpendicular to the plane,

so I can conclude that ax + by + cz + d = 0 will represent a straight line in space but with the above modification what I got as (y + z ) x1 = (y1 + z1 ) x.

Just wanted to know if my approach is correct ?

thanks.

(y + z ) x1 = (y1 + z1) x is also an equation of a straight line in 3 dimensional space,

where (x1,y1,z1) and (x,y,z) are the coordinates of a given point and a variable point respectively on a 3D line that passes through the origin,

have seen equation of a straight line in 3 dimensional space as,

(x-x1)/a = (y-y1)/b = (z-z1)/c

where (x1,y1,z1) is a known point in the line and a,b,c are the direction ratios (drs) of the line,

I am trying to modify the equation into another form which one will look like the equation of a line in a 2D case like ax + by + c = 0,

My approach,

Let (x1,y1,z1) and (x,y,z) be the coordinates of any given and a variable point respectively on a 3D line that passes through the origin, as direction ratios are just the length of the bases of the triangles formed in the space,

a = x1, b = y1 , c = z1,

now, instead of working on direction ratios measured by the cosine functions, I am taking the tan functions and using tan θ = opposite side/ base, along the y axis,

tan β = y1/x1 = y/x,

along the z axis,

tan γ = z1/x1 = z/x,

where β and γ are the angles made by the line with y and z axes respectively, so,we are getting,

y = (y1/x1) x,

z = (z1/x1) x,

y + z = (y1/x1) x + (z1/x1) x,

y + z = (y1+ z1) x/x1,

(y + z ) x1 = (y1 + z1 ) x,

which resembles the equation of a 2D line passing through the origin ax + by = 0,

my approach is to simply the equation of a 3D line into simple linear equation of three variables,

also in books it is given ax + by + cz + d = 0 is not the equation of a straight line but is the equation of a plane what I am finding is not the case as there a,b,c are not related to the plane but on the properties of a line what is perpendicular to the plane,

so I can conclude that ax + by + cz + d = 0 will represent a straight line in space but with the above modification what I got as (y + z ) x1 = (y1 + z1 ) x.

Just wanted to know if my approach is correct ?

thanks.

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