Equation form of a line in 3-D

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SUMMARY

The equation of the line L that is perpendicular to the plane Q defined by -x + 3y + 2z = 1 and passes through the origin is derived using the parametric form of a line. The line can be expressed as L = (0,0,0) + s(-1,3,2), leading to the equations x = -s, y = 3s, and z = 2s. By eliminating the parameter s, the relationship between x, y, and z can be established as x + y + z = 4s, confirming that for different values of s, different points on the line are obtained.

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Homework Statement



Fine the equation form of the line L that is perpendicular to the plane Q with equation
-x+3y+2z=1 and passing through the origin.

Homework Equations



Parametric form of a line:
L=(0,0,0)+s(-1,3,2)
so x/-1=y/3=z/2

The Attempt at a Solution


x+y+z=4s
but then what is the value of s?
 
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s is just a parameter, for different values of s you will get different points on the line.

Remember L is also (x,y,z) so get s in terms of each x,y and z and they will all be equal to one another.
 
rock.freak667 said:
s is just a parameter, for different values of s you will get different points on the line.

Remember L is also (x,y,z) so get s in terms of each x,y and z and they will all be equal to one another.

So will the equation form of this line just be x+y+z=4s?
 
yvonnars said:
So will the equation form of this line just be x+y+z=4s?

No, from L=(0,0,0)+s(-1,3,2) = (0-s,0+3s,0+2s)

L = (x,y,z) = (0-s,0+3s,0+2s)

x = 0-s and so on. Just make s the subject for each one and then put them all equal to each other.
 

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