Calc III. Consider the line which passes through the point P(-3,1,-2), and which is parallel to the line x=1+1t,y=2+5t,z=3+2t find the point of intersection of this new line with each of the coordinate planes: xy-plane:(x,y,z) xz-plane:(x,y,z) yz-plane: (x,y,z)

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Calc III. Consider the line which passes through the point P(-3,1,-2), and which is parallel to the line x=1+1t,y=2+5t,z=3+2t find the point of intersection of this new line with each of the coordinate planes: xy-plane:(x,y,z) xz-plane:(x,y,z) yz-plane: (x,y,z)

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As a start, write out the equation of the line in parametric form
<1,2,3>+t<1,5,2> would be the parametric form of the equation right?
`<1,2,3>+t<1,5,2> would be the parametric form of the equation right?` looks good now find the parallel line

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How would I go about that exactly?
btw it's a vector equation not parametric http://tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfLines.aspx
well you know how the line is pointing so to speak based on this vector `<1,5,2>` after the t
all parallel lines to this line will also point in the same direction
r = <1,2,3>+t<1,5,2> anything parallel to line r will be in the form s = P + t<1,5,2> where P is any point you want s to run through
you can think of "<1,5,2>" as the "slope" of the line it's not really the slope, but it plays a similar role
is s just a another name for the parallel line
I just made up the label
imagine we're in R2 and we have r = <1,1> + t*<3,4> so we would start at (1,1) and go up 4, to the right 3 |dw:1444017803516:dw|
anything parallel to line r in R2 is going to have the same "slope" so we're going to go up 4, to the right 3 if we want to go through say (2,-3), then we just start there and move along following the vector <3,4> |dw:1444017996477:dw|

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