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Equation of line parallel to plane and intersaction with other line

  • Thread starter Theofilius
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  • #1
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Homework Statement



Hello!

I have one problem which seems not so difficult:

-Find the equation of line which passes throught the point M(1,0,7), parallel of the plane 3x-y+2z-15=0 and it intersects the line [tex]\frac{x-1}{4}=\frac{y-3}{2}=\frac{z}{1}[/tex]

Homework Equations





The Attempt at a Solution



The equation of the line will be: [tex]\frac{x-1}{a_1}=\frac{y}{a_2}=\frac{z-7}{a_3}[/tex]

So we need to find [tex]\vec{a}(a_1,a_2,a_3)[/tex] and we need three conditions in the system.

The first condition is [tex]\vec{a} \circ \vec{n}=0[/tex] or [tex](a_1,a_2,a_3)(3,-1,2)=0[/tex] or [tex]3a_1-a_2+2a_3=0[/tex].

The second condition is the intersection of two lines, and it is:

[tex]-17a_1+28a_2+12a_3=0[/tex]

What about the third condition?
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
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There is no third condition. You cannot determine a1, a2, and a3 uniquely. Any multiple of a given a1, a2, and a3 will also determine the same line. From the two equations you have you can solve for two of the numbers as functions of the third. Choose that third as you please, say equal to 1, and solve for the other 2.
 
  • #3
378
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Other way (I think it is very similar to what you did):
r = (1,3,0) + t (4,2,1)
P = (1,0,7)
find direction vector: r - P
and you know n.(r-P) = 0

So, only one condition .. (and only one unknown)
 
  • #4
rootX what is r, and what is P?
 
  • #5
378
2
rootX what is r, and what is P?
r is (1,3,0) + t (4,2,1) ... a line
P is (1,0,7) .. a point

:smile:

Should work, shouldn't it?
 
  • #6
It wouldn't work, since by your opinion you will find point, product of the intersection of line and plane... And we'll need to find parallel vector to the line which satisfies the above conditions...
 

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