Line of intersection of two planes

AI Thread Summary
The discussion focuses on finding the equation of the line of intersection between two planes defined by the equations r.(3i-j+k)=1 and r.(i+4j-2k)=2. Participants emphasize the need to determine a point on the line and a direction vector, which can be obtained by taking the cross product of the normal vectors of the planes. They suggest choosing a variable arbitrarily, such as x=t, to express y and z in terms of t, leading to a parametric equation for the line. The method of setting one variable to zero for simplification is also discussed. Overall, the conversation clarifies the approach to solving the problem effectively.
Krushnaraj Pandya
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Homework Statement


Find equation of line of intersection of planes r.(3i-j+k)=1 and r.((i+4j-2k)=2

Homework Equations


a x b gives a perpendicular vector to a and b...(i)

The Attempt at a Solution


to write equation of a line r=a+tb we need a point a on the line and a parallel vector b. Taking the cross product of 3i-j+k and i+4j-2k gives us b, assuming r to be xi+yj+zk, we get 3x-y+z=1 and x+4y-2z=2, putting...2 equations in three variables, how do we end up finding a and is there a faster method?
 
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You can pick any ## x ## you like, and then solve for ## y ## and ## z ## to get the point ## a ##.
 
Charles Link said:
You can pick any ## x ## you like, and then solve for ## y ## and ## z ## to get the point ## a ##.
So my method is correct? That's basically what I did but I picked z=0 and found x and y
 
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Charles Link said:
You can pick any ## x ## you like, and then solve for ## y ## and ## z ## to get the point ## a ##.
so I got 14/39 , 5/13, 0 as a. r=a+t(-2i+7j+13k) is this a correct final answer
 
Krushnaraj Pandya said:

Homework Statement


Find equation of line of intersection of planes r.(3i-j+k)=1 and r.((i+4j-2k)=2

Homework Equations


a x b gives a perpendicular vector to a and b...(i)

The Attempt at a Solution


to write equation of a line r=a+tb we need a point a on the line and a parallel vector b. Taking the cross product of 3i-j+k and i+4j-2k gives us b, assuming r to be xi+yj+zk, we get 3x-y+z=1 and x+4y-2z=2, putting...2 equations in three variables, how do we end up finding a and is there a faster method?
As you wrote, you have two equations 3x-y+z=1 and x+4y-2z=2 and want the solution.You can choose one variable arbitrarily, as parameter. Say, x=t. Write y and z in terms of t and you have the parametric equation of the line of intersection.
 
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ehild said:
As you wrote, you have two equations 3x-y+z=1 and x+4y-2z=2 and want the solution.You can choose one variable arbitrarily, as parameter. Say, x=t. Write y ynd z in terms of t and you have the parametric equation of the line of intersection.
Seems like I'm on the right track. Thank you :D
 
Krushnaraj Pandya said:

Homework Statement


Find equation of line of intersection of planes r.(3i-j+k)=1 and r.((i+4j-2k)=2

Homework Equations


a x b gives a perpendicular vector to a and b...(i)

The Attempt at a Solution


to write equation of a line r=a+tb we need a point a on the line and a parallel vector b. Taking the cross product of 3i-j+k and i+4j-2k gives us b, assuming r to be xi+yj+zk, we get 3x-y+z=1 and x+4y-2z=2, putting...2 equations in three variables, how do we end up finding a and is there a faster method?
For equations:
##3x-y+z=1##
##x+4y-2z=2##​
Set ##\ x = 0\,, ## then divide the second equation by 2 & add the equations.
 
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SammyS said:
For equations:
##3x-y+z=1##
##x+4y-2z=2##​
Set ##\ x = 0\,, ## then divide the second equation by 2 & add the equations.
Got it! Thank you :D
 
Just to add on to what I think has already been said, the solution for a line is often written in the form ## \frac{x-x_1}{A}=\frac{y-y_1}{B}=\frac{z-z_1}{C} ## which can be set equal to ## t ##. If you pick ## x_1=0 ## it simplifies the first term. You can even multiply out the ## A ## and then set ## x=t ## as @ehild suggested.
 
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Charles Link said:
Just to add on to what I think has already been said, the solution for a line is often written in the form ## \frac{x-x_1}{A}=\frac{y-y_1}{B}=\frac{z-z_1}{C} ## which can be set equal to ## t ##. If you pick ## x_1=0 ## it simplifies the first term. You can even multiply out the ## A ## and then set ## x=t ## as @ehild suggested.
Crystal clear now! Thank you very much :D
 
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