# Finding the length of an intersection Of 2 planes

## Homework Statement

Line of intersection between P1: x+y+z=7 and P2:2x-3y-z=-8 crosses the XZ plane at point A and crosses the YZ plane at point B

Find the length Of AB

Okay so first of all im having trouble with understanding crossing the XZplane or YZplane.

does this mean that A x=z=0 A(0,something,0) and B(something,0,0)

What im thinking to do with this problem, is finding the points A and B, an then finding vector AB and finding the magnitude, and that should be the length.

So the first thing i did was get the equation of the line..

1 x+y+z=7
2 2x-3y-z=-8

subtract eq1-eq2

-x+4y+0 = 15

x= 4y-15 (3)

sub (3) in 1

4y-15 +y + z =7
z= 22-4y
y=t

x=-15 +4t
y=t
z=22-4t

r=(-15,0,22) + t(4,1,-5) (equation of line of intersection)

so that means that the direction vector is paralell to the line segment AB

but now im clueless as to how to solve for A or for B... can some one give me a hintÉ

HallsofIvy
Homework Helper

## Homework Statement

Line of intersection between P1: x+y+z=7 and P2:2x-3y-z=-8 crosses the XZ plane at point A and crosses the YZ plane at point B

Find the length Of AB

Okay so first of all im having trouble with understanding crossing the XZplane or YZplane.

does this mean that A x=z=0 A(0,something,0) and B(something,0,0)
No, the "xz plane" has points with all values for x and z but y= 0. And the "yz plane has x= 0

What im thinking to do with this problem, is finding the points A and B, an then finding vector AB and finding the magnitude, and that should be the length.
Yep, good plan!

So the first thing i did was get the equation of the line..

1 x+y+z=7
2 2x-3y-z=-8

subtract eq1-eq2

-x+4y+0 = 15
No, z- (-z)= 2z not 0. To eliminate z, add the two equations.

x= 4y-15 (3)

sub (3) in 1

4y-15 +y + z =7
z= 22-4y
y=t

x=-15 +4t
y=t
z=22-4t

r=(-15,0,22) + t(4,1,-5) (equation of line of intersection)

so that means that the direction vector is paralell to the line segment AB

but now im clueless as to how to solve for A or for B... can some one give me a hintÉ
Again, once you have found the correct equation for the line of intersection, to find A, set y= 0 to solve for the value of the parameter and the find x and y. To find B, set x= 0, solve for the parameter and use that to find y and z.

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Again, once you have found the correct equation for the line of intersection, to find A, set y= 0 to solve for the value of the parameter and the find x and y. To find B, set x= 0, solve for the parameter and use that to find y and z.

AH THATS GENIUS!!!! Thank you kindly sir all of your help is much appreciated, and it really means a lot to me that there are people like you in the world who are willing to spread the knowledge of math. It`s quite beautiful actually.

Again, once you have found the correct equation for the line of intersection, to find A, set y= 0 to solve for the value of the parameter and the find x and y. To find B, set x= 0, solve for the parameter and use that to find y and z.

Okay so

EQ1 + EQ2
3x-2y + 0 =-1

(3) x=(2y-1)/3

plug 3 in eq1

(2Y-1)/3 + Y + Z =7
z=(22-5y)/3

let y=t

x=(2t-1)/3
y=t
z=(22-5t)/3

r= (-1/3,0,22/3) + t(2/3,1,-5/3)

so therefore for point A

y=0

therefore t=0

x=-1/3
z=22/3

A(-1/3,0,22/3)

for point B

0=(2t-1)/3

t=1/2

x=0
y=1/2
z=13/2

So B(0,1/2,13/2)

so AB=(1/3,1/2,-5/6)

mAGNITUDEofAB= √((1/3)sqrd +(1/2)sqrd + (-5/6)sqrd) = √(38)/6

How does this look, personally the only thing i'm concerned with is the fact that i had an ugly parametric equation with the fractions.. but it would seem everything else is correct to me, from how you described to do this.

* i made an error earlier, but fixed it.

Last edited:
HallsofIvy