Find parametric equations for the line of intersection of the planes 3x-2y+z=1,

CheemnCatelvew

CheemnCatelvew

Answered question

2021-10-23

Find parametric equations for the line of intersection of the planes 3x2y+z=1, 2x+y3z=3

Answer & Explanation

Laith Petty

Laith Petty

Skilled2021-10-24Added 103 answers

3x2y+z=1
2x+y3z=3
FInd a point of intersection. Set z=0, solve for x and y.
3x2y+0=1x=13(2y+1)
2x+y0=3
213(2y+1)+y=3
2(2y+1)+3y=9
7y+2=9
7y=7
y=1
x=13(2(1)+1)=1
Point 1: (x0,y0,z0)=(1,1,0)
We can use the cross product to find a direction vector for the line.
The cross product of the normals to the planes results in a vector that is orthogonal to both, and would be parallel to the line of intersection.
<a,b,c>×<d,e,f<bfce,cdaf,aebd>
<3,2,1>×<2,1,3<2(3)1(1),1(2)3(3),3(1)(2)(2)>
=<5,11,7>
Plug the point into (x0,y0,z0) and the vector found into <a,b,c>
x=x0+at
y=y0+bt
z=z0+ct
x=1+(5)t
y=1+(11)t
z=0+(7)t
x=1+5t
y=1+11t
z=7t
Result: x=1+5t y=1+11t z=7t

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