[FIXED] Linear Algebra Problem

[µ]

I got a few answers already, will take little time myself to check

Thanks

[b!z]

Cool username bro

[cryptworld]

you take the vectors of L1 and L2
L1 vector: (1,-3,1)
L2 vector: (-1,1,2)

so we have your plane equation:

x=1+λ*1+μ*-1
y=2-3*λ+1*μ
z=-2+1*λ+2*μ

1CXJRbrgWXe9i889TzfSHED6mER52RR4x9

[fsb4000]

If I remember correctly and didn't make arithmetic errors
(P.S. English is not my native)

1)Make the equation of the plane by the point and two non-collinear vectors:
|x - 1    1   -1 |
|y - 2   -3   1  | = 0
|z-(-2)  1    2  |

(x-1)*|-3 1| - (y-2)*|1 -1| +(z+2)*|1 -1|= 0
         |1 2 |             |1  2|             |-3 1|

(x-1)*(-6-2) - (y-2)*(2+1)+(z+2)*(1-3)=0

-8x+8 -3y + 6 -2z-4 =0

-8x -3y -2z +10=0

 8x +3y +2z -10 =0

2)Make the equation to the normal form
To find the normal equation of the plane defined by the equation Ax + By + Cz + D = 0, it suffices to divide both sides of this equation by:

where the upper sign is taken when D> 0, and lower when D <0; if D = 0, then we can take any sign.

So divide both sides by:
  ____________       __
\/8*8+3*3+2*2 = \/77

          8___ x  +  3 ___ y + 2___ z - 10___ = 0
Answer:\/77           \/77        \/77       \/77

BTC address: 1LqXE9xzjLS3HcoRe6AvxAw9zhMimDWCjP

[nsimmons]

i already answered this, pm me for btc

https://bitcointalk.org/index.php?topic=815697.0

[Gleb Gamow]

If I remember correctly and didn't make arithmetic errors
(P.S. English is not my native)

1)Make the equation of the plane by the point and two non-collinear vectors:
|x - 1    1   -1 |
|y - 2   -3   1  | = 0
|z-(-2)  1    2  |

(x-1)*|-3 1| - (y-2)*|1 -1| +(z+2)*|1 -1|= 0
         |1 2 |             |1  2|             |-3 1|

(x-1)*(-6-2) - (y-2)*(2+1)+(z+2)*(1-3)=0

-8x+8 -3y + 6 -2z-4 =0

-8x -3y -2z +10=0

 8x +3y +2z -10 =0

2)Make the equation to the normal form
To find the normal equation of the plane defined by the equation Ax + By + Cz + D = 0, it suffices to divide both sides of this equation by:

where the upper sign is taken when D> 0, and lower when D <0; if D = 0, then we can take any sign.

So divide both sides by:
  ____________       __
\/8*8+3*3+2*2 = \/77

          8___ x  +  3 ___ y + 2___ z - 10___ = 0
Answer:\/77           \/77        \/77       \/77

BTC address: 1LqXE9xzjLS3HcoRe6AvxAw9zhMimDWCjP

Boy, this brings back fond memories of when I used to do shit like the above as a hobby till my mother told me to go to bed because I had school in the morning.

[b!z]

If I remember correctly and didn't make arithmetic errors
(P.S. English is not my native)

1)Make the equation of the plane by the point and two non-collinear vectors:
|x - 1    1   -1 |
|y - 2   -3   1  | = 0
|z-(-2)  1    2  |

(x-1)*|-3 1| - (y-2)*|1 -1| +(z+2)*|1 -1|= 0
         |1 2 |             |1  2|             |-3 1|

(x-1)*(-6-2) - (y-2)*(2+1)+(z+2)*(1-3)=0

-8x+8 -3y + 6 -2z-4 =0

-8x -3y -2z +10=0

 8x +3y +2z -10 =0

2)Make the equation to the normal form
To find the normal equation of the plane defined by the equation Ax + By + Cz + D = 0, it suffices to divide both sides of this equation by:

where the upper sign is taken when D> 0, and lower when D <0; if D = 0, then we can take any sign.

So divide both sides by:
  ____________       __
\/8*8+3*3+2*2 = \/77

          8___ x  +  3 ___ y + 2___ z - 10___ = 0
Answer:\/77           \/77        \/77       \/77

BTC address: 1LqXE9xzjLS3HcoRe6AvxAw9zhMimDWCjP

Boy, this brings back fond memories of when I used to do shit like the above as a hobby till my mother told me to go to bed because I had school in the morning.

I am sure you had an interesting childhood. Do tell me more

[µ]

Thank you for your answers to my question cryptworld, fsb4000 and nsimmons.

I will still take a look myself at this problem and see if i can get this to work, and if your replies was correct step by step with right answer i also want to give you a few parts of BTC for the help. After this post i don't need any more help with this anymore. Thanks so far

Greetings from µ and we stay in touch cryptworld, fsb4000 and nsimmons.