Question

Find an equation of the plane.
The plane that passes through the pt (-1,2,1) and contains line of intersection of the planes x+y-z=2
and 2x-y+3z=1 Find an equation of the plane.
The plane that passes through the pt (-1,2,1) and contains line of intersection of the planes x+y-z=2
and 2x-y+3z=1 @Mathematics

Answer 1

well, first we have to determine the line of intersection for the other two planes

to do this we solve the augmented matrix into a row reduced echelon form

Answer 2

Matrix:
1 1 -1 | 2
2 -1 3 | 1
Echelon form: take row 2 and subtract 2 of row 1
1 1 -1 | 2
0 -3 5 | -3
divide the second row by -3 in order to get a 1 in the diagonal
1 1 -1 | 2
0 1 -5/3 | 1
Subtract row 2 from row 1
1 0 2/3 | 1
0 1 -5/3 | 1

assuming i didn't mess up in arithmatic, we now have RREF matrix with 1 free variable: z
so now we write all three variable in terms of t do get a parametricized line.
z=t
y-5/3 t=1 => y=1+5/3 t
x+2/3 t=1 => x=1-2/3 t

so our line is the parametric of that.

Answer 3

thats true, how did you do that?
the answer was x-2y+4z=-1, so its true but what math class is that in

Answer 4

From what i did, what i would do is then find the orthogonal vector from the line to the point, an write the linear combination to get a parametricized plan, and then implicitize, but that could get messy...

i will think of a better way unless someone posts it first.

what i did is all linear algebra, which at my school is part of Calc 2

Answer 5

easiest method i can think of: we then take two points on this line, lets go with t=0 & t=1, or <1,1,0> an <-1,6,3> and find two of the vectors differencces two of our three points (those two and the one stated)
our first will be the difference the two i just named: <1,1,0>-<-1,6,3>=<2,-5,-3>
The second two will be <1,1,0> an <-1,2,1>: <1,1,0>-<-1,2,1>=<2,-1,-1>

we now use these two vectors to find the normal line to our plane by taking the cross product: and we get <2,-4,8> (this takes a bit of math, but its the determinant of the matrix with row 1 being i,j,k, & the other two rows being our vectors)
these are our coefficients.
we then plug in 2(x-r)-4(y-s)+8(z-v)=0 where (r,s,v) is one of our points
2(x+1)-4(y-2)+8(z-1)=0
2x-4y+8z+2=0
2x-4y+8z=-2
x-2y+4z=-1

Answer 6

this is true. linear algebra is not part of calulus 2 at A&M.
this was a cal 3 ?
i guess i need to do more problems to understand better. let me ask you this how many hrs a day do or did u study for a math class?

Answer 7

I go to GT where our math system is screwed up. Um i dont really study for my math classes, but mainly cause i self-teach until i am proficient at the material. in addition, i have some natural math ability.

Answer 8

how many hours a day do you self teach?

Answer 9

for lets say one section out of a chapter

Answer 10

um, as much as i feel needed. its very inconsitatnt