k, Math question. Point of intersection ? huh?

Question

when you have 3 systems.. for ex
x + 2y +3z =5
2x - y + 2z = 7
3x - 3y + 2z = 10

ok and they say "find the coordinates of the point of intersection of the planes"

that just means you have to solve for x y z right ? and each value is a coordinate of intersection ? right? im not sure.

so if im right. and lets say X was = to 2, then one coordinate would be {2, 0, 0} right? i dont know

thanks

OK SO I WORKED OUT THE ANSWER. SO ARE MY VALUES FOR x ,y, z THE POINT OF INTERSECTION?? so should i just right "the point of intersection is {x,y,z} "

thanks for the help

Answer 1

yes. you can have a single point, a line or all the planes can be co-planar. Since they say it is a"point of intersection" and assuming they are truthful then there is one point you need to find. You have three equations and three unknowns so you've got what you need.

1. eliminate x between eqs 1 and 2
2. eliminate x between eqs 1 and 3
3. eliminates either y or z from the results of 1 and 2
4. If you get rid of y, solve for z. If you got rid of z solve for y.
5. Use the value from 4 to get either y or z from eqs 1 or 2.
6. then use y and z in one of the original eqs to get x

Answer 2

there are 2 procedures; substitution and removing. with substitution, you sparkling up for between the variables in between the equations, and then replace. with removing, you subtract or upload the equations so as that between the variables receives eradicated substitution: forty-one + 9y = 8x 8x = forty-one + 9y x = (forty-one + 9y) / 8 now replace this into the different equation 4x = 3 - 3y 4((forty-one + 9y) / 8) = 3 - 3y (forty-one + 9y) / 2 = 3 - 3y forty-one + 9y = 2(3 - 3y) forty-one + 9y = 6 - 6y forty-one + 15y = 6 15y = --35 y = -35 / 15 y = -7 / 3 now replace this into between the equations and sparkling up for x forty-one + 9y = 8x forty-one + 9(-7 / 3) = 8x forty-one - 21 = 8x 20 = 8x x = 20 / 8 x = 5 / 2 x = 5 / 2 , y = -7 / 3, as a result, they intersect at (5/2 , -7/3) removing: forty-one + 9y = 8x 3 - 3y = 4x multiply the 2d equations by way of 2 so as that it could have 8x on the ultimate area, it relatively is the comparable element because of the fact the precise equation 2(3 - 3y) = 2(4x) 6 - 6y = 8x now subtract those equations on the two elements (forty-one + 9y) - (6 - 6y) = (8x) - (8x) forty-one + 9y - 6 + 6y = 0 35 + 15y = 0 15y = -35 y = -35 / 15 y = -7 / 3 now you merely plug this in to between the equations the comparable way you do with the substitution approach.