The problem is: The points (2,-1,-2), (1,3,12), and (4,2,3) lie on a unique plane. Where does

OIPLZLqm6

OIPLZLqm6

Answered question

2022-02-25

The problem is:
The points (2,-1,-2), (1,3,12), and (4,2,3) lie on a unique plane. Where does this plane cross the z-axis?
I can easily solve this problem by calculus and cross product: the equation of a plane is
22x+33x11z=55.
Hence, this plane crosses the z-axis at z=5. However, the problem requires to be sold by the system of linear equations transformed into matrix. Even vectors have not been introduces yet. Just a matrix in echelon form and a back substitution. Hence, when I am writing three equations:
2ab2c=d
a+3b+12c=d
4a+2b+3c=d
I end up with 4 variables to find. What am I missing? Is there one more linear equation possible to add?

Answer & Explanation

Arif Coates

Arif Coates

Beginner2022-02-26Added 6 answers

By Gaussian elimination,
(2ab2c=d),(a+3b+12c=d),(4a+2b+3c=d):
(72b+13c=d2),(4b+7c=d):
552c=112d
and the requested point satisfies
a0+b0+cz=d.
So z=5.
mastifo5h

mastifo5h

Beginner2022-02-27Added 6 answers

It helped me a lot, thanks!

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