Thread: determinant can represent a plane?

Why the above determinant can represent a plane?
Can anyone interpret it?
Thank you.

Hey happymatthematics.

Hint: Try expanding it and relate the cross product of vectors to the plane normal.

You ask only if it represents a plane, but I think you can say more. It actually represents the plane through , , and , doesn't it?

- Hollywood

Hello chiro and hollywood,
Yes I realized that it represents a plane passing through (x1,y1,z1) , (x2, y2, z2) and (x3, y3, z3).
but is it a must that the last number in each row is 1 in this determinant?

Yes, the last column must be all the same non-zero value if you want the plane to pass through each of the points (x1,y1,z1), etc. So you might as well make this common value 1.

To see why you get a plane, expand the determinant along the first row and you get ax +by +cz + d =0, where a, b, c and d are cofactors. Now to see (x1,y1,z1) satisfies this equation, replace the first row x y z 1 by x1 y1 z1 1, and you have a determinant with two rows the same, so the determinant is 0. Notice though if the three points are collinear, you don't get a plane at all; the determinant is identically 0!

Thank you. Then How to interpret This three point form equation of a plane when all the 1 on the rightmost is converted to 2?

Multiplying a column by 2 multiplies the determinant by 2, so when you set it equal to 0 and cancel 2, you end up with the same equation.

- Hollywood