How does one show that given 3 non-parallel vectors in RR^3, a curve with parameterisation: s(t)=(x_1 t^2+y_1 t+c_1,x_2t^2+y_2t+c_2, x_3t^2+y_3t+c_3) lies in a plane?

Brooklyn Farrell

Brooklyn Farrell

Open question

2022-08-14

How to show a curve lies in a plane given 3 constant vectors
How does one show that given 3 non-parallel vectors in R 3 , a curve with parameterisation:
s ( t ) = ( x 1 t 2 + y 1 t + c 1 , x 2 t 2 + y 2 t + c 2 , x 3 t 2 + y 3 t + c 3 )
lies in a plane?
Additionally, how do we find the equation of the plane?

Answer & Explanation

Siena Bennett

Siena Bennett

Beginner2022-08-15Added 17 answers

Note that s ( t ) = ( x 1 y 1 c 1 x 2 y 2 c 2 x 3 y 3 c 3 ) ( t 2 t 1 ) , and the curve ( t 2 t 1   ) lies in a plane, and the image of a plane under a linear transformation is a plane.
Pader6u

Pader6u

Beginner2022-08-16Added 4 answers

s ( t ) is a plane curve iff det ( s ( t ) , s ( t ) , s ( t ) ) = 0

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