A=(-5,1,3),B=(-2,6,-3),C=(-4,3,4) I also have a line with points: D=(-2,7,6),E=

horrupavinvo

horrupavinvo

Answered question

2022-02-23


A=(5,1,3),B=(2,6,3),C=(4,3,4)
I also have a line with points:
D=(2,7,6),E=(1,8,2)
The vector equation form of the line:
((2),(7),(6))+r((1),(1),(8)):|rRR
The vector equation form of the plane:
((5),(1),(3))+s((3),(5),(6))+t((1),(2),(1)):|s,tRR
I am supposed to compute the intersection between these two.
What I know about intersection is that if I have a linear equation with no solution, then there is no intersection. If i have a Linear equation with one solution, then there is one intersection. If i have a linear equation with infinitely many solutions, then the line is contained within the plane.
I the answer is that the line is contained within the plane. I know the answer, but how is the solution squired is beyond me.

Answer & Explanation

tardanetkd2

tardanetkd2

Beginner2022-02-24Added 9 answers

Hint: We have that x lies in the intersection iff:
x=((2),(7),(6))+r((1),(1),(8)) and
x=((5),(7),(3))+s((3),(5),(6))+t((1),(2),(1)).
This leads to:
((2),(7),(6))+r((1),(1),(8))=((5),(1),(3))+s((5),(1),(3))+t((1),(2),(1))
This is a system of 3 linear equations with 3 unknowns. Can you take it from here?

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