3-D Geometry & Direction Ratios

Question

Question:

Find the condition for the two lines to be coplanar.

a₁x + b₁y + c₁z + d₁ =0 = a₂x + b₂y + c₂z + d₂

a₃x + b₃y + c₃z + d₃ =0 = a₄x + b₄y + c₄z + d₄

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My work so far:

I know that the general line passing through the first lines can be represented as a₁x + b₁y + c₁z + d₁ +k(a₂x + b₂y + c₂z + d_{2 ),} for real number k

Also, the general line passing through the second lines can be represented as a₃x + b₃y + c₃z + d₃ +d(a₄x + b₄y + c₄z + d₄)_, for real number r

How do we then, prove the required result?

Thank you!

Answer 1

first principles kind of

given equation of two planes, can use variable elimination to eliminate "x"

then set "z"=0 to solve for "y=y₀" which provides x=x₀, now have point (x₀,y₀,0)

for direction of line,

if plane 1 is a₁x+b₁y+c₁z+d₁=0

and plane 2 is a₂x+b₂y+c₂z+d₂=0

then the normal vector to plane 1 is (a₁,b₁,c₁)

and normal vector to plane 2 is (a₂,b₂,c₂)

the crossproduct of the two normal vectors points in intersection direction (or minus direction)

recall crossproduct would be calculating determinant

row1=i,,,j,,,k

row2=a₁,,,b₁,,,c₁

row3=a₂,,,b₂,,,,c₂

the factor of i is x component of direction, factor of j is y component of direction and factor of k the z direction

line defined by (x-x₀)/(b₁c₂-b₂c₁)=-(y-y₀)/(a₁c₂-a₂c₁)=(z)/(a₁b₂-a₂b₁)

is the line of intersection between plane 1 and plane 2.

this is usually the best way to find the line.

next find line for plane 3 intersecting with plane 4

if two lines are coplanar, they either intersect or are parallel.