3-D Geometry, Coplanarity & Direction Ratios

Question

Find the condition for the two lines to be coplanar.

a₁ + b₁x + c₁x + d₁ =0 = a₂ + b₂x + c₂x + d₂

a₃ + b₃x + c₃x + d₃ =0 = a₄ + b₄x + c₄x + d₄

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I know that the general line passing through the first lines can be represented as a₁ + b₁x + c₁x + d₁ +k(a₂ + b₂x + c₂x + d_2), for real number k

Also, the general line passing through the second lines can be represented as a₃ + b₃x + c₃x + d₃ +d(a₄ + b₄x + c₄x + d_{4 ),} for real number r

How do we then, prove the required result?

Thank you!

Answer 1

it becomes very contorted with all the subscripts. but doable.

there is a proof concluding having two lines

L1,,,, (x-x₁)/a=(y-y₁)/b=(z-z₁)/c

L2,,,,(x-x₂)/d=(y-y₂)/e=(z-z₂)/f

then the determinant =0 of

row 1 = (x₁-x₂),,,(y₁-y₂),,,(z₁-z₂)

row2 = a,,,,b,,,c

row3= d,,,,e,,,,f

means the two lines either intersect or are parallel - just the conditions wanted

for the two lines to be coplanar.

for your problem, finding a,b,c means taking crossproduct with matrix

row1=i,,,j,,,k

row2=a₁,,,,b₁....c₁

row3=a₂,,,b₂,,,,,c₂

a is the factor multiplying i, etc.

finding d,e,f is same just with other two planes.

Finding x₁,y₁,z₁ requires solving the first tow equations as system. eliminating x, will

have equation y in terms of z. set z=z₁=0 to find y₁, then compute x₁

Do the same for x₂,y₂,z_2. Note: z₂ also set to 0

multiply the determinant specified above and set to zero

and some contorted relationship between the a's, and the b's and the c's and the d's

ensures the lines are coplanar