Question

Again 3-d!
;)

Answer 1

question?

Answer 2

Consider two lines:
L1: x-1/1=y-2/2=z-3/3 and
L2:x-2/3=y-3/1=z-1/2
Then
If π denotes the plane x+by+cz+d=0 parallel to lines L1,L2 and which is equidistant from L1 and L2 then

Answer 3

A).1+b^2=c^2+d^2
B).d=√bc
C).b=cd
D).2b+c+d=0

Answer 4

@ganeshie8 can you please help here?

Answer 5

One key idea:
Since plane is parallel to the given lines, the normal vector of the plane must be perpendicular to both the lines. Recall cross product.

Answer 6

i think your last "3-d" question really amounted to proving the scalar triple product equalled zero (or not). fwiw, i had a crack at it and could get beyond some horrible algebra.

this one, i'm guessing, involves using the vector triple product and relying upon that being zero. more horrible algebra.

which is another way of looking at ganesh's advice. plane parallel to 2 lines, so the plane's normal runs parallel to the cross product of the direction of those 2 lines.

so the triple cross is zero.

Answer 7

Well, sir i literally didn't have a trouble finding the plane's equation in termz of a,b, and c since i gathered its normal vector dcs' so i could easily have it .
I came across with the difficulty in finding out the condition of d from the second part of the question which says that lines are equidistant from the planes.
Can you please explain that ?

Answer 8

@ganeshie8 pls help .. :(