3d geometry

[umeh pandey]

find the eqn of the plane through the points (1,0,-1) & (3,2,2) & parallel to the line passing to the line (x-1)/1=(1-y)/2=(z-2)/3...help me out :-|

 

[stapel]

find the eqn of the plane through the points (1,0,-1) & (3,2,2) & parallel to the line passing to the line (x-1)/1=(1-y)/2=(z-2)/3...help me out :-|

What is the meaning of "parallel to the line passing to the line"?

Also, what have you tried? Where are you stuck? Or are you unable even to get started, so you need lessons with worked-out examples?

Thank you!

 

[HallsofIvy]

I presume you mean "parallel to the line (x- 1)/1= (1- y)/2= (z- 2)/3". The first thing I would do is convert that line equation to "parametric" form which I find simpler. Setting t= (x- 1)= (1- y)/2= (z- 3)/3, x= t+ 1, y= 1- 2t, and z= 3+ 3t. A vector in that direction is <1, -2, 3> (which, admittedly, you could have gotten directly from the original form!) and since the plane is to be parallel to the line, that vector must lie in the plane. Another vector which must lie in the plane is the vector from (1,0,-1) to (3,2,2), <3- 1, 2- 0, 2- (-1)>= <2, 2, 3>.

The cross product of those two vectors is normal to the plane and you can use either of the two given points with the normal vector to give the equation of the plane: the equation of the plane having normal vector <A, B, C> and containing point \(\displaystyle (x_0, y_0, z_0)\) is \(\displaystyle A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0\).