Solving Equations: Perpendicular Lines and Planes in 3D Space

[mohdfasieh]

Hello

Plz please tell me the solutions of these question :

I have tried the questions many time but not getting the exact logic please help me

1): Find equation of straight line perpendicular to x-1=(y-1)/2=(z+2)/3 and (x+2)/2=5-y=(z+3)/2 and passing through their point of intersection .


2):Find equation of plane passing through the point (1,0.-1) and (3,2,2) and parallel to the line x-1=(1-y)/2=(z-2)/3

3):Find equation of straight line passing through midpoint of AB ,perpendicular to AB and lies in the plane x=z where A(1,-1,1) and B(2,1,2)

 

[HallsofIvy]

Geniuses? Will you settle for me?

1): Find equation of straight line perpendicular to x-1=(y-1)/2=(z+2)/3 and (x+2)/2=5-y=(z+3)/2 and passing through their point of intersection .

The line x-1= (y-1)/2= (z+2)/3 has parametric equation x= 1+ t, y= 1+ 2t, z= -2+ 3t (setting t to be the common value above).
The line (x+2)/2= 5- y= (z+3)/2 has parametric equation x= -2+ 2s, y=5- s, z= -3+ 2s.

Set x= x, y= y, z= z in the two above to determine where they intersect. (That's three equations in two unknown variables, s and t. In general two lines do not intersect in 3 dimensions but you are told that these do.) Vectors in the direction of the two lines are i+ 2j+ 3k and 2i- j+ 2k. The cross product of those will give you a vector perpendicular to both.

2):Find equation of plane passing through the point (1,0.-1) and (3,2,2) and parallel to the line x-1=(1-y)/2=(z-2)/3

To find the equation of a plane, you need a vector perpendicular to that plane. A vector in the the plane is (3-1)i+ (2-0)j+ (2-(-1))k= 2i+ 2j+ 3k. Another vector in the plane is i+ 2j+ 3z (from the given line). Their cross product is perpendicular to the plane. Use either of the given points to determine the equation of the plane.

3):Find equation of straight line passing through midpoint of AB ,perpendicular to AB and lies in the plane x=z where A(1,-1,1) and B(2,1,2)

The midpoint of AB is ((1+2)/2, (-1+ 1)/2, (1+ 3)/2)= (3/2, 1, 3/2) and a vector in the direction of AB is (2-1)i+ (1-(-1))j+ (2-1)k= i+ 2j+ k. The plane perependicular to that vector, containing that point (x- 3/2)+ 2(y- 1)+ (z- 3/2)= 0. The line you are looking for is the intersection of that plane and z= x. You can solve those two equations for 2 of x, y, z in terms of the third and then use that third as a parameter.

 

[mohdfasieh]

thz HallsofIvy for helping me thz a lot

 

[wtee85]

Hey hallsoivy.. check out my post.. wana help me out?