Love these kind of questions.
1. normals are (2,3,-1) and (1,2,0)
a∙b = │a││b│cosß
2+6+0 = (√14)(√5)cosß
cosß=8/√70
ß will be the angle between the normals, can you take it from here?
2. The line has direction (0,1,2) and the plane has a normal of (2,-10,5)
take the dot product.
If the normal is perpendicular to the line, shouldn't the line be paralle to the plane?
3. vector AB = (3,-2,-2)
this would be a normal to the plane, so the equation of the plane is
3x - 2y - 2z = C
Shouldn't this plane pass through the midpoint of AB ??
I am sure you can take it from there.
Someone please explain these!
The angle between two planes is defined as the angle between their normals. Determine the angle theta between 0 and 90 degrees between the given planes.
a) 2x+3y-z+9 = 0 and x+2y+4 = 0
------------------------------------------
If the positive z-axiz points up, show that the line x=0, y=t, z=2t is parallel to and below the plane 2x-10y+5z-1 = 0.
-----------------------------------------
Find an equation for the set of points P(x,y,z) that are equidistant from the points A(1,2,3) and B(4,0,1)
1 answer
3 answers