Posted by Alex on Thursday, April 17, 2008 at 9:41pm.
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.
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