Submitted by maths123 on Thu, 04/30/2015 - 20:14
I am struggling with understanding the geometric interpretation of this question and hence the last part 'describe the surface...'
5i - If L1 is inclined equally to OA and to OB then surely it is already an angle bisector of AOB for all values of m,n,p? - if it is inclined equally it must make an angle of $ \alpha $ to OA and to OB
How can you deduce that the plane equation in the last part describes a double cone? and how can a plane equation describe a set of lines - surely the equation of a plane describes a set of points?
In addition the mark scheme states the axis is OA of cone - surely it can also be OB since the angle with OB is also $ \alpha $ ?
How is it that there are an infinite number of lines inclined at an angle to a specific line?
For part $\mathrm{(i)}$, when
For part $\mathrm{(i)}$, when the question says 'angle bisector', I think it wants you to force $L_{1}$ to be in the same plane as $A$, $O$ and $B$.
For part $\mathrm{(ii)}$, try and imagine the locus of points as $L_{2}$ (which goes through the origin $O$) rotates around $OA$ for fixed $\alpha$
Let me know if this is unclear and I'll try and come up with a better explanation. It's quite difficult to explain without showing you in person!
Angle bisectors in 3d
Does that mean that in 3d, a line equally inclined to two lines makes the same angle with each one - angle x as defined by a.b = mod(a)mod(b) cosx
However, in 3d a line that is an angle bisector of 2 lines must make the same angle with each line and lie in the same plane as both of the lines?
Yes, I believe that's what
Yes, I believe that's what this question wants. This means that there is a plane (namely, the plane perpendicular to the line $AB$ through $O$) of lines (i.e. infinitely many) that are equally inclined to $OA$ and $OB$ but there is only one line that is the angle bisector of $\angle AOB$.
Valid alternative method?
Surely a line that is an angle bisector means that the line is equally inclined to both planes?
What is the need to give the initial information that the line is equally inclined to OA and OB?
We can just separately work out the equation of the planes $ \Pi_1 $ and $ \Pi_2 $ which contain all lines making an angle $ \alpha $ with OA and OB respectively? Then we can solve to get intersection of the two planes $ \Pi_1 $ and $ \Pi_2 $ to get the equation of line which is the angular bisector of OA and OB?
Yes, $L_{1}$ is an angle
Yes, $L_{1}$ is an angle bisector $\implies$ $L_{1}$ is equally inclined to the lines $OA$ and $OB$, but the converse ($L_{1}$ is equally inclined to the lines $OA$ and $OB$ $\implies$ $L_{1}$ is an angle bisector) is not necessarily true.
I guess it's just another step in the question and you can use the relationship between $m$, $n$ and $p$ derived in the first part of $\mathrm{(i)}$ to determine the values of $m$, $n$ and $p$ for which $L_{1}$ is an angle bisector.
Bear in mind that the surface containing all lines through $O$ that make an angle $\alpha$ with $OA$ will not be a plane but rather a double cone as in part $\mathrm{(ii)}$ (and similarly for the surface containing all lines through $O$ that make an angle $\alpha$ with $OB$) and solving for the intersection of 2 double cones probably won't be that easy.