Problem involving vectors

[Vexxon]

Been thinking about this problem for a couple of hours, and can't figure out how to do it:

The axis of a light in a lighthouse is tilted. When the light points east, it is inclined upward at 1 degree(s). When it points north, it is inclined upward at 8 degree(s). What is its maximum angle of elevation (in degrees)?

Seems like it has something to do with finding the normal to the plane formed by the beam of the lighthouse, but I have no idea how to do that from the information given. Any advice on how to proceed would be welcome.

 

[Deleted member 4993]

Been thinking about this problem for a couple of hours, and can't figure out how to do it:

The axis of a light in a lighthouse is tilted. When the light points east, it is inclined upward at 1 degree(s). When it points north, it is inclined upward at 8 degree(s). What is its maximum angle of elevation (in degrees)?

Seems like it has something to do with finding the normal to the plane formed by the beam of the lighthouse, but I have no idea how to do that from the information given. Any advice on how to proceed would be welcome.

First draw a picture (sketch) - stare at it - light will dawn on you!!!

Remember, when it comes back to east - it should come back to the same point (or line of ray).

 

[Vexxon]

Heh, part of my problem is that I can't draw (or visualize) the problem in 3 dimensions.

I'm kind of thinking of it like this:

I think the angle between the normal of the light-house-beam plane, and the horizontal plane would be the same as the maximum angle of elevation. I can't figure out how to calculate that angle, though...

 

[galactus]

I don't want to give away the punchline, but if x and y are the angles they are tilted, then \(\displaystyle \sqrt{x^{2}+y^{2}}\).

For small values of x, then \(\displaystyle x\sim tan(x)\)

What are the directional derivatives?.

See where I am going?. It's actually very easy if you see it.

 

[Vexxon]

Oh ho, I get it. It was that simple? Just finding the diagonal of the two... interesting.

So in that case, sqrt(65) should be the answer.

Thanks, guys.

 

[galactus]

Well, no, not quite. That is close, though.

Remember I said x is asymptotic to tan(x) when x is small.

\(\displaystyle tan^{-1}\left(\sqrt{tan^{2}(1)+tan^{2}(8)}\right)\)

See now?.

You can see this is very close to sqrt(65)