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Math Help - Find the acute angle between the plane 2(x-1)+16(y-2)+2(z-1) and the xy-plane

  1. #1
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    Find the acute angle between the plane 2(x-1)+16(y-2)+2(z-1) and the xy-plane

    Problem:

    Find an equation of the tangent plane to the ellipsoid x^2+4y^2+z^2=18 at the point (1,2,1) and determine the acute angle that this plane makes with the xy-plane.

    Work:

    Equation of Tangent Plane is: 2(x-1)+16(y-2)+2(z-1)


    Now, this is how I tried to find the acute angle between this plane and the xy-plane:

    I used the equation cosθ= n1Ěn2/|n1||n2|, where n1 and n2 are the normal vector (one of the tangent plane and one of the xy plane).

    The normal vector I used for the xy plane was <0,0,1> and the normal vector for the tangent plane is <2,16,2>

    Plugging everything into the equation I found that the acute angle is 1.44739 radians or 82.929 degrees.


    Can someone verify that this technique is correct? Mainly, I think I need to know if that was the correct normal vector to use for the xy-plane, and if the method I used in general is correct.

    Thanks!
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  2. #2
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    Given two intersecting planes with normals N_1~\&~N_2 the acute angle between them is found by:
    \arccos \left( {\frac{{\left| {N_1  \cdot N_2 } \right|}}<br />
{{\left\| {N_1 } \right\|\left\| {N_2 } \right\|}}} \right) .
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  3. #3
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    1. What do those lines mean in the numerator?

    2. Did I use the right vector for the xy-plane, namely, <0,0,1>?
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  4. #4
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    Many of us do not like to use \left| N \right| for the length of a vector therefore we use {\left\| {N } \right\|}.
    It is becoming fairly standard to do that.
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  5. #5
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    1. You can't normalize the dot product of 2 vectors as it's a scalar; that's why I was asking. Are you sure the numerator should have those lines?

    2. Can you verify for me that I used the correct normal vector for the xy-plane? It would be much appreciated.
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  6. #6
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    Quote Originally Posted by divinelogos View Post
    1. You can't normalize the dot product of 2 vectors as it's a scalar; that's why I was asking. Are you sure the numerator should have those lines?
    The lines in the numerator are absolute value. The problem asked for the acute angle between the two planes and, for an actute angle, cosine is positive. Taking the absolute value guarentees you get the acute angle and not its complement.

    Can you verify for me that I used the correct normal vector for the xy-plane? It would be much appreciated.
    Yes, the z-axis is perpendicular to the xy-plane and so < 0, 0, 1> is a normal vector. Or: the xy-plane is defined by the equation z= 0 or 0x+ 0y+ 1z= 0 so a normal vector is < 0, 0, 1>.
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  7. #7
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    Quote Originally Posted by HallsofIvy View Post
    The lines in the numerator are absolute value. The problem asked for the acute angle between the two planes and, for an actute angle, cosine is positive. Taking the absolute value guarentees you get the acute angle and not its complement.


    Yes, the z-axis is perpendicular to the xy-plane and so < 0, 0, 1> is a normal vector. Or: the xy-plane is defined by the equation z= 0 or 0x+ 0y+ 1z= 0 so a normal vector is < 0, 0, 1>.
    Awesome! Thanks for your help I really appreciate it. That's very interesting about the absolute value I hadn't noticed that.
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