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acedstud
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Asked: 2021-05-06T11:41:39+00:00 In: ,

Find the angle between the normals to the surface xy=z^2 at (1, 4, 2) and (-3, -3, 3)

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Find the angle between the normals to the surface xy=z^2 at (1, 4, 2) and (-3, -3, 3).

angle between normal to a curveangle between two vectorsnormal to a curvevector calculusvectors

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  2. acedstud
    This answer was edited.

    First, we express the curve in the form F(x, y, z) = c, where c is a constant.

    Thus, F(x, y, z) = xy - z^2 = 0

    Next, we find the normal vector to F, which is the gradient of F, that is \nabla{F}.

    Thus, \nabla{F}=\left(\dfrac{\partial}{\partial{x}}i+\dfrac{\partial}{\partial{y}}j+\dfrac{\partial}{\partial{z}}k\right)(xy-z^2)=yi+xj-2zk.

    The normal vector at (1, 4, 2) is 4i+j-4k, and the normal vector at (-3, -3, 3) is -3i-3j-6k.

    Finally, the angle between the two vectors 4i+j-4k and -3i-3j-6k is given by \theta=\cos^{-1}\left(\frac{(4i+j-4k)\cdot(-3i-3j-6k)}{|4i+j-4k||-3i-3j-6k|}\right)

    =\cos^{-1}\left(\frac{9}{\sqrt{33}\times\sqrt{54}}\right)=\cos^{-1}(0.2132)\approx77.7^o.

    Therefore, the angle between the normals to the surface xy=z^2 at (1, 4, 2) and (-3, -3, 3) is approximately 77.7^o.

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