What does the angle between a line and an ellipsoid mean? Does it mean the angle between the line and a plane tangent to the ellipsoid at the point of intersection?

By making this assumption, I end up with the wrong answer.

First I want to find the point of intersection. Substituting one independent variable for another I get $\pm (1, 2, 1)$. The ellipsoid and line both look symmetric about the origin, so I'll assume the angles at both points are the same. I'll throw away the negative intersection.

Then I determine the vector that represents the line. I get $\frac{1}{2}\vec{i} + \vec{j} + \frac{1}{2} \vec{k}$.

The magnitude is $\sqrt{\frac{3}{2}}$.

So far so good.

Since I'm looking for a tangent plane, I want a vector normal to that plane. Since this ellipsoid looks symmetric about the origin, I believe the gradient will be that normal vector I'm looking for.

$grad f = 4x \vec{i} + 2 y \vec{j} + 4 z \vec{k}$

For the point of intersection, we have

$grad f = 4 \vec{i} + \vec{j} + 4 \vec{k}$

The magnitude is $\sqrt{33}$.

Let $\theta$ be the angle between the line and the gradient.

To get the cosine of $\theta$ between the line and the gradient, I do

$\cos \theta = \frac{<4, 1, 4> \cdot < \frac{1}{2}, 1, \frac{1}{2} >}{\sqrt{33} \sqrt{\frac{3}{2}}}$

$\cos \theta = \frac{2 + 1 + 2}{\sqrt{33} \sqrt{\frac{3}{2}}}$

$\cos \theta = \frac{5}{\sqrt{33} \sqrt{\frac{3}{2}}}$

$\cos \theta = \frac{5}{\sqrt{\frac{99}{2}}}$

$\cos \theta = \frac{5 \sqrt{2}}{3 \sqrt{11}}$

But we're actually looking for the angle between the line and the tangent plane. Let $\phi$ be that angle. It follows that $\phi=\frac{\pi}{2} - \theta$.

So $\sin \phi = \frac{5 \sqrt{2}}{3 \sqrt{11}}$.

Then $\phi = sin^{-1} \frac{5 \sqrt{2}}{3 \sqrt{11}}$.

However, that's the wrong answer. Was my interpretation of the problem statement incorrect?

  • $\begingroup$ The points of intersection are $\pm (1,2,1)$. $\endgroup$ – Paracosmiste Oct 9 '15 at 18:32
  • $\begingroup$ Oh yes, that's what I have on my paper. The problem still remains. $\endgroup$ – Marty B. Oct 9 '15 at 18:43
  • $\begingroup$ $grad f=4(1,1,1)$ $\endgroup$ – Paracosmiste Oct 9 '15 at 18:48
  • $\begingroup$ No, you're right, that was entirely the problem. My understanding of the problem was correct. It was the arithmetic that got me. Thank you for pointing it out. $\endgroup$ – Marty B. Oct 9 '15 at 18:49

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