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Following Given 3 orthogonal vectors how to calculate the ellipsoid induced by them?

Assume an arbitrary point $x_0$ and an ellipsoid defined as $(x-v)^t A (x-v) = 1$

Where the ellipsoid's principal axis matrix is given by: $W = [w_1 | w_2 | w_3]$ s.t. $W^t W = I$.

  1. How to project $x_0$ on the ellipsoid surface?
  2. How to project $x_0$ on the ellipsoid's principal axes?
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    You need to define what you mean by *projection*, there are plenty different projections. If you're not sure please provide an example. – 2017-01-27
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    @flawr can you please say explicitly some of the plenty possible projections? – 2017-01-27
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    Projections onto the axes of the ellipsoid? Projections onto the surface of the ellipsoid? The list goes on – 2017-01-27
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    How can I project on it surface for example? On principal axes it's obvious, simply multiple by the matrix of column eigen vectors – 2017-01-27
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    Let $u$ be an arbitrary vector, define the projection $v = (u^tAu)^{-1/2} u$. Then $v$ is a projection of $u$ onto the ellipse. – 2017-01-27
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    @flawr, was I right by saying that in order to project a point $x_0$ on the ellipsoid principal axes one needs to take $W = [w_1 | w_2 | w_3]$ the matrix of the row principal direction vectors and calculate $W x_0$. I am still wondering what happens if I take $A x_0$, using your notation in http://math.stackexchange.com/a/2115252/24403 – 2017-02-13
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    @flawr how did you come to that conclusion? – 2018-08-21
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    @SridharThiagarajan I think you can use the ansatz $v = tu$, and solve for $t$. But as I said, this is *one* possible projection that projects along the lines through the origin. – 2018-08-21
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    @flawr I just saw that the orthogonal projection of a point onto a ellipse cannot be done in closed form generally. We need to solve a QP. What sort of projection is the one you suggest? – 2018-08-21
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    That is right, the one I suggested projects by moving the point closer to (or further away from) the origin, until it is on the ellipse. – 2018-08-21

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