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Find the equations of the tangent planes from the line:

$r=(1,-2,3)+λ(2,3,4)$

to the surface:

$(x^2 /3) + (y^2 /2 ) - z^2 -1 =0$

I didn't get what he means by a tangent plane “from a line”, but I know I'm supposed to have a tangency point & then easily I'll be able to form the equation of the tangent plane. So I thought I might just try to solve the equation of the line with the surface & then I'll have $2$ points of intersection, then use them as tangency points but it got really messy.

This is my first time asking, so I hope I've got the tags right.

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    The plane contains the line. – 2017-01-13
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    The tangent plane contains the line? – 2017-01-13
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    Yes, that's correct. – 2017-01-13
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    So I'm supposed to solve the line equation with the surface & it'll give me one value for λ right? – 2017-01-13
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    A plane is determined by three distinct points. You're already given the first two, and are further told that the third lies somewhere on a certain surface, so that the resultant plane is tangent to that specific surface $($as opposed to, say, intersecting it$).$ – 2017-01-13
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    I find very complicated coefficients with $\sqrt{85}$ everywhere for the 2 solution planes. Have you checked the coefficients' values you have provided ? – 2017-01-13
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    Yes I did , I solved it 3 times now & still get 2 values for λ so there's no way this line lies on the tangent plane :/ ? – 2017-01-13
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    I am surprised you have such a homework: the tangent planes to surfaces like this one which is non-convex (hyperboloid with one sheet) are so counter intuitive... I would have understood for an ellipsoid like $(x^2 /3) + (y^2 /2 ) + z^2 -1 =0$... – 2017-01-13
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    @JeanMarie The numbers get pretty messy for an ellipsoid with these axis lengths, too. For my part, I’m wondering if the equation is meant to be $(x/3)^2+(y/2)^2\dots$ instead. – 2017-01-14
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    @amd It might be the explanation... – 2017-01-14

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