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Can someone please explain how the direction vector was found in problem $2$ of this worksheet?

Below is an image of the problem $2$ of the worksheet.

enter image description here

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Dear Shankar, Welcome to the website. Kindly do not write out your entire question in the title. The title should be short and capture the essence of your question. – user17762 Jun 13 '12 at 23:39
Sure, sorry about that. I'm wondering where <1, 0, -2> came from. Please help..! – Shankar Kumar Jun 13 '12 at 23:42
The $2x^2$ in the first line of Berkeley's solution set is a typo. – math-visitor Jun 14 '12 at 0:03
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You want to know the tangent line to the ellipse at the given point along the plane $y=2$, so go ahead and plug in $y=2$ to obtain $4x^2 + 8 + z^2 = 16$. Take the total differential to obtain
$$ 8x dx + 2 z dz = 0. $$ This means $$ 2z dz = -8x dx, \mbox{ or } \dfrac{dz}{dx} = -\dfrac{8x}{2z} = -\dfrac{4x}{z}. $$ Since $$ \dfrac{dz}{dx}|_{(x,z)=(1,2)} = -\dfrac{4}{2} = -2, $$ the line tangent to the intersection of the ellipsoid and the plane at the point $(1,2,2)$ has the direction vector $$ (1,0,-2). $$

Note that the tangent line is parallel to the $y=2$ plane so the vector should not be changing in the $y$-direction. Moreover, any nonzero scalar multiple of $(1,0,-2)$ will also work.

Thus the tangent line is $$ l(t) = (1+t, 2, 2-2t), $$ or you could also write it as $$ l(t) = (1,2,2)+ t(1,0,-2), $$

where $t$ varies over the reals.

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Firstly by the observation direction vector should lie at the $zx$ plane so $y$ in the direction vector should be $0$. Secondly after that he is thinking of the slope of the direction vector as the direction vector of the fuction $ z(x)$ where the direction vector is that case is $(1,z'(x_0))$

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