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I am looking at the following exercise:

Show that, if a quadric contains three points on a straight line, it contains the whole line.

Deduce that, if $L_1$, $L_2$ and $L_3$ are nonintersecting straight lines in $\mathbb{R}^3$, there is a quadric containing all three lines.

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A straight line is of the form $\gamma (t)=a+tb$, right?

Do we use the following equation that defines the quadric?

$$v^tAv+b^tv+c=0$$

What does it mean that the quadric contains three points on a straight line?

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EDIT:

I am looking also at the next exercise:

enter image description here

I have the following:

Let $L_1, L_2, L_3$ be three nonintersecting straight lines of the first family. From the previous Exercise we have that there is a quadric that contains all the three lines. We have that each line of the second family , with at most a finite number of exceptions, intersects each line of the first family.

Let $\tilde{L}$ such a line of the second family. So $\tilde{L} $ intersects the lines $L_1, L_2, L_3$.

Since the above quadric contains $L_1, L_2, L_3$ we have that the quadric contains three points on $\tilde{L}$. Therefore the quadric contains the whole $\tilde{L}$. So the quadric contains all the lines of the second family, with at most a finite number of exceptions. So a doubly ruled surface is a quadric surface, or part of a quadric surface.

Is this correct?

Which quadric surfaces are doubly ruled?

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    $\begingroup$ May be your question should be quadric rather than quadratic as quadratic has two roots it can never contain three points of the same line where equation is $0$ $\endgroup$ – Archis Welankar Jan 10 '16 at 11:36
  • $\begingroup$ Oh sorry... It should be quadric. Do you have an idea what we are supposed to do? @ArchisWelankar $\endgroup$ – Mary Star Jan 10 '16 at 11:43
  • $\begingroup$ I have added an edit to my initial post... Could you take a look at it? Do you have an idea? @ArchisWelankar $\endgroup$ – Mary Star Jan 11 '16 at 11:44
  • $\begingroup$ Ya sure i will look $\endgroup$ – Archis Welankar Jan 11 '16 at 11:48
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    $\begingroup$ Can you consider a rectangle as a reference and then go further $\endgroup$ – Archis Welankar Jan 11 '16 at 11:53
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For the first part, you are on the good track. Plug in $\gamma(t)$ in the equation of a quadric. As an equation in $t$ it is quadratic and by the assumption it has 3 different zeros. That means it is identically zero, or in other words every point on a line is on a quadric too.

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  • $\begingroup$ To plug $\gamma (t)$ in the equation of a quadric do we substitute $v$ with $\gamma$ in the equation of a quadratic I wrote in my initial post? $\endgroup$ – Mary Star Jan 10 '16 at 12:07
  • $\begingroup$ exactly. Then consider it as a quadratic equation in $t$ $\endgroup$ – user26977 Jan 10 '16 at 12:08
  • $\begingroup$ So then we have $$\gamma^tA\gamma+b^t\gamma+c=0$$ Considering that the quadric can be written also as $$a_1x^2 + a_2y^2 + a_3z^2 + 2a_4xy + 2a_5yz + 2a_6xz + b_1x + b_2y + b_3z + c = 0$$ we have that this will contain only the variable $t$, since $x,y,z$ are functions of $t$, right? When it has $3$ different zeros, does it mean that it is identically zero, because the highest power of $t$ is $2$, so it has at most $2$ different zeros? $\endgroup$ – Mary Star Jan 10 '16 at 13:04
  • $\begingroup$ I added also an other question at the edit part of my initial post... Could you take a look at it? $\endgroup$ – Mary Star Jan 10 '16 at 15:30
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a) If you choose three points $P_i,Q_i, R_i$ on each line $L_i$ and force a quadric to pass through these nine points, the quadric will contain each line by the first part of the question, and you will be done.

b) But can you force a quadric to pass through these points?

c) Sure: passing through one of them gives a homogeneous linear condition on the ten coefficients $a_i,b_j,c$ of the quadric (in the notations of your comment to user26977's answer).
Since your have 9 homogeneous equations for ten unknowns, linear algebra tells you that you can find a non trivial solution $a_i^0, b_j^0,c^0$, and this will give you the coefficients of the equation of the required quadric.

An illustration
In order to show how simple the procedure is, notice that the condition for a quadric to go through $Q_3=(0,-1,0)$ (say) is $a_2-b_2+c=0$.

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  • $\begingroup$ I added also an other question at the edit part of my initial post... Could you take a look at it? $\endgroup$ – Mary Star Jan 10 '16 at 15:30
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    $\begingroup$ No, I will not. What you should do is acknowledge the answers you got and think hard about them instead of piling up new questions in Edits or in new threads, as you tend to do. $\endgroup$ – Georges Elencwajg Jan 10 '16 at 18:40

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