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Solving a plane geometry problem. This is what I did, and I had a little doubt at around the end. Is this procedure correct?

Determine the equation of the plane that contains the point $(4,2,-1)$ and the line $$L : (x,y,z) = (1,1,1) + t(-2,0,3), t \in \mathbb{R}$$

I guess it means the vectorial equation which determines all the points $X$ that belong to the plane.

If I recall correctly, such equation is of the form

$$X = P + a\cdot\vec{u} + b \cdot\vec{v}$$

For some point $P$ in the plane, two non-parallel arrows $\vec{u}$ and $\vec{v}$ in it and some $a,b\in\mathbb{R}$.


Well, I already have such point $P$, which is $(4,2,-1)$.

I need only to find two non-parallel direction arrows, right?


So we have a line in this plane. Clearly the direction arrow in the line $L$ is a direction arrow of the whole plane.

So, yay, we have one of the two direction arrows:

$$(-2,0,3)$$

Now we need a direction arrow that is not parallel to that one...


Looks like it is time to use the point we were given at the beginning, $(4,2,-1)$.

I wonder, first, if this point belongs to the line $L$?

$$(4,2,-1) = (1,1,1) + t(-2,0,3)$$

$$\begin{cases} 4 = 1-2t\\ 2 = 1\\ -1 = 1+3t \end{cases}$$

Which is inconsistent and therefore this point $P$ does not belong to the line $L$.

How can we use this point to get the second direction arrow of the plane, which cannot be parallel to $(-2,0,3)$?

I guess that, since this point $P$ doesn't belong to $L$, any direction arrow that begins from a point in $L$ and moves towards $P$ will NOT be parallel to $L$.

So if we make an arrow from $(1,1,1)$ (which belongs to $L$) and it moves to $(4,2,-1)$, we have

$$(4,2,-1) - (1,1,1) = (3,1,-2)$$

This new arrow must be another direction arrow, which is not parallel to $(-2,0,3)$.


So we have one point, and two non-parallel direction arrows. This must be enough:

$$X = (4,2,-1) + a \cdot (-2,0,3) + b \cdot (3,1,-2)$$

For some $a,b \in \mathbb{R}$

This is a vectorial equation of this plane.

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  • $\begingroup$ I didn't check the working but the reasoning is correct. If you take a point $P$ not on a line $L$ and any point $Q$ on $L$, then $PQ$ cannot be parallel to $L$ because two parallel lines never intersect unless they are identical (in Euclidean geometry of course). $\endgroup$
    – user21820
    May 24 '14 at 5:34
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Equation of plane is :

$$A(x-4)+B(y-2)+C(z+1)=0$$

Also :

$$-2A+0B+3C=0$$

Also $(1,1,1)$ lies on it.

$$-3A-B+2C=0$$

Divide by $C$ and solve for $A/C,B/C$.

Substitute in equation of place and cancel $C$. This was done assuming $C\ne0$

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  • $\begingroup$ Thank you, I will check this out soon. Goddamn, so did everything wrong? D: $\endgroup$ May 24 '14 at 4:58
  • $\begingroup$ @ZolTunKul so did you checked this? Two months have passed. $\endgroup$ Jul 26 '14 at 13:09
  • $\begingroup$ @user31683 Most likely, he would have since he has not come back again for any doubts... Btw its check not checked... $\endgroup$
    – evil999man
    Jul 26 '14 at 14:21

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