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Show that the lines $x=-2+t,y=3+2t,z=4-t$ and $x=3-t,y=4-2t,z=t$ are parallel. Find the equation of the plane they determine.

Here what is the meaning of "they determine"?

snulty
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Raisul RONI
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  • Two lines in general position in space are skew lines, which means that usually you cannot find any plane containing two given lines. However, when the two lines are parallel and distinct, that is a special position, and the two lines span a unique plane in that case. Try to comprehend the difference between skew lines and parallel lines, then you will see. "Determine" from the problem text is to be understood in the light of this. – Jeppe Stig Nielsen Sep 30 '16 at 15:01
  • @MikeEarnest: True, but in the OP's problem, the lines are parallel, not intersecting. – Brian Tung Sep 30 '16 at 20:00

6 Answers6

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If two lines in 3D space ($\Bbb R^3$) intersect or are parallel there is a plane in that 3D space that contains those two lines.

So you can define a plane by defining two lines that intersect or are parallel.

Ruts
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  • Two lines, in three dimensions, either are parallel or intersect or ate skew. In the first two cases there exist exactly one plane containing both lines. The lines "determine" the plane. – user247327 Sep 30 '16 at 12:18
  • In the degenerate case where the two lines are coincident (in which situation people might say that the lines are both parallel and intersecting, maybe), there are an infinitude of planes, so "the" plane cannot be determined in that case. – Jeppe Stig Nielsen Sep 30 '16 at 14:50
  • @JeppeStigNielsen - In geometry, by common convention elements are assumed to be distinct unless it is explicitly stated otherwise. This is because most results in geometry break down in degenerate cases, so it is far easier to state explicitly the results that hold even in degenerate cases, than to repeat "unique" ad nauseum for all the results that don't. So just by saying "two lines", Ruts excluded the degenerate case. – Paul Sinclair Sep 30 '16 at 16:42
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Rewrite the lines in vector form: $$l_1=(-2,3,4)+\lambda(1,2,-1)$$ $$l_2=(3,4,0)+\mu(-1,-2,1)$$ where $\lambda,\mu\in\Bbb R$.

Since $(1,2,-1)=-1\cdot(-1,-2,1)$, either the two lines are coincident or they are parallel. Since $(3,4,0)$ does not lie on $l_1$, they are parallel.

To find the plane containing both lines, take the vector connecting their initial points: $(3,4,0)-(-2,3,4)=(5,1,-4)$. Now do the cross product of this vector with either of the line's direction vectors: $(5,1,-4)×(1,2,-1)=(7,1,9)$. This gives the plane's normal, and all that remains is to compute the scalar of the plane equation: $(7,1,9)\cdot(3,4,0)=25$.

Therefore the plane containing both lines has the equation $$\mathbf r\cdot(7,1,9)=25$$ or in Cartesian form $$7x+y+9z=25.$$

Parcly Taxel
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  • "take the vector connecting their initial points" and if you want the plane in vector form, you can stop right there. $$p = (-2,3,4) + k(1,2,-1) + m(5,1,-4)$$ with $k, m \in \mathbb R$. – null Sep 30 '16 at 16:38
  • @null I know that's possible, but I find this form a little inelegant. Although this was the first form introduced in my mathematics class, I found the normal form much more powerful and simpler to use. – Parcly Taxel Sep 30 '16 at 16:42
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    That's true. However, I see some elegance in simply not performing a calculation if it's not necessary. The vector form of the equation for the plane is an equation no less and that's what the question asks for. I have no doubt that you know the vector form @Parcly ;-), my comment was aimed at Raisul, with the intent to show that there's a shortcut to the solution bypassing any more calculations that might be confusing. – null Sep 30 '16 at 16:57
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For $t=0$ and $t=1$ we get two points in each line.

In $L_1$ we get $(-2,3,4)$ and $(-1,5,3)$.

In $L_2$ we get $(3,4,0)$ and $(2,2,1)$.

It follows the a same director vector, $\vec u=\vec i+2\vec j-\vec k$, for both lines which means the lines are parallel and obviously distinct so its determine the plan containing both lines.

enter image description here

Piquito
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the parametric equations of two lines are

For first line $$\frac{x+2}{1}=\frac{y-3}{2}=\frac{z-4}{-1}$$ For second line $$\frac{x-3}{-1}=\frac{y-4}{-2}=\frac{z-0}{1}$$

multiply the second parametric equation by $(-1)$ $$\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-0}{-1}$$

the Denominators for both equations are equal, so the two lines are parallel.

the general equation of plane is $$A(x-x_0)+B(y-y_0)+C(z-z_0)=0$$

so that the $A$,$B$, and $C$ are the components of normal vector on plane

to find the normal vector: $$\overrightarrow{v_1}=i+2j-k$$ $$\overrightarrow{v_2}=(-2-(3))i+(3-(4))j+(4-0)k$$ $$\overrightarrow{v_2}=-5i-j+4k$$ the normal vector is $$\overrightarrow{n}=\overrightarrow{v_1}\times \overrightarrow{v_2}$$ then comlplete the solution

E.H.E
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1947898
Coëfficients of $t\text{ in first line:}\{1,2,-1\}$
Coëfficients of $t\text{ in second line:}\{-1,-2,1\}$
These are the top two rows of a determinant of order 3.
The third row is: $\{1,1,1\}.$
The determinant

$\left|\begin{array}{rrr} 1&2&-1\\ -1&-2&1\\ 1&1&1 \end{array}\right|=0,$
so the two lines are parallel.
Compute two points from the first line, say,
$(-2\mid 3\mid 4)\text{ and }(-3\mid 1\mid 5)$
and one point from the second, say, $(3\mid 4\mid 0).$
The equation

$\left|\begin{array}{rrrr} x&y&z&1\\ -2&3&4&1\\ -3&1&5&1\\ 3&4&0&1 \end{array}\right|=7x+y+9z-25=0$
is the equation you seek.

$7x+y+9z=25$

Senex Ægypti Parvi
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take any one point from two lines and find the equation of line for this and find the perpendicular vector by cross product of the give any line and the new line this will be enough to find the plane

Spiker
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