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Is there a plane that contains both $L_1$ and $L_2$? If so find its equation

These are lines.


$L_1$ is the line through points $(0,1,1)$ and $(-1,2,1)$

The direction vector is $(-1,2,1)-(0,1,1)=(-1,1,0)$

So $(x,y,z)=(0,1,1)+t(-1,1,0),t\in\mathbb{R}$


$L_2$ is a line that is orthogonal to the plane $\pi$, which has the normal vector $(-2,2,2)$, and $L_2$ passes through the point $(4,0,1)$

And so we have that $(x,y,z)=(4,0,1)+k(-2,2,2),k\in\mathbb{R}$


The question says that if there is a plane that contains both $L_1$ and $L_2$, find its equation.


For a plane we need $3$ points. How do I found $3$ points that are on both lines?

If I equate I get that:

$(0,1,1)-(4,0,1)=t(-1,1,0)-k(-2,2,2)$

which means

$(-4,1,0)=t(-1,1,0)-k(-2,2,2)$

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Two lines can intersect in zero places, one place, or infinitely many. In this case, it's impossible to find three intersections.

To define a plane, we need three points, and they need to be non-collinear. Imagine I gave you the points $(0, 0, 1), (0, 0, 2), (0, 0, 3)$ in $\mathbb{R}^3$ and told you to find a plane which contained them. Then any plane which contains the $z$-axis would work -- for example, both $x = 0$ and $y = 0$, and many other planes.

You can find three non-collinear points by taking two points from one line and the third from the other.

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Basically we need to find weather a plane exists, that pass through the given 4 points. If yes then it is the required plane.

Proof. Since the plane contains $2$ points of $l_1$, so we have $n_1.d_1=0$ where $n_1$ is normal of the plane and $d_1$ is the direction ratios of line. Same is true for $l_2$. So the plane either contain $l_1$ and $l_2$ or parallel to them, since the plane contains $4$ points passing through the line, it contains $l_1$ and $l_2$

To find answer find the plane passing through $3$ of the $4$ given points, and check if $4^{th}$ point lies on it or not

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  • $\begingroup$ How would I check if the 4th point lies on plane or not? After computing the equation of the plane for the 3 points, I have an equation like $5x+4y+z=d$ (I just made up numbers), if I plug in $(x,y,z)$ as the forth point, that will only give me value of $d$, it will not say anything regarding whether the fourth point in the plane or not? $\endgroup$ – K Split X Oct 13 '17 at 3:12
  • $\begingroup$ For $3$ non collinear points we can get a unique plane. Assume of the equation of plane to be $ax+by+cz=d$ divide whole equation by $a$ the we have just $3$ variables, which then would be solvable $\endgroup$ – Raghukul Raman Oct 13 '17 at 3:16
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You have two points on one of the lines and one point on the other. If a plane is to contain both lines, it must span these points. We can write the equation of a plane as $ax+by+cz+d=0$, so the condition that these points lie on the plane can be written in matrix form as $$\begin{bmatrix} 0&1&1&1 \\ -1&2&1&1 \\ 4&0&1&1 \end{bmatrix}\begin{bmatrix}a\\b\\c\\d\end{bmatrix}=\mathbf 0.$$ This says that $[a,b,c,d]^T$ lies in the null space of the matrix on the left. Using the usual methods, we can find that the null space is spanned by $[0,0,1,-1]^T$, so our candidate plane is $z=1$. The direction vector of $L_2$ is the normal to $\mathbf\pi$, $(-2,2,2)^T$, which is obviously not orthogonal to this plane’s normal, $(0,0,1)^T$, so there is no plane that contains both lines.

We could also have gotten there in one step by including the “point at infinity” that corresponds to $L_2$’s direction vector in the above matrix: $$\begin{bmatrix} 0&1&1&1 \\ -1&2&1&1 \\ 4&0&1&1 \\ -2&2&2&0 \end{bmatrix}.$$ Row-reduction reveals that this matrix has full rank, so its null space is trivial and there is no plane that contains both lines.

If you’re familiar with homogeneous coordinates or projective geometry, you should recognize the above as working with the homogeneous vector representations of the points, vectors and planes.

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