# Intersection of a Line and a Plane - Mathematics

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A given line and a given plane may or may not intersect. If the line does intersect with the plane, it's possible that the line is completely contained in the plane as well. How can we differentiate between these three possibilities?

Example (PageIndex{8}): Finding the intersection of a Line and a plane

Determine whether the following line intersects with the given plane. If they do intersect, determine whether the line is contained in the plane or intersects it in a single point. Finally, if the line intersects the plane in a single point, determine this point of intersection.

[egin{align*} ext{Line:}quad x &=2 - t & ext{Plane:} quad 3x - 2y + z = 10 [5pt] y &= 1 + t [5pt] z &= 3t end{align*} onumber]

Solution

Notice that we can substitute the expressions of (t) given in the parametric equations of the line into the plane equation for (x), (y), and (z).

[3(2-t) - 2(1+t) + 3t = 10 onumber]

Solving this equation for (t):

[6 - 3t -2 - 2t + 3t = 10 onumber]

[4 - 2t = 10 onumber]

[-2t = 6 onumber]

[t = -3 onumber]

Since we found a single value of (t) from this process, we know that the line should intersect the plane in a single point, here where (t = -3). So the point of intersection can be determined by plugging this value in for (t) in the parametric equations of the line.

Here: (x = 2 - (-3) = 5,quad y = 1 + (-3) = -2, , ext{and}quad z = 3(-3) = -9).

So the point of intersection of this line with this plane is (left(5, -2, -9 ight)). We can verify this by putting the coordinates of this point into the plane equation and checking to see that it is satisfied.

Check: (3(5) - 2(-2) + (-9) = 15 + 4 - 9 = 10quadcheckmark)

Now that we have examined what happens when there is a single point of intersection between a line and a point, let's consider how we know if the line either does not intersect the plane at all or if it lies on the plane (i.e., every point on the line is also on the plane).

Example (PageIndex{9}): Other relationships between a line and a plane

Determine whether the following line intersects with the given plane. Finally, if the line intersects the plane in a single point, determine this point of intersection.

[egin{align*} ext{Line:}quad x &=1 + 2t & ext{Plane:} quad x + 2y - 2z = 5 [5pt] y &= -2 + 3t [5pt] z &= -1 + 4t end{align*} onumber]

Solution

Substituting the expressions of (t) given in the parametric equations of the line into the plane equation gives us:

[(1+2t) +2(-2+3t) - 2(-1 + 4t) = 5 onumber]

Simplifying the left side gives us:

[1 + 2t -4 + 6t + 2 - 8t = 5 onumber]

Collecting like terms on the left side causes the variable (t) to cancel out and leaves us with a contradiction:

[-1 = 5 onumber]

Since this is not true, we know that there is no value of (t) that makes this equation true, and thus there is no value of (t) that will give us a point on the line that is also on the plane. This means that this line does not intersect with this plane and there will be no point of intersection.

How can we tell if a line is contained in the plane?

What if we keep the same line, but modify the plane equation to be ( x + 2y - 2z = -1)? In this case, repeating the steps above would again cause the variable (t) to be eliminated from the equation, but it would leave us with an identity, (-1 = -1), rather than a contradiction. This means that every value of (t) will produce a point on the line that is also on the plane, telling us that the line is contained in the plane whose equation is ( x + 2y - 2z = -1).

## Intersection of a Line and a Plane - Mathematics

A line is a collection of points along a straight path extending indefinitely in both directions. This is indicated by the arrows at each end, shown in the figure below.

A line has one dimension, its length. Two non-overlapping points determine a unique line and we can name the line with those two points or any other two points on the line.

line AB or BA defined by points A and B

Lines are used in shapes, angles, and many other geometric contexts.

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## Parametric Line Intersecting a Plane

DAVID JORDAN: Hello, and welcome back to recitation. The problem I'd like to work with you right now is we have a line which goes through two points that are given to us explicitly, and we have a plane which is given to us by an equation. And what we want to know is where does this line intersect this plane? And so one thing I would suggest to get started is we need to give a parametrization of our line to get started. OK, so why don't you work on that, pause the tape, and we'll come back in a moment and work it out together.

OK, welcome back. Let's get started. So let's start off by drawing a cartoon of what's going on here.

So we have this plane sitting in space. And we have some line kind of going through space. So maybe it's like this. And there's this single point of intersection. So even from the cartoon, we can kind of, sort of see two things which are going on.

Which is that we would expect a point of intersection, or we would expect exactly one, if we choose kind of a generic line and generic plane. In order for there to be no points of intersection, we would have to have a line which was parallel to the plane, which is very unlikely. And then otherwise, we expect exactly just one point of intersection.

So I want to break this sort of into two components. So we have an equation for the plane. And when I see an equation describing a plane, I think of that as a sort of test for membership. We can plug in a point (x, y, z) to the equation, and we can ask, does this point make the equation true, or doesn't it? And if it does, then that point (x, y, z) is in the plane, and otherwise it's not in the plane.

For the line, what we're going to need to do in a second is we're going to need to come up with a parametrization. And a parametrization is a different kind of thing than an equation describing a line. A parametrization, rather than being a test for membership, is really a way of listing all the points on the line. So when we give a parametrization-- in a second-- then we're going to be able to list all the points in the line. And then we're going to be able to plug our list into the equation for the plane, and find out which point on our list is actually in the plane. Which one satisfies the membership equation.

So why don't we get started first with parametrizing the line. So the general kind of picture here is we have a point P_1 in space, and we have another point P_2 in space, and we want to parametrize the line which goes between them. And there's actually a very simple way to do this.

What we do is we want to take our original point P_1, and we want to add a variable t times the vector P_2 minus P_1 which connects them. So that's this one here. OK. So this is a reasonable thing to do, because if we plug in t equals 0, then we just get P_1. And if we plug in t equals 1, then we get P_1 plus P_2 minus P_1 we just get P_2. So this line definitely goes through those two points, and that's all that we really need.

So in our specific problem here, we have-- P_1 we can take to be the first point, 0, minus 1, 1. And then we have t times-- so we have 2 minus 0 is 2-- 3 minus a negative 1 is 4-- and 3 minus 1 is 2. And this is the vector connecting those. And so we can write, we can just combine these two and we get 2t, 4t minus 1, and 2t plus 1. OK? So this here is a parametrization of the line.

So as we vary t-- now, walking back over to our picture-- as we vary t, we're going to just be listing all the points on the line. And we're going to ask, for which point are we actually contained in the plane? So let's go over to the board over here and solve that.

So what we want to know is does this point satisfy the equation for the plane? And our plane was given to us by the equation 2x plus y minus z equals 1. So x on our line is 2t. So we have 2 times 2t, plus-- y is 4t minus 1, minus-- z is 2t plus 1. And all this is meant to equal 1.

OK. So if we expand this out, we get 4t plus another 4t minus 2t and we get minus 1 minus another 1-- so we get minus 2-- equals 1. So altogether we get 6t equals 3, so that tells us that t is 1/2. OK?

And finally, to get our answer, we need to go back over to our parametrization of the line and plug in t equals 1/2. So coming back over here, plugging in t equals 1/2, we get 1-- 2 minus 1 is 1-- and 1 plus 1 is 2. OK, so we get the point of intersection, (1, 1, 2).

So that was quite a few steps, so let's review what we did. So to begin with, we needed to understand that the equation for a plane is a test for membership. It's not a list of all the points in the plane, it's a test for membership.

The parametrization of the line, on the other hand, is a way of listing all of the points on the line. And so if our goal is to find which particular point on the line is contained in the plane, then we need to parametrize our line, and then we need to plug in our parametrization to our equation for the plane, and then solve for the value of t which makes it true. Finding that t then we've-- that's is equivalent to finding a point on our line.

## Can the intersection of a plane and a line segment be a ray ?

A line is a one-dimensional term in geometry that has only length but not width.It extends up to infinity.A plane is a two-dimensional flat surface that extends up to infinity.

A plane is like a paper sheet , when we intersect a sheet by other sheet , we will observe that the intersection is a line ( as we given in the attachment).

Hence, the intersection of two planes is a line.

Therefore , the the correct answer is Option A - a line

10. If you are asked to give the intersection of any plane and any line not in that plane, your answer will always be be a point

11. If you are asked to give the intersection of two non-parallel planes, your answer will always be a line

12. If you are asked to give the intersection of two lines, your answer will always be a point

10. In space, a plane and a line that is not lying on this plane or parallel to it will always meet at point.

For instance, line PN intersects plane A at point N

11. The intersection of two planes in space is a line.

For instance, the intersection of planes QRST and PQRN is line RQ.

12. Two lines that are not parallel in the same plane will always intersect at a point.

For instance, lines TQ and PQ intersects at point Q

However, if two parallel lines are identical and coincide with each other, they intersect at infinitely many points.

16. By saying that a line intersects one plane and then another, you are saying that a line is existing on two planes. This is a direct contradiction to the statement.

17. The triangle is equilateral because syllogism is basically connecting the dots. If the angles in the triangle are all equal, it has all equal sides, and if it has all equal sides, then it is equilateral, therefore, it is D, not C.

A plane is a one dimensional figure

False. Line segments are one dimensional, therefore any interaction they have with a plane will be a point.

If two planes would intersect then it would make a line.

For a plane namely and a line intersection each other the intersection will be defined as a point and is known as the point of intersection. In other words the point of intersection for any two line, two plane or a line and a plane is a point where the value for both lines, planes or line and plane will be same.

In this case we can define the point as where the co-ordinates of point line and plane will be same say

When you have two planes whose intersection is not the empty set, you have two posibilites:

- The two planes are the same. In this case you don't have a line as the intersection of them, but it is again a plane.

-The two plane s are not the same . Then the intersection must be a line by an incidence axiom.

I suspect that by two vectors, you really mean two points, and want to intersect the line connecting those two points with the plane defined by Y=0 .

If that's the case, then you could use the definition of a line between two points:

<A + (D - A)*u, B + (E - B)*u, C + (F - C)*u>

Where <A,B,C> is one of your points and <D,E,F> is the other point. u is an undefined scalar that is used to calculate the points along this line.

Since you're intersecting this line with the plane Y=0 , you simply need to find the point on the line where the "Y" segment is 0.

Specifically, solve for u in B + (E - B)*u = 0 , and then feed that back into the original line equation to find the X and Z components.

The simplest (and very generalizable) way to solve this is to say that

which gives you 3 equations in 3 variables. Solve for x, y and z, and then substitute back into either of the original equations to get your answer. This can be generalized to do complex things like find the point that is the intersection of two planes in 4 dimensions.

For an alternate approach, the cross product N of (P2-P1) and (P3-P1) is a vector that is at right angles to the plane. This means that the plane can be defined as the set of points P such that the dot product of P and N is the dot product of P1 and N . Solving for x such that (L1 + x*(L2 - L1)) dot N is this constant gives you one equation in one variable that is easy to solve. If you're going to be intersecting a lot of lines with this plane, this approach is definitely worthwhile.

Written out explicitly this gives:

Note that that cross product trick only works in 3 dimensions, and only for your specific problem of a plane and a line.

## Intersection of a Line and a Plane - Mathematics

For problems 1 – 3 write down the equation of the plane.

1. The plane containing the points (left( <4, - 3,1> ight)), (left( < - 3, - 1,1> ight)) and (left( <4, - 2,8> ight)). Solution
2. The plane containing the point (left( <3,0, - 4> ight)) and orthogonal to the line given by (vec rleft( t ight) = leftlangle <12 - t,1 + 8t,4 + 6t> ight angle ). Solution
3. The plane containing the point (left( < - 8,3,7> ight)) and parallel to the plane given by (4x + 8y - 2z = 45). Solution

For problems 4 & 5 determine if the two planes are parallel, orthogonal or neither.

1. The plane given by (4x - 9y - z = 2) and the plane given by (x + 2y - 14z = - 6). Solution
2. The plane given by ( - 3x + 2y + 7z = 9) and the plane containing the points (left( < - 2,6,1> ight)), (left( < - 2,5,0> ight)) and (left( < - 1,4, - 3> ight)). Solution

For problems 6 & 7 determine where the line intersects the plane or show that it does not intersect the plane.

## Contents

There can be more than one primitive object, such as points (pictured above), that form an intersection. The intersection can be viewed collectively as all of the shared objects (i.e., the intersection operation results in a set, possibly empty), or as several intersection objects (possibly zero).

The intersection of two sets A and B is the set of elements which are in both A and B. In symbols,

For example, if A = <1, 3, 5, 7>and B = <1, 2, 4, 6>then AB = <1>. A more elaborate example (involving infinite sets) is:

As another example, the number 5 is not contained in the intersection of the set of prime numbers <2, 3, 5, 7, 11, …>and the set of even numbers <2, 4, 6, 8, 10, …>, because although 5 is a prime number, it is not even. In fact, the number 2 is the only number in the intersection of these two sets. In this case, the intersection has mathematical meaning: the number 2 is the only even prime number.

Intersection is denoted by the U+2229 ∩ INTERSECTION from Unicode Mathematical Operators.

Peano also created the large symbols for general intersection and union of more than two classes in his 1908 book Formulario mathematico. [4] [5]

## Contents

Euclid set forth the first great landmark of mathematical thought, an axiomatic treatment of geometry. [1] He selected a small core of undefined terms (called common notions) and postulates (or axioms) which he then used to prove various geometrical statements. Although the plane in its modern sense is not directly given a definition anywhere in the Elements, it may be thought of as part of the common notions. [2] Euclid never used numbers to measure length, angle, or area. In this way the Euclidean plane is not quite the same as the Cartesian plane.

This section is solely concerned with planes embedded in three dimensions: specifically, in R 3 .

### Determination by contained points and lines Edit

In a Euclidean space of any number of dimensions, a plane is uniquely determined by any of the following:

• Three non-collinear points (points not on a single line).
• A line and a point not on that line.
• Two distinct but intersecting lines.
• Two distinct but parallel lines.

### Properties Edit

The following statements hold in three-dimensional Euclidean space but not in higher dimensions, though they have higher-dimensional analogues:

• Two distinct planes are either parallel or they intersect in a line.
• A line is either parallel to a plane, intersects it at a single point, or is contained in the plane.
• Two distinct lines perpendicular to the same plane must be parallel to each other.
• Two distinct planes perpendicular to the same line must be parallel to each other.

### Point–normal form and general form of the equation of a plane Edit

In a manner analogous to the way lines in a two-dimensional space are described using a point-slope form for their equations, planes in a three dimensional space have a natural description using a point in the plane and a vector orthogonal to it (the normal vector) to indicate its "inclination".

Specifically, let r0 be the position vector of some point P0 = (x0, y0, z0) , and let n = (a, b, c) be a nonzero vector. The plane determined by the point P0 and the vector n consists of those points P , with position vector r , such that the vector drawn from P0 to P is perpendicular to n . Recalling that two vectors are perpendicular if and only if their dot product is zero, it follows that the desired plane can be described as the set of all points r such that

The dot here means a dot (scalar) product.
Expanded this becomes

which is the point–normal form of the equation of a plane. [3] This is just a linear equation

In mathematics it is a common convention to express the normal as a unit vector, but the above argument holds for a normal vector of any non-zero length.

Conversely, it is easily shown that if a, b, c and d are constants and a, b , and c are not all zero, then the graph of the equation

is a plane having the vector n = (a, b, c) as a normal. [4] This familiar equation for a plane is called the general form of the equation of the plane. [5]

Thus for example a regression equation of the form y = d + ax + cz (with b = −1 ) establishes a best-fit plane in three-dimensional space when there are two explanatory variables.

### Describing a plane with a point and two vectors lying on it Edit

Alternatively, a plane may be described parametrically as the set of all points of the form

where s and t range over all real numbers, v and w are given linearly independent vectors defining the plane, and r0 is the vector representing the position of an arbitrary (but fixed) point on the plane. The vectors v and w can be visualized as vectors starting at r0 and pointing in different directions along the plane. The vectors v and w can be perpendicular, but cannot be parallel.

### Describing a plane through three points Edit

#### Method 1 Edit

The plane passing through p1 , p2 , and p3 can be described as the set of all points (x,y,z) that satisfy the following determinant equations:

#### Method 2 Edit

To describe the plane by an equation of the form a x + b y + c z + d = 0 , solve the following system of equations:

This system can be solved using Cramer's rule and basic matrix manipulations. Let

If D is non-zero (so for planes not through the origin) the values for a, b and c can be calculated as follows:

These equations are parametric in d. Setting d equal to any non-zero number and substituting it into these equations will yield one solution set.

#### Method 3 Edit

This plane can also be described by the "point and a normal vector" prescription above. A suitable normal vector is given by the cross product

and the point r0 can be taken to be any of the given points p1 , p2 or p3 [6] (or any other point in the plane).

### Distance from a point to a plane Edit

Another vector form for the equation of a plane, known as the Hesse normal form relies on the parameter D. This form is: [5]

### Line–plane intersection Edit

In analytic geometry, the intersection of a line and a plane in three-dimensional space can be the empty set, a point, or a line.

### Line of intersection between two planes Edit

The remainder of the expression is arrived at by finding an arbitrary point on the line. To do so, consider that any point in space may be written as r = c 1 n 1 + c 2 n 2 + λ ( n 1 × n 2 ) >=c_<1><oldsymbol >_<1>+c_<2><oldsymbol >_<2>+lambda (<oldsymbol >_<1> imes <oldsymbol >_<2>)> , since < n 1 , n 2 , ( n 1 × n 2 ) >>_<1>,<oldsymbol >_<2>,(<oldsymbol >_<1> imes <oldsymbol >_<2>)>> is a basis. We wish to find a point which is on both planes (i.e. on their intersection), so insert this equation into each of the equations of the planes to get two simultaneous equations which can be solved for c 1 > and c 2 > .

#### Dihedral angle Edit

In addition to its familiar geometric structure, with isomorphisms that are isometries with respect to the usual inner product, the plane may be viewed at various other levels of abstraction. Each level of abstraction corresponds to a specific category.

At one extreme, all geometrical and metric concepts may be dropped to leave the topological plane, which may be thought of as an idealized homotopically trivial infinite rubber sheet, which retains a notion of proximity, but has no distances. The topological plane has a concept of a linear path, but no concept of a straight line. The topological plane, or its equivalent the open disc, is the basic topological neighborhood used to construct surfaces (or 2-manifolds) classified in low-dimensional topology. Isomorphisms of the topological plane are all continuous bijections. The topological plane is the natural context for the branch of graph theory that deals with planar graphs, and results such as the four color theorem.

The plane may also be viewed as an affine space, whose isomorphisms are combinations of translations and non-singular linear maps. From this viewpoint there are no distances, but collinearity and ratios of distances on any line are preserved.

Differential geometry views a plane as a 2-dimensional real manifold, a topological plane which is provided with a differential structure. Again in this case, there is no notion of distance, but there is now a concept of smoothness of maps, for example a differentiable or smooth path (depending on the type of differential structure applied). The isomorphisms in this case are bijections with the chosen degree of differentiability.

In the opposite direction of abstraction, we may apply a compatible field structure to the geometric plane, giving rise to the complex plane and the major area of complex analysis. The complex field has only two isomorphisms that leave the real line fixed, the identity and conjugation.

In the same way as in the real case, the plane may also be viewed as the simplest, one-dimensional (over the complex numbers) complex manifold, sometimes called the complex line. However, this viewpoint contrasts sharply with the case of the plane as a 2-dimensional real manifold. The isomorphisms are all conformal bijections of the complex plane, but the only possibilities are maps that correspond to the composition of a multiplication by a complex number and a translation.

In addition, the Euclidean geometry (which has zero curvature everywhere) is not the only geometry that the plane may have. The plane may be given a spherical geometry by using the stereographic projection. This can be thought of as placing a sphere on the plane (just like a ball on the floor), removing the top point, and projecting the sphere onto the plane from this point). This is one of the projections that may be used in making a flat map of part of the Earth's surface. The resulting geometry has constant positive curvature.

Alternatively, the plane can also be given a metric which gives it constant negative curvature giving the hyperbolic plane. The latter possibility finds an application in the theory of special relativity in the simplified case where there are two spatial dimensions and one time dimension. (The hyperbolic plane is a timelike hypersurface in three-dimensional Minkowski space.)

The one-point compactification of the plane is homeomorphic to a sphere (see stereographic projection) the open disk is homeomorphic to a sphere with the "north pole" missing adding that point completes the (compact) sphere. The result of this compactification is a manifold referred to as the Riemann sphere or the complex projective line. The projection from the Euclidean plane to a sphere without a point is a diffeomorphism and even a conformal map.

The plane itself is homeomorphic (and diffeomorphic) to an open disk. For the hyperbolic plane such diffeomorphism is conformal, but for the Euclidean plane it is not.

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