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Two Planes Intersect At A

Written By Tuesday, November 1, 2022 Add Comment Edit

In this explainer, we will learn how to find the points and lines of intersection betwixt lines and planes in space.

Definition: The Full general Grade of the Equation of a Airplane

A plane in 3D space, ℝ , can exist described in many dissimilar ways. For example, the general equation of a plane is given by 𝑎 𝑥 + 𝑏 𝑦 + 𝑐 𝑧 + 𝑑 = 0 .

This airplane has a normal vector ⃑ 𝑛 = ( 𝑎 , 𝑏 , 𝑐 ) , which defines the plane's orientation in 3D space. This normal vector is non unique. Any other nonzero scalar multiple of this vector, 𝜆 ⃑ 𝑛 , is also normal to the plane.

The additional constant 𝑑 has no effect on the plane'southward orientation but translates the aeroplane 𝑑 units in the management of the normal vector ⃑ 𝑛 .

For example, for the plane described by the equation 𝑥 + ii 𝑦 + 3 𝑧 + i 0 = 0 , 1 𝑥 + ii 𝑦 + 3 𝑧 + 1 0 = 0 , a normal vector ⃑ 𝑛 to the plane is ( one , 2 , iii ) . Whatsoever nonzero scalar multiple of this vector is also a normal vector to the airplane, for example, ( − 1 , − two , − 3 ) or ( 5 , 1 0 , 1 5 ) .

Definition: Intersection of Planes

Whatever two planes in ℝ with nonparallel normal vectors will intersect over a direct line.

This line is the ready of solutions to the simultaneous equations of the planes: 𝑎 𝑥 + 𝑏 𝑦 + 𝑐 𝑧 + 𝑑 = 0 , 𝑎 𝑥 + 𝑏 𝑦 + 𝑐 𝑧 + 𝑑 = 0 .

This system of 2 equations has three unknowns: 𝑥 , 𝑦 , and 𝑧 . Therefore, the arrangement will have either infinitely many solutions or no solutions at all. The quondam describes the example where the ii planes intersect, and the latter describes the example where the two planes are parallel and never intersect.

How To: Finding the Full general Equation of a Line of Intersection betwixt 2 Planes

  1. Eliminate one of the three variables (it does not affair which ane, but choose 𝑧 for instance) from the 2 equations, and express one of the two remaining variables explicitly in terms of the other, for example, 𝑥 = 𝑓 ( 𝑦 ) .
  2. Eliminate the dependent variable, 𝑦 , from the original two equations and express the contained variable, 𝑥 , in terms of the remaining variable, 𝑧 , so 𝑥 = 𝑔 ( 𝑧 ) .
  3. The general equation of the line of intersection is so given by 𝑥 = 𝑓 ( 𝑦 ) = 𝑔 ( 𝑧 ) .

Permit'due south consider an example of finding the line of intersection between ii planes:

𝑥 − 4 𝑦 + 3 𝑧 − 4 = 0 , 2 𝑥 + 2 𝑦 − ix 𝑧 + vii = 0 . ( 1 ) ( two )

Showtime, nosotros need to eliminate one of the three variables. We tin eliminate 𝑧 by multiplying equation (1) past 3 and calculation information technology to equation (two) , which gives 3 𝑥 − 1 2 𝑦 + 9 𝑧 − 1 2 = 0 + 2 𝑥 + 2 𝑦 − 9 𝑧 + 7 = 0 5 𝑥 − 1 0 𝑦 − 5 = 0 .

We can rearrange for 𝑥 by adding 1 0 𝑦 + 5 to both sides and dividing past five, which gives

Now nosotros need to eliminate the dependent variable, 𝑦 , from the original two equations to find an expression for 𝑥 in terms of 𝑧 . We can multiply the second equation by two and add together it to the first, which gives 𝑥 − iv 𝑦 + 3 𝑧 − 4 = 0 + 4 𝑥 + 4 𝑦 − i viii 𝑧 + 1 4 = 0 5 𝑥 − 1 5 𝑧 + 1 0 = 0 .

We can rearrange for 𝑥 by adding ( ane 5 𝑧 − 1 0 ) to both sides and dividing by 5, which gives 𝑥 = 3 𝑧 − 2 .

Together with equation (3), nosotros now accept two expressions for 𝑥 , one in terms of 𝑦 and 1 in terms of 𝑧 : 𝑥 = two 𝑦 + 1 , 𝑥 = iii 𝑧 − 2 .

These two equations can exist rewritten as ane equation with two equalities: 𝑥 = 2 𝑦 + 1 = 3 𝑧 − two .

This is the general equation of a line in 3D infinite.

We cannot reduce the organization of equations any farther than this, or observe values for 𝑥 , 𝑦 , and 𝑧 that uniquely solve the equations, because we take i more unknown than the number of equations. However, we are free to choose whatsoever value for ane variable, which has corresponding values to the other 2 variables that solve the equations.

For case, setting 𝑥 = 1 in the equation to a higher place gives 1 = 2 𝑦 + 1 = three 𝑧 − ii .

From the outset part of the equation, nosotros can rearrange to requite 𝑦 = 0 , and from the second office, we can rearrange to give 𝑧 = 1 .

Therefore, from setting 𝑥 = 1 , nosotros accept 𝑦 = 0 and 𝑧 = 1 , which gives one signal of intersection between the 2 planes: ( 1 , 0 , 1 ) .

Likewise, we could set 𝑥 = 2 , from which we would obtain 𝑦 = one two and 𝑧 = 4 3 , giving another point of intersection betwixt the two planes 2 , i 2 , 4 3 .

We do non, of course, demand to choose 𝑥 for the variable to prepare as a parameter. We could only as freely choose 𝑦 or 𝑧 . For example, choosing 𝑦 = 1 in the main equation to a higher place gives 𝑥 = two ( ane ) + ane = 3 𝑧 − 2 , which in plow tin be rearranged to give 𝑥 = three and 𝑧 = v 3 , so 3 , 1 , 5 3 is another point on the line of intersection between the 2 planes.

These are just some of the infinitely many solutions to the system of equations that form the line of intersection betwixt the 2 planes.

The general form is non the only mode of describing a line of intersection between ii planes. Some other way is with a set up of parametric equations—using an external parameter that defines the iii variables 𝑥 , 𝑦 , and 𝑧 separately.

Definition: The Parametric Form of the Equation of a Line in 3D Space

A line in 3D infinite may be divers past the full general set of parametric equations 𝑥 = 𝑓 ( 𝑡 ) = 𝑥 + 𝑎 𝑡 , 𝑦 = 𝑔 ( 𝑡 ) = 𝑦 + 𝑏 𝑡 , 𝑧 = ℎ ( 𝑡 ) = 𝑧 + 𝑐 𝑡 , where 𝑡 is a parameter; 𝑥 , 𝑦 , and 𝑧 are the coordinates of a indicate lying on the line; and 𝑎 , 𝑏 , and 𝑐 are the components of the direction vector of the line or parallel to the line.

Since there are infinitely many points on the line, in that location are infinitely many choices of ( 𝑥 , 𝑦 , 𝑧 ) for the parametric equation of the line.

How To: Finding the Parametric Equation of a Line of Intersection between Two Planes

  1. Express one of the three variables in the equations of the two planes as a linear function of a parameter, 𝑡 , for example, 𝑥 = 𝑥 + 𝑎 𝑡 .
  2. Substitute this expression into the original equations of the planes, and solve the arrangement of equations to limited the other ii variables in terms of the parameter, 𝑡 .

Let'due south expect at an example of amalgam a set of parametric equations for a line of intersection given the general equations of ii planes.

Example 1: Finding the Parametric Equation of the Line of Intersection of Two Planes

Detect the parametric equations of the line of intersection between the 2 planes 𝑥 + 𝑧 = three and 2 𝑥 − 𝑦 − 𝑧 = − two .

  1. 𝑥 = iii + 𝑡 , 𝑦 = iv + 3 𝑡 , 𝑧 = − 𝑡 a n d
  2. 𝑥 = 3 + 𝑡 , 𝑦 = 8 − 3 𝑡 , 𝑧 = − 𝑡 a northward d
  3. 𝑥 = 3 + 𝑡 , 𝑦 = 8 + 3 𝑡 , 𝑧 = − 𝑡 a n d
  4. 𝑥 = 1 + 𝑡 , 𝑦 = 3 + 3 𝑡 , 𝑧 = ii − 𝑡 a n d
  5. 𝑥 = 3 + 𝑡 , 𝑦 = 4 − 3 𝑡 , 𝑧 = − 𝑡 a northward d

Reply

One approach to solving this question is to choose a parametric equation to correspond one of our variables. We can do this as we do, in fact, accept a "free variable." In terms of how we get about choosing the parametric equation for the variable, we tin can practice this in a couple of different ways. We tin can either cull a general parametrization, for example, 𝑥 = 𝑥 + 𝑎 𝑡 , and and then fix values for 𝑥 and 𝑎 at a stage of the calculation that is convenient, or we tin fix the parametrization at the start of our calculation, for example, 𝑧 = 𝑡 , and adjust our reply at the end every bit required. We will demonstrate both methods here.

Method one: Directly Fixing a Parametrization

If we reference the options presented in the question, it might seem sensible to set 𝑧 = − 𝑡 equally it seems likely that we would and so country on the correct answer. However, we volition fix 𝑧 = 𝑡 and so demonstrate that this will in fact give us an equivalent line of intersection, which we can then "tweak" to determine the correct respond from the provided options.

If we substitute our chosen parameter for 𝑧 into the equation for the kickoff plane, nosotros go 𝑥 + 𝑡 = 3 , which gives 𝑥 = 3 − 𝑡 . If nosotros now substitute 𝑥 and 𝑧 into the equation for our 2d plane, we get ii ( 3 − 𝑡 ) − 𝑦 − 𝑡 = − two .

If nosotros distribute over the parentheses and simplify, we get 6 − 2 𝑡 − 𝑡 + 2 = 𝑦 , which simplifies to 𝑦 = 8 − 3 𝑡 . Therefore, the parametric equations for 𝑥 , 𝑦 , and 𝑧 are 𝑥 = 3 − 𝑡 , 𝑦 = viii − 3 𝑡 , 𝑧 = 𝑡 . a n d

As we can run across from the question, this is not really one of our options, simply information technology must be equivalent to ane of the options. We have described a line that passes through the point ( three , eight , 0 ) with management vector ( − 1 , − iii , i ) and need to identify which of the options is equivalent to this course of the line. To do this, we can first compare the direction vectors of each of the lines and then identify which of the points described besides lies on our line.

In this particular case, this is not too difficult to do. We can quickly discount options B and E due to the inconsistent signs of 3 𝑡 and 𝑡 when compared with the direction vector of our line. The remaining options have a direction vector that is a multiple of − one of our line and is, therefore, equivalent.

We can and so discount option A equally it passes through the same 𝑥 -coordinate but a different 𝑦 -coordinate, which leaves options C and D. Choice C is described by the same indicate, ( 3 , 8 , 0 ) , and then it must be a solution.

Finally, we demand to check whether option D is also a solution. We can do this by checking whether ( 3 , 8 , 0 ) is a point on this detail line: substituting the value 𝑡 = two into each of the parametric equations leads us to the betoken ( iii , 9 , 0 ) . Therefore, this is not a valid equation for the line of intersection.

Therefore, the answer is pick C.

Method 2: Using a Full general Parametrization

Call up that the full general form for the gear up of parametric equations for a line in 3D space is given by 𝑥 = 𝑓 ( 𝑡 ) = 𝑥 + 𝑎 𝑡 , 𝑦 = 𝑔 ( 𝑡 ) = 𝑦 + 𝑏 𝑡 , 𝑧 = ℎ ( 𝑡 ) = 𝑧 + 𝑐 𝑡 , where 𝑡 is a parameter; 𝑥 , 𝑦 , and 𝑧 are the coordinates of a point lying on the line; and 𝑎 , 𝑏 , and 𝑐 are the components of the direction vector of the line or parallel to the line.

To find the set of parametric equations for the line of intersection, we fix an expression for one variable in terms of the parameter, substitute this expression into the equations of the planes, and and then rearrange the resulting equations to observe expressions for the other two variables in terms of the parameter.

Allow 𝑥 = 𝑥 + 𝑎 𝑡 .

Substituting this expression into the equations of the planes gives

𝑥 + 𝑎 𝑡 + 𝑧 = three , ii ( 𝑥 + 𝑎 𝑡 ) − 𝑦 − 𝑧 = − 2 . ( 4 ) ( 5 )

We now have ii simultaneous equations for 𝑦 and 𝑧 , which can be "solved" to give expressions for 𝑦 and 𝑧 in terms of 𝑡 .

From equation (4), we tin can rearrange to give an expression for 𝑧 in terms of 𝑡 : 𝑧 = 3 − 𝑥 − 𝑎 𝑡 .

And substituting this expression for 𝑧 into equation (v) gives two ( 𝑥 + 𝑎 𝑡 ) − 𝑦 − ( iii − 𝑥 − 𝑎 𝑡 ) = − 2 .

Distributing over the parentheses and rearranging for 𝑦 gives an expression for 𝑦 in terms of 𝑡 : 𝑦 = iii ( 𝑥 + 𝑎 𝑡 ) − 1 .

We tin can now choose values for 𝑥 and 𝑎 at our convenience to make the equations every bit simple every bit possible.

We cannot choose 𝑎 = 0 , because the parameter would so be constant and not uniquely define every indicate on the line, but we can choose any value of 𝑥 we like.

From the listing of possible answers, four of them have the parametric equation for 𝑥 as 𝑥 = 3 + 𝑡 , so let'south try this 1. This ways that we take 𝑥 = three , 𝑎 = 1 .

Substituting these values of 𝑥 and 𝑎 into the expressions for 𝑦 and 𝑧 gives 𝑦 = three ( iii + 𝑡 ) − ane = 8 + 3 𝑡 .

And 𝑦 = 3 − 3 − 𝑡 = − 𝑡 .

We then have one possible set of parametric equations for 𝑥 , 𝑦 , and 𝑧 : 𝑥 = 3 + 𝑡 , 𝑦 = viii + three 𝑡 , 𝑧 = − 𝑡 , a n d which matches with answer C.

This confirms the answer that we constitute in method i, option C.

A concluding mode of describing the line of intersection betwixt 2 planes is with a vector equation.

Definition: The Vector Form of the Equation of a Line in 3D Infinite

A line in 3D space may be defined in vector form by the general equation ⃑ 𝑟 = ⃑ 𝑟 + 𝑡 ⃑ 𝑑 , where ⃑ 𝑟 = ( 𝑥 , 𝑦 , 𝑧 ) is the position vector of a known betoken on the line, ⃑ 𝑑 is a nonzero vector parallel to the line, and 𝑡 is a scalar.

How To: Finding the Vector Equation of a Line of Intersection between Two Planes

  1. Notice the position vector, ⃑ 𝑟 , of a single point that lies in both planes. This can be done by setting the value of i variable, for case, 𝑥 = 𝑥 , and solving the equations of the two planes to detect the corresponding values of the other 2 variables, 𝑦 = 𝑦 and 𝑧 = 𝑧 .
  2. Decide normal vectors to each plane, ⃑ 𝑛 and ⃑ 𝑛 , past reading off the coefficients from their equations.
  3. Take the cross product of the normal vectors, ⃑ 𝑑 = ⃑ 𝑛 × ⃑ 𝑛 , to give a vector, ⃑ 𝑑 , parallel to the line of intersection between the planes.

  4. The vector equation of the line of intersection is then given past ⃑ 𝑟 = ⃑ 𝑟 + 𝑡 ⃑ 𝑑 , where 𝑡 is a scalar.

Let's expect at an example of using the cross product to observe the direction vector of the line of intersection between ii planes, and then the vector equation of that line.

Instance two: Finding the Vector Equation of the Line of Intersection of Two Planes

Find the vector equation of the line of intersection between the ii planes 𝑥 + iii 𝑦 + ii 𝑧 − 6 = 0 and 2 𝑥 − 𝑦 + 𝑧 + 2 = 0 .

  1. ⃑ 𝑟 = ( 0 , 2 , − 1 2 ) + 𝑡 ( 5 , 3 , − 7 )
  2. ⃑ 𝑟 = ( 0 , 1 4 , one 2 ) + 𝑡 ( 2 , − iii , 2 )
  3. ⃑ 𝑟 = ( 0 , 2 , 0 ) + 𝑡 ( 2 , − 3 , two )
  4. ⃑ 𝑟 = ( 0 , 2 , 0 ) + 𝑡 ( 5 , 3 , − 7 )
  5. ⃑ 𝑟 = ( 0 , 1 iv , i two ) + 𝑡 ( v , 3 , − 7 )

Answer

To find the vector equation of the line of intersection between the ii planes, we demand to notice the position vector, ⃑ 𝑟 , of a point that lies in both planes and and so observe a nonzero direction vector ⃑ 𝑑 parallel to the line of intersection. The vector equation of the line is then given by ⃑ 𝑟 = ⃑ 𝑟 + 𝑡 ⃑ 𝑑 , where 𝑡 is a scalar.

Allow'south start with finding the position vector, ⃑ 𝑟 , of a point that lies in both planes. Nosotros begin by choosing ane variable every bit a parameter and setting it to a value of our choice.

Since all of the possible answers given have a constant vector with an 𝑥 component of zero, information technology makes sense to set 𝑥 = 0 .

Let 𝑥 = 0 .

In the equations of the ii planes, this gives 3 𝑦 + ii 𝑧 − 6 = 0 , − 𝑦 + 𝑧 + two = 0 .

If we practice non have given possible answers, it is possible that our pick of value for a variable volition be invalid. For instance, if the line of intersection lies parallel to the 𝑦 𝑧 -plane, the value of 𝑥 will exist constant along the line and probably not equal to the value chosen. If this is the case, even so, it will exist obvious on replacing the value we accept chosen in the equations of the 2 planes, since there will be no solutions for a betoken in both planes with a value that prevarication on the line of intersection.

This is not the case here, so we at present have two equations for 𝑦 and 𝑧 that can be solved simultaneously. From the equation of the second aeroplane, 𝑦 = 𝑧 + 2 .

Substituting this expression for 𝑦 into the equation for the first plane gives 3 ( 𝑧 + 2 ) + 2 𝑧 − 6 = 0 .

Distributing over the parentheses and rearranging for 𝑧 gives 𝑧 = 0 .

From the equation in a higher place, 𝑦 = 𝑧 + ii , and then we have 𝑦 = 2 .

Then, the position vector of 1 point on the line of intersection between the planes is ⃑ 𝑟 = ( 0 , 2 , 0 ) .

We now need to observe a direction vector parallel to the line of intersection betwixt the two planes. We can do this past taking the cantankerous product (or cross product) of the normal vectors of each plane.

We can find normal vectors to the two planes merely by reading off the coefficients of the variables in their equations 1 𝑥 + 3 𝑦 + 2 𝑧 − 6 = 0 , 2 𝑥 − 1 𝑦 + 1 𝑧 + 2 = 0 .

Therefore, two normal vectors to the planes are ⃑ 𝑛 = ( 1 , 3 , 2 ) and ⃑ 𝑛 = ( two , − 1 , 1 ) respectively.

We can now evaluate the cantankerous production ⃑ 𝑛 × ⃑ 𝑛 by taking the determinant of the matrix: ⃑ 𝑖 ⃑ 𝑗 ⃑ 𝑘 1 3 ii 2 − i 1 .

Evaluating the determinant, | | | | ⃑ 𝑖 ⃑ 𝑗 ⃑ 𝑘 1 3 two 2 − 1 i | | | | = ⃑ 𝑖 | | 3 2 − 1 1 | | − ⃑ 𝑗 | | 1 2 2 i | | + ⃑ 𝑘 | | 1 3 2 − 1 | | = ⃑ 𝑖 ( three ⋅ one − ii ⋅ ( − ane ) ) − ⃑ 𝑗 ( ane ⋅ 1 − 2 ⋅ 2 ) + ⃑ 𝑘 ( 1 ⋅ ( − 1 ) − 3 ⋅ 2 ) = ⃑ 𝑖 ( 3 + 2 ) − ⃑ 𝑗 ( 1 − iv ) + ⃑ 𝑘 ( − 1 − 6 ) = 5 ⃑ 𝑖 + iii ⃑ 𝑗 − vii ⃑ 𝑘 = ( 5 , 3 , − 7 ) .

Thus, we have the direction vector for the line of intersection between the ii planes: ⃑ 𝑑 = ( five , three , − 7 ) .

Hence, the vector equation of the line of intersection betwixt the two planes is given by ⃑ 𝑟 = ⃑ 𝑟 + 𝑡 ⃑ 𝑑 = ( 0 , two , 0 ) + 𝑡 ( five , 3 , − 7 ) .

This is option D.

Definition: Point of Intersection betwixt a Line and a Plane

A line and a nonparallel plane will intersect at a unmarried indicate.

This point is the unique solution of the equation of the line and the equation of the airplane.

The equation of the aeroplane, 𝑎 𝑥 + 𝑏 𝑦 + 𝑐 𝑧 + 𝑑 = 0 , is one equation, and the equation of the line, 𝑎 𝑥 + 𝑥 = 𝑏 𝑦 + 𝑦 = 𝑐 𝑧 + 𝑧 , can be rewritten equally ii distinct equations: 𝑎 𝑥 + 𝑥 = 𝑏 𝑦 + 𝑦 , 𝑎 𝑥 + 𝑥 = 𝑐 𝑧 + 𝑧 .

This is a organisation of three distinct equations for three unknowns and therefore volition have either no solutions (if the line and plane are parallel and practise not intersect), one unique solution (if the line and airplane are not coplanar and intersect), or infinitely many solutions (if the line and aeroplane are coplanar).

Every bit with any system of 𝑛 equations for 𝑛 unknowns, there are multiple methods of solution.

Example 3: Finding the Intersection of a Line and a Plane given Their General Equations

Find the point of intersection of the straight line − 3 𝑥 = 4 𝑦 − two = 𝑧 + 1 and the aeroplane − three 𝑥 + 𝑦 + 𝑧 = 1 3 .

Answer

The point of intersection ( 𝑥 , 𝑦 , 𝑧 ) between a line and a plane volition exist given by the unique solution to the organisation of equations of the straight line and the aeroplane. In that location are multiple methods of solution. For this example, we will solve the equations algebraically.

We brainstorm by rewriting the equation of the line equally two distinct equations, both involving 𝑧 : − 3 𝑥 = 𝑧 + i , four 𝑦 − 2 = 𝑧 + 1 .

Rearranging these ii equations gives 𝑥 and 𝑦 explicitly in terms of 𝑧 : 𝑥 = − 1 3 ( 𝑧 + 1 ) , 𝑦 = ane 4 ( 𝑧 + three ) .

Substituting these expressions for 𝑥 and 𝑦 into the equation for the plane gives an equation merely in 𝑧 , which we can solve for 𝑧 : − 3 − i three ( 𝑧 + i ) + 1 4 ( 𝑧 + three ) + 𝑧 = 1 iii .

Distributing over the parentheses and simplifying gives 𝑧 + 1 + 𝑧 4 + 3 4 + 𝑧 = 1 3 nine 𝑧 four = 4 five iv 𝑧 = 5 .

Substituting this value for 𝑧 into the equations for 𝑥 and 𝑦 , 𝑥 = − i 3 ( 5 + 1 ) = − two . 𝑦 = 1 4 ( 5 + 3 ) = 2 .

Therefore, the point of intersection between the line and the plane is ( − two , ii , 5 ) .

The point of intersection between a line and a plane may also be found given their vector equations.

Definition: The Vector Form of the Equation of a Aeroplane

A airplane may be divers by a vector equation of the form ⃑ 𝑛 ⋅ ⃑ 𝑟 = 𝑐 , where ⃑ 𝑟 is the position vector of a general signal on the plane, ⃑ 𝑛 is a constant vector that is normal to the aeroplane, and 𝑐 is a abiding scalar.

Also recall that the vector equation of a line in ℝ is given past ⃑ 𝑟 = ⃑ 𝑟 + 𝑡 ⃑ 𝑑 , where ⃑ 𝑟 is the position vector of a betoken on the line, ⃑ 𝑑 is any nonzero vector parallel to the line, and 𝑡 is a scalar.

The value of the scalar parameter 𝑡 uniquely defines every point on the line, then the point of intersection between the line and the plane will be given by a unique value of 𝑡 . This value of 𝑡 may be found by setting the general position vector ⃑ 𝑟 in the equation of the airplane equal to the general position vector ⃑ 𝑟 = ⃑ 𝑟 + 𝑡 ⃑ 𝑑 in the equation of the line, since at the point of intersection (if it exists) the position vectors will be the aforementioned.

Therefore, we need to find the value of 𝑡 that solves the equation: ⃑ 𝑛 ⋅ ⃑ 𝑟 + 𝑡 ⃑ 𝑑 = 𝑐 .

Let'southward look at an example of using this method to find the betoken of intersection between a line and a plane in 3D infinite given their vector equations.

Example 4: Finding the Coordinates of the Intersection Point of a Straight Line and a Airplane

Notice the coordinates of the point of intersection of the straight line ⃑ 𝑟 = ( eight , 2 , − five ) + 𝑡 ( − 7 , − 9 , one 3 ) with the plane ( 9 , 4 , − five ) ⋅ ⃑ 𝑟 = − v 9 .

Answer

If the line and the plane intersect, at that place must be a unique value of 𝑡 for which the vector ⃑ 𝑟 is equal in both the equation of the line and the plane.

We brainstorm by rewriting the vector equation of the line in terms of ane vector: ⃑ 𝑟 = ( viii − 7 𝑡 , 2 − 9 𝑡 , − v + 1 3 𝑡 ) .

At the betoken of intersection, the position vector ⃑ 𝑟 will be the same in both equations, and then we can substitute the vector ⃑ 𝑟 from the equation of the line into the equation of the plane. This gives ( nine , iv , − five ) ⋅ ( eight − 7 𝑡 , 2 − 9 𝑡 , − 5 + 1 iii 𝑡 ) = − 5 9 .

Expanding the scalar production, ix ( 8 − 7 𝑡 ) + 4 ( 2 − 9 𝑡 ) − 5 ( − 5 + 1 3 𝑡 ) = − 5 9 .

Simplifying and solving for 𝑡 , vii 2 − half-dozen 3 𝑡 + 8 − three 6 𝑡 + ii 5 − 6 5 𝑡 = − five ix − 1 6 iv 𝑡 = − 1 6 iv 𝑡 = 1 .

This is the value of 𝑡 at the signal of intersection between the line and the plane. Substituting this into the equation of the line, ⃑ 𝑟 = ( viii − seven , 2 − ix , − 5 + 1 3 ) = ( 1 , − 7 , 8 ) .

Therefore, the point of intersection betwixt the line and the plane is ( 1 , − 7 , eight ) .

For three singled-out planes in 3D space, there is a much broader range of possible scenarios.

  1. If all three planes are parallel, there is no intersection between whatever of them.

  2. If two planes are parallel to each other and a 3rd is not, then this third plane will intersect the other ii planes over two separate lines of intersection.

  3. If all iii planes are nonparallel to each other, they may intersect at a single point.

  4. Also, if all the planes are non-parallel, they may intersect along a line.

  5. If all iii planes are nonparallel, the third airplane may too intersect with the other two planes separately, giving iii lines of intersection that are parallel to each other.

Let'south expect at an case of finding the single point of intersection between three planes in scenario 𝑐 above.

Example v: Finding the Point of Intersection of 3 Planes

Find the point of intersection of the planes − 5 𝑥 − 2 𝑦 + 6 𝑧 − 1 = 0 , − 7 𝑥 + eight 𝑦 + 𝑧 − 6 = 0 , and 𝑥 − 3 𝑦 + iii 𝑧 + 1 1 = 0 .

Respond

In this case, it is given that there is a single indicate of intersection between the three planes. Since a bespeak of intersection satisfies the equations of all three planes, there is a unique solution to the system of iii equations.

Like any system of linear equations, there are multiple methods of solution.

Method one: Geometric Approach

Ane method to notice the betoken of intersection between the three planes is to first find the line of intersection between the starting time ii planes and so discover the point of intersection between this line and the third plane.

We tin can practice this by finding the parametric equation for the line of intersection between the starting time 2 planes, expressing 𝑥 , 𝑦 , and 𝑧 in terms of a parameter, 𝑡 . Nosotros tin then substitute these expressions for 𝑥 , 𝑦 , and 𝑧 into the equation for the third plane and solve the resulting equation to requite the value of 𝑡 . Substituting this value of 𝑡 into the parametric equation for the line volition give the 𝑥 -, 𝑦 -, and 𝑧 -coordinates of the point of intersection between all iii planes.

Consider the full general equations for the first 2 planes: − v 𝑥 − 2 𝑦 + 6 𝑧 − i = 0 , − 7 𝑥 + 8 𝑦 + 𝑧 − 6 = 0 .

Nosotros can discover the parametric equation for the line of intersection between these two planes by setting one variable equal to parameter 𝑡 and and then solving the resulting equations to give expressions for the other two variables in terms of 𝑡 .

Permit 𝑧 = 𝑡 .

Substituting this expression for 𝑧 into the equations of the two planes gives

− 5 𝑥 − 2 𝑦 + 6 𝑡 − ane = 0 , − seven 𝑥 + viii 𝑦 + 𝑡 − vi = 0 . ( 6 ) ( vii )

Nosotros at present need to eliminate one variable from the equations. Multiplying equation (6) past four and adding it to equation (7) gives − two vii 𝑥 + two five 𝑡 − one 0 = 0 .

Solving for 𝑥 , 𝑥 = two five 𝑡 − 1 0 2 seven .

Now, we can substitute this expression for 𝑥 into equation (6) and solve for 𝑦 : − five 2 5 𝑡 − 1 0 ii 7 − 2 𝑦 + 6 𝑡 − 1 = 0 𝑦 = − 5 + half-dozen 𝑡 − ane 2 𝑦 = three seven 𝑡 + 2 3 v 4 .

And then, we now have the set up of 𝑥 , 𝑦 , and 𝑧 values that lie on the line of intersection between the first two planes expressed in terms of parameter 𝑡 . If we now substitute these expressions for 𝑥 , 𝑦 , and 𝑧 into the equation of the third plane, we tin can solve for 𝑡 , giving the value of 𝑡 at the point of intersection between all 3 planes.

The equation of the third aeroplane is given past 𝑥 − 3 𝑦 + 3 𝑧 + 1 1 = 0 .

Substituting in the parametric expressions for 𝑥 , 𝑦 , and 𝑧 , two 5 𝑡 − 1 0 2 seven − three 3 7 𝑡 + 2 3 5 4 + 3 𝑡 + 1 1 = 0 .

Now, solving for 𝑡 , 5 0 𝑡 − ii 0 v four − 1 1 1 𝑡 + half dozen ix v 4 + 1 six 2 𝑡 v 4 + v nine 4 5 iv = 0 5 0 𝑡 − 2 0 − ( 1 1 one 𝑡 + half dozen nine ) + 1 6 ii 𝑡 + 5 9 4 = 0 1 0 i 𝑡 + five 0 v = 0 𝑡 = − v .

Substituting this value of 𝑡 into the parametric equations for 𝑥 , 𝑦 , and 𝑧 gives 𝑥 = 2 5 𝑡 − 1 0 2 7 = − 1 2 5 − 1 0 ii 7 = − v , 𝑦 = iii seven 𝑡 + 2 3 5 4 = − 1 viii five + 2 3 five 4 = − iii , 𝑧 = 𝑡 = − 5 .

Therefore, the point of intersection between all three planes is ( − 5 , − 3 , − five ) .

Method 2: Cramer's Rule

We begin by rewriting the system of equations every bit a matrix equation of the form 𝐴 𝑋 = 𝐵 : − five 𝑥 − 2 𝑦 + 6 𝑧 − 1 = 0 , − 7 𝑥 + eight 𝑦 + 𝑧 − 6 = 0 , 𝑥 − iii 𝑦 + 3 𝑧 + one 1 = 0 .

Taking the constants − i , − vi , and eleven to the right-hand side and rewriting the left-manus side as the production of a matrix 𝐴 and the solution matrix, 𝑋 = 𝑥 𝑦 𝑧 , we and so take − 5 − 2 six − seven viii 1 1 − 3 3 𝑥 𝑦 𝑧 = ane half dozen − 1 one .

At present, Cramer's dominion tells us that 𝑥 = Δ Δ , 𝑦 = Δ Δ , 𝑧 = Δ Δ is the unique solution to this system of equations, where Δ is the determinant of the matrix of coefficients, 𝐴 and Δ is the determinant of the matrix formed by replacing the column of 𝐴 associated with 𝑥 (the first column) with matrix 𝐵 .

It is worth noting here that the three planes will intersect at a single point if and only if the determinant of the matrix, Δ , is nonzero. This is equivalent to the existence of a unique solution to the system of equations.

Since Δ , the determinant of the unchanged matrix 𝐴 , is common to all three equations, let'southward evaluate this showtime: Δ = | | | | − v − 2 6 − 7 8 ane ane − 3 3 | | | | = − 5 | | 8 1 − 3 iii | | + 2 | | − vii 1 1 three | | + six | | − seven 8 ane − three | | = − 5 ( 2 iv − ( − 3 ) ) + two ( − 2 1 − 1 ) + 6 ( ii one − viii ) = − 1 3 5 − iv iv + 7 8 = − one 0 1 .

Although this was given in the question, we have now confirmed that the 3 planes must intersect at a single point, since the determinant Δ is nonzero.

At present, to find Δ , we find the determinant of the matrix formed by replacing the column of 𝐴 associated with 𝑥 with matrix 𝐵 on the correct-hand side: Δ = | | | | 1 − 2 6 6 8 1 − 1 i − 3 3 | | | | = 1 | | 8 one − 3 iii | | + 2 | | half-dozen 1 − 1 1 three | | + 6 | | 6 eight − 1 i − 3 | | = ( 8 ⋅ iii − 1 ⋅ ( − 3 ) ) + two ( 6 ⋅ 3 − 1 ⋅ ( − 1 one ) ) + vi ( 6 ⋅ ( − 3 ) − viii ⋅ ( − 1 1 ) ) = 2 7 + 5 8 + 4 2 0 = 5 0 5 .

Substituting this value of Δ into Cramer's rule, 𝑥 = Δ Δ = 5 0 5 − 1 0 one = − v .

Nosotros tin can follow the same procedure for 𝑦 and 𝑧 : Δ = | | | | − 5 1 6 − 7 six 1 1 − one 1 iii | | | | = − 5 | | 6 1 − 1 one 3 | | − ane | | − 7 i 1 iii | | + six | | − 7 6 1 − i one | | = − 5 ( 6 ⋅ three − ane ⋅ ( − 1 1 ) ) − ( − 7 ⋅ 3 − 1 ⋅ 1 ) + 6 ( − 7 ⋅ ( − i 1 ) − half dozen ⋅ one ) = − 1 4 5 + 2 2 + four ii half dozen = 3 0 3 .

Substituting this value of Δ into Cramer's rule, 𝑦 = Δ Δ = 3 0 iii − 1 0 one = − iii .

And finally, for 𝑧 , Δ = | | | | − 5 − 2 1 − 7 viii 6 1 − iii − ane ane | | | | = − v | | eight 6 − 3 − 1 1 | | + 2 | | − 7 vi i − 1 1 | | + ane | | − vii eight 1 − iii | | = − v ( 8 ⋅ ( − one 1 ) − 6 ⋅ ( − 3 ) ) + 2 ( − seven ⋅ ( − 1 ane ) − 6 ⋅ 1 ) + ( − 7 ⋅ ( − three ) − 8 ⋅ 1 ) = iii 5 0 + 1 4 2 + 1 iii = 5 0 five .

Substituting this value of Δ into Cramer'southward rule, 𝑧 = Δ Δ = v 0 five − ane 0 1 = − five .

So, we have 𝑥 = − 5 , 𝑦 = − 3 , and 𝑧 = − five . This is the unique solution to the equations of the iii planes. Therefore, the signal of intersection betwixt the three planes is ( − v , − 3 , − 5 ) .

We conclude our discussion of points and lines of intersection between lines and planes in ℝ by noting some primal points.

Key Points

  • Two nonparallel planes in ℝ will intersect over a straight line, which is the one-dimensionally parametrized fix of solutions to the equations of both planes.
  • The direction vector, ⃑ 𝑑 , of the line of intersection of two planes may be given by the cross product of the normal vectors of the planes, ⃑ 𝑛 × ⃑ 𝑛 .
  • A line and a nonparallel plane in ℝ will intersect at a unmarried point, which is the unique solution to the equation of the line and the equation of the airplane.
  • Three nonparallel planes volition intersect at a unmarried signal if and but if there exists a unique solution to the system of equations of the three planes. When written as a matrix equation, this equates to the determinant of the coefficient matrix beingness invertible, that is, Δ ≠ 0 . If the determinant of the coefficient matrix is cipher, then the planes exercise non intersect at a unique bespeak, if at all.

Two Planes Intersect At A,

Source: https://www.nagwa.com/en/explainers/435186018068/

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