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how to find where two planes intersect

Written By Saturday, May 21, 2022 Add Comment Edit

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

Definition: The General Form of the Equation of a Plane

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

This airplane has a normal vector ⃑ 𝑛 = ( 𝑎 , 𝑏 , 𝑐 ) , which defines the aeroplane's orientation in 3D space. This normal vector is not unique. Whatsoever other nonzero scalar multiple of this vector, 𝜆 ⃑ 𝑛 , is also normal to the plane.

The additional constant 𝑑 has no upshot on the plane's orientation only translates the plane 𝑑 units in the direction of the normal vector ⃑ 𝑛 .

For instance, for the plane described by the equation 𝑥 + 2 𝑦 + 3 𝑧 + i 0 = 0 , ane 𝑥 + 2 𝑦 + three 𝑧 + 1 0 = 0 , a normal vector ⃑ 𝑛 to the plane is ( 1 , 2 , iii ) . Any nonzero scalar multiple of this vector is besides a normal vector to the airplane, for example, ( − one , − 2 , − 3 ) or ( v , 1 0 , 1 5 ) .

Definition: Intersection of Planes

Any two planes in ℝ with nonparallel normal vectors volition intersect over a directly line.

This line is the set 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 two equations has three unknowns: 𝑥 , 𝑦 , and 𝑧 . Therefore, the system will have either infinitely many solutions or no solutions at all. The erstwhile describes the instance where the two planes intersect, and the latter describes the case where the ii planes are parallel and never intersect.

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

  1. Eliminate one of the iii variables (it does non matter which one, only choose 𝑧 for example) from the two equations, and express ane of the 2 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 limited the contained variable, 𝑥 , in terms of the remaining variable, 𝑧 , so 𝑥 = 𝑔 ( 𝑧 ) .
  3. The full general equation of the line of intersection is then given by 𝑥 = 𝑓 ( 𝑦 ) = 𝑔 ( 𝑧 ) .

Permit's consider an case of finding the line of intersection between ii planes:

𝑥 − 4 𝑦 + 3 𝑧 − 4 = 0 , ii 𝑥 + 2 𝑦 − 9 𝑧 + 7 = 0 . ( 1 ) ( ii )

Offset, nosotros need to eliminate i of the three variables. We tin eliminate 𝑧 by multiplying equation (1) past 3 and adding it to equation (2) , which gives 3 𝑥 − one 2 𝑦 + 9 𝑧 − ane ii = 0 + two 𝑥 + 2 𝑦 − 9 𝑧 + seven = 0 five 𝑥 − 1 0 𝑦 − 5 = 0 .

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

Now we need to eliminate the dependent variable, 𝑦 , from the original two equations to find an expression for 𝑥 in terms of 𝑧 . We tin can multiply the 2nd equation past ii and add information technology to the first, which gives 𝑥 − 4 𝑦 + three 𝑧 − 4 = 0 + 4 𝑥 + 4 𝑦 − 1 viii 𝑧 + one 4 = 0 five 𝑥 − 1 5 𝑧 + ane 0 = 0 .

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

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

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

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

Nosotros cannot reduce the system of equations any further than this, or find values for 𝑥 , 𝑦 , and 𝑧 that uniquely solve the equations, considering we take ane more unknown than the number of equations. However, we are gratis to choose whatever value for one variable, which has respective values to the other ii variables that solve the equations.

For example, setting 𝑥 = 1 in the equation above gives i = 2 𝑦 + ane = 3 𝑧 − 2 .

From the beginning part of the equation, we tin rearrange to requite 𝑦 = 0 , and from the second function, we can rearrange to give 𝑧 = 1 .

Therefore, from setting 𝑥 = i , we have 𝑦 = 0 and 𝑧 = i , which gives one point of intersection betwixt the 2 planes: ( one , 0 , 1 ) .

Likewise, we could prepare 𝑥 = 2 , from which we would obtain 𝑦 = 1 2 and 𝑧 = four 3 , giving another betoken of intersection between the two planes two , 1 two , 4 3 .

We practice not, of class, need to cull 𝑥 for the variable to set as a parameter. We could merely equally freely choose 𝑦 or 𝑧 . For example, choosing 𝑦 = 1 in the chief equation above gives 𝑥 = 2 ( 1 ) + 1 = 3 𝑧 − ii , which in turn tin be rearranged to requite 𝑥 = three and 𝑧 = 5 iii , then 3 , 1 , 5 iii is another betoken on the line of intersection between the 2 planes.

These are only some of the infinitely many solutions to the system of equations that course the line of intersection between the ii planes.

The general form is not the only fashion of describing a line of intersection between two planes. Another way is with a set of parametric equations—using an external parameter that defines the three variables 𝑥 , 𝑦 , and 𝑧 separately.

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

A line in 3D space may be divers by the 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 point lying on the line; and 𝑎 , 𝑏 , and 𝑐 are the components of the management vector of the line or parallel to the line.

Since there are infinitely many points on the line, there 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 betwixt Two Planes

  1. Express one of the 3 variables in the equations of the ii 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 organisation of equations to express the other ii variables in terms of the parameter, 𝑡 .

Permit's look at an case of constructing a set of parametric equations for a line of intersection given the general equations of 2 planes.

Example 1: Finding the Parametric Equation of the Line of Intersection of 2 Planes

Detect the parametric equations of the line of intersection between the two planes 𝑥 + 𝑧 = 3 and 2 𝑥 − 𝑦 − 𝑧 = − 2 .

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

Answer

1 approach to solving this question is to cull a parametric equation to represent i of our variables. Nosotros can practise this every bit we exercise, in fact, take a "free variable." In terms of how we go about choosing the parametric equation for the variable, we can do this in a couple of different means. We can either cull a general parametrization, for example, 𝑥 = 𝑥 + 𝑎 𝑡 , and then set values for 𝑥 and 𝑎 at a stage of the calculation that is convenient, or we can fix the parametrization at the first of our adding, for case, 𝑧 = 𝑡 , and adjust our respond at the end as 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 as it seems likely that nosotros would then state on the correct reply. However, we will set 𝑧 = 𝑡 and so demonstrate that this volition in fact give us an equivalent line of intersection, which we can then "tweak" to determine the correct answer from the provided options.

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

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

Equally we can see from the question, this is not actually one of our options, but it must be equivalent to one of the options. We have described a line that passes through the betoken ( three , viii , 0 ) with direction vector ( − one , − iii , 1 ) and demand to identify which of the options is equivalent to this form of the line. To do this, we can first compare the direction vectors of each of the lines then place which of the points described as well lies on our line.

In this particular instance, this is not also hard to do. We can chop-chop discount options B and E due to the inconsistent signs of three 𝑡 and 𝑡 when compared with the direction vector of our line. The remaining options have a direction vector that is a multiple of − ane of our line and is, therefore, equivalent.

We tin so discount choice A as it passes through the same 𝑥 -coordinate but a unlike 𝑦 -coordinate, which leaves options C and D. Choice C is described by the aforementioned point, ( iii , viii , 0 ) , then it must be a solution.

Finally, we need to bank check whether pick D is also a solution. We can do this by checking whether ( 3 , 8 , 0 ) is a indicate on this detail line: substituting the value 𝑡 = 2 into each of the parametric equations leads us to the signal ( three , 9 , 0 ) . Therefore, this is not a valid equation for the line of intersection.

Therefore, the answer is selection C.

Method 2: Using a General Parametrization

Recall that the full general course for the set up of parametric equations for a line in 3D infinite is given past 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 management vector of the line or parallel to the line.

To find the set of parametric equations for the line of intersection, we set up an expression for one variable in terms of the parameter, substitute this expression into the equations of the planes, and and so 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 = iii , 2 ( 𝑥 + 𝑎 𝑡 ) − 𝑦 − 𝑧 = − ii . ( 4 ) ( v )

Nosotros at present take two simultaneous equations for 𝑦 and 𝑧 , which can exist "solved" to give expressions for 𝑦 and 𝑧 in terms of 𝑡 .

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

And substituting this expression for 𝑧 into equation (5) gives 2 ( 𝑥 + 𝑎 𝑡 ) − 𝑦 − ( three − 𝑥 − 𝑎 𝑡 ) = − two .

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

We can now cull values for 𝑥 and 𝑎 at our convenience to brand the equations equally uncomplicated as possible.

We cannot cull 𝑎 = 0 , because the parameter would then be constant and non uniquely define every signal on the line, but nosotros tin can choose whatever value of 𝑥 we like.

From the list of possible answers, four of them have the parametric equation for 𝑥 as 𝑥 = 3 + 𝑡 , so let's try this ane. This means that we have 𝑥 = iii , 𝑎 = 1 .

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

And 𝑦 = 3 − three − 𝑡 = − 𝑡 .

We then have one possible set up of parametric equations for 𝑥 , 𝑦 , and 𝑧 : 𝑥 = iii + 𝑡 , 𝑦 = 8 + three 𝑡 , 𝑧 = − 𝑡 , a n d which matches with answer C.

This confirms the answer that we plant in method 1, option C.

A final manner of describing the line of intersection between 2 planes is with a vector equation.

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

A line in 3D space may be divers in vector form by the general equation ⃑ 𝑟 = ⃑ 𝑟 + 𝑡 ⃑ 𝑑 , where ⃑ 𝑟 = ( 𝑥 , 𝑦 , 𝑧 ) is the position vector of a known point 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 betwixt Two Planes

  1. Detect the position vector, ⃑ 𝑟 , of a single point that lies in both planes. This can exist washed by setting the value of one variable, for example, 𝑥 = 𝑥 , and solving the equations of the two planes to notice the corresponding values of the other ii variables, 𝑦 = 𝑦 and 𝑧 = 𝑧 .
  2. Determine normal vectors to each aeroplane, ⃑ 𝑛 and ⃑ 𝑛 , by reading off the coefficients from their equations.
  3. Have the cross product of the normal vectors, ⃑ 𝑑 = ⃑ 𝑛 × ⃑ 𝑛 , to requite a vector, ⃑ 𝑑 , parallel to the line of intersection betwixt the planes.

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

Let's await at an example of using the cantankerous product to find the direction vector of the line of intersection betwixt two planes, and so the vector equation of that line.

Example 2: Finding the Vector Equation of the Line of Intersection of Two Planes

Find the vector equation of the line of intersection between the 2 planes 𝑥 + 3 𝑦 + two 𝑧 − 6 = 0 and 2 𝑥 − 𝑦 + 𝑧 + 2 = 0 .

  1. ⃑ 𝑟 = ( 0 , 2 , − ane 2 ) + 𝑡 ( 5 , 3 , − 7 )
  2. ⃑ 𝑟 = ( 0 , i 4 , i 2 ) + 𝑡 ( two , − three , 2 )
  3. ⃑ 𝑟 = ( 0 , ii , 0 ) + 𝑡 ( two , − iii , two )
  4. ⃑ 𝑟 = ( 0 , 2 , 0 ) + 𝑡 ( v , 3 , − vii )
  5. ⃑ 𝑟 = ( 0 , 1 4 , 1 ii ) + 𝑡 ( 5 , 3 , − seven )

Respond

To find the vector equation of the line of intersection between the ii planes, we need to find the position vector, ⃑ 𝑟 , of a point that lies in both planes and then notice a nonzero management 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.

Let'southward start with finding the position vector, ⃑ 𝑟 , of a bespeak that lies in both planes. We begin past choosing i variable as 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 aught, information technology makes sense to ready 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 nosotros do not take given possible answers, it is possible that our choice of value for a variable will be invalid. For instance, if the line of intersection lies parallel to the 𝑦 𝑧 -plane, the value of 𝑥 will be constant along the line and probably not equal to the value chosen. If this is the case, however, information technology will be obvious on replacing the value we take chosen in the equations of the 2 planes, since there will be no solutions for a point in both planes with a value that lie on the line of intersection.

This is not the instance here, and so we at present take two equations for 𝑦 and 𝑧 that can be solved simultaneously. From the equation of the 2d plane, 𝑦 = 𝑧 + two .

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

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

From the equation above, 𝑦 = 𝑧 + two , so nosotros have 𝑦 = 2 .

And so, the position vector of one point on the line of intersection between the planes is ⃑ 𝑟 = ( 0 , 2 , 0 ) .

We now need to find a direction vector parallel to the line of intersection between the ii planes. Nosotros can exercise this by taking the cross product (or cantankerous product) of the normal vectors of each airplane.

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

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

We tin now evaluate the cross product ⃑ 𝑛 × ⃑ 𝑛 by taking the determinant of the matrix: ⃑ 𝑖 ⃑ 𝑗 ⃑ 𝑘 1 3 2 2 − one 1 .

Evaluating the determinant, | | | | ⃑ 𝑖 ⃑ 𝑗 ⃑ 𝑘 i 3 two 2 − i i | | | | = ⃑ 𝑖 | | 3 2 − one 1 | | − ⃑ 𝑗 | | i 2 two 1 | | + ⃑ 𝑘 | | one 3 2 − one | | = ⃑ 𝑖 ( three ⋅ i − 2 ⋅ ( − 1 ) ) − ⃑ 𝑗 ( ane ⋅ 1 − 2 ⋅ two ) + ⃑ 𝑘 ( 1 ⋅ ( − 1 ) − 3 ⋅ two ) = ⃑ 𝑖 ( 3 + 2 ) − ⃑ 𝑗 ( ane − 4 ) + ⃑ 𝑘 ( − i − half dozen ) = 5 ⃑ 𝑖 + iii ⃑ 𝑗 − vii ⃑ 𝑘 = ( 5 , iii , − 7 ) .

Thus, we take the direction vector for the line of intersection between the two planes: ⃑ 𝑑 = ( five , 3 , − 7 ) .

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

This is option D.

Definition: Point of Intersection between a Line and a Airplane

A line and a nonparallel plane will intersect at a single bespeak.

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

The equation of the plane, 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 as two singled-out equations: 𝑎 𝑥 + 𝑥 = 𝑏 𝑦 + 𝑦 , 𝑎 𝑥 + 𝑥 = 𝑐 𝑧 + 𝑧 .

This is a organisation of three distinct equations for iii unknowns and therefore will take either no solutions (if the line and aeroplane are parallel and exercise not intersect), one unique solution (if the line and aeroplane are non coplanar and intersect), or infinitely many solutions (if the line and plane are coplanar).

As with any arrangement of 𝑛 equations for 𝑛 unknowns, there are multiple methods of solution.

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

Find the point of intersection of the straight line − three 𝑥 = 4 𝑦 − 2 = 𝑧 + i and the airplane − three 𝑥 + 𝑦 + 𝑧 = one 3 .

Reply

The bespeak of intersection ( 𝑥 , 𝑦 , 𝑧 ) between a line and a airplane volition be given past the unique solution to the system of equations of the straight line and the plane. At that place are multiple methods of solution. For this example, we will solve the equations algebraically.

We begin by rewriting the equation of the line as 2 singled-out equations, both involving 𝑧 : − 3 𝑥 = 𝑧 + 1 , 4 𝑦 − 2 = 𝑧 + one .

Rearranging these two equations gives 𝑥 and 𝑦 explicitly in terms of 𝑧 : 𝑥 = − ane iii ( 𝑧 + one ) , 𝑦 = one 4 ( 𝑧 + three ) .

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

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

Substituting this value for 𝑧 into the equations for 𝑥 and 𝑦 , 𝑥 = − 1 three ( v + one ) = − 2 . 𝑦 = 1 four ( 5 + iii ) = 2 .

Therefore, the point of intersection between the line and the plane is ( − two , 2 , 5 ) .

The point of intersection betwixt a line and a plane may as well exist establish given their vector equations.

Definition: The Vector Form of the Equation of a Airplane

A plane may be defined past a vector equation of the course ⃑ 𝑛 ⋅ ⃑ 𝑟 = 𝑐 , where ⃑ 𝑟 is the position vector of a general betoken on the plane, ⃑ 𝑛 is a constant vector that is normal to the plane, and 𝑐 is a constant scalar.

Besides recall that the vector equation of a line in ℝ is given by ⃑ 𝑟 = ⃑ 𝑟 + 𝑡 ⃑ 𝑑 , where ⃑ 𝑟 is the position vector of a signal 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, so the point of intersection betwixt 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 plane equal to the general position vector ⃑ 𝑟 = ⃑ 𝑟 + 𝑡 ⃑ 𝑑 in the equation of the line, since at the bespeak of intersection (if it exists) the position vectors will be the same.

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

Let'south await at an example of using this method to find the point of intersection between a line and a plane in 3D space given their vector equations.

Example 4: Finding the Coordinates of the Intersection Point of a Directly Line and a Plane

Observe the coordinates of the betoken of intersection of the directly line ⃑ 𝑟 = ( viii , two , − 5 ) + 𝑡 ( − seven , − 9 , 1 3 ) with the aeroplane ( 9 , 4 , − 5 ) ⋅ ⃑ 𝑟 = − 5 9 .

Answer

If the line and the plane intersect, there 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 begin by rewriting the vector equation of the line in terms of one vector: ⃑ 𝑟 = ( 8 − 7 𝑡 , 2 − 9 𝑡 , − 5 + 1 3 𝑡 ) .

At the point of intersection, the position vector ⃑ 𝑟 will exist the same in both equations, so we tin substitute the vector ⃑ 𝑟 from the equation of the line into the equation of the airplane. This gives ( 9 , 4 , − 5 ) ⋅ ( 8 − 7 𝑡 , 2 − 9 𝑡 , − 5 + 1 three 𝑡 ) = − v 9 .

Expanding the scalar product, nine ( 8 − 7 𝑡 ) + iv ( 2 − 9 𝑡 ) − 5 ( − 5 + one 3 𝑡 ) = − 5 ix .

Simplifying and solving for 𝑡 , seven ii − vi 3 𝑡 + 8 − three 6 𝑡 + 2 five − 6 v 𝑡 = − 5 nine − 1 6 four 𝑡 = − i 6 4 𝑡 = 1 .

This is the value of 𝑡 at the bespeak of intersection between the line and the plane. Substituting this into the equation of the line, ⃑ 𝑟 = ( 8 − seven , 2 − 9 , − 5 + one 3 ) = ( 1 , − 7 , viii ) .

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

For iii distinct 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 whatsoever of them.

  2. If two planes are parallel to each other and a third is non, then this third plane volition intersect the other two planes over two separate lines of intersection.

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

  4. Also, if all the planes are not-parallel, they may intersect forth a line.

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

Permit'southward look at an case of finding the single indicate of intersection between three planes in scenario 𝑐 to a higher place.

Case five: Finding the Point of Intersection of Three Planes

Find the signal of intersection of the planes − v 𝑥 − ii 𝑦 + 6 𝑧 − 1 = 0 , − 7 𝑥 + 8 𝑦 + 𝑧 − 6 = 0 , and 𝑥 − iii 𝑦 + iii 𝑧 + 1 1 = 0 .

Answer

In this example, information technology is given that in that location is a unmarried point of intersection between the 3 planes. Since a point of intersection satisfies the equations of all iii planes, there is a unique solution to the arrangement of three equations.

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

Method 1: Geometric Approach

1 method to find the point of intersection between the three planes is to first notice the line of intersection betwixt the offset ii planes then find the betoken of intersection between this line and the third airplane.

We can exercise this by finding the parametric equation for the line of intersection between the start 2 planes, expressing 𝑥 , 𝑦 , and 𝑧 in terms of a parameter, 𝑡 . Nosotros can so substitute these expressions for 𝑥 , 𝑦 , and 𝑧 into the equation for the third airplane and solve the resulting equation to give the value of 𝑡 . Substituting this value of 𝑡 into the parametric equation for the line will give the 𝑥 -, 𝑦 -, and 𝑧 -coordinates of the bespeak of intersection between all three planes.

Consider the general equations for the first two planes: − 5 𝑥 − 2 𝑦 + vi 𝑧 − 1 = 0 , − seven 𝑥 + viii 𝑦 + 𝑧 − six = 0 .

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

Let 𝑧 = 𝑡 .

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

− 5 𝑥 − 2 𝑦 + 6 𝑡 − 1 = 0 , − 7 𝑥 + viii 𝑦 + 𝑡 − 6 = 0 . ( half-dozen ) ( 7 )

We now demand to eliminate i variable from the equations. Multiplying equation (half-dozen) by 4 and adding it to equation (7) gives − ii 7 𝑥 + 2 5 𝑡 − ane 0 = 0 .

Solving for 𝑥 , 𝑥 = two v 𝑡 − one 0 two seven .

Now, we tin substitute this expression for 𝑥 into equation (half dozen) and solve for 𝑦 : − 5 two v 𝑡 − 1 0 ii 7 − two 𝑦 + six 𝑡 − i = 0 𝑦 = − 5 + half dozen 𝑡 − 1 2 𝑦 = 3 seven 𝑡 + ii iii 5 4 .

And then, we now have the set of 𝑥 , 𝑦 , and 𝑧 values that lie on the line of intersection between the beginning 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 aeroplane, we can solve for 𝑡 , giving the value of 𝑡 at the point of intersection between all iii planes.

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

Substituting in the parametric expressions for 𝑥 , 𝑦 , and 𝑧 , ii 5 𝑡 − 1 0 2 7 − 3 3 vii 𝑡 + ii 3 5 four + three 𝑡 + 1 i = 0 .

At present, solving for 𝑡 , v 0 𝑡 − 2 0 five four − 1 one i 𝑡 + 6 9 5 4 + one half-dozen two 𝑡 5 4 + v 9 4 5 4 = 0 5 0 𝑡 − ii 0 − ( one 1 one 𝑡 + vi 9 ) + ane 6 2 𝑡 + 5 nine 4 = 0 i 0 1 𝑡 + 5 0 5 = 0 𝑡 = − 5 .

Substituting this value of 𝑡 into the parametric equations for 𝑥 , 𝑦 , and 𝑧 gives 𝑥 = 2 v 𝑡 − ane 0 2 vii = − 1 2 v − 1 0 2 vii = − 5 , 𝑦 = 3 7 𝑡 + 2 3 5 4 = − ane 8 v + 2 three 5 iv = − 3 , 𝑧 = 𝑡 = − 5 .

Therefore, the point of intersection betwixt all three planes is ( − 5 , − three , − 5 ) .

Method 2: Cramer'southward Rule

We begin past rewriting the system of equations as a matrix equation of the course 𝐴 𝑋 = 𝐵 : − 5 𝑥 − ii 𝑦 + 6 𝑧 − 1 = 0 , − 7 𝑥 + viii 𝑦 + 𝑧 − 6 = 0 , 𝑥 − three 𝑦 + 3 𝑧 + ane 1 = 0 .

Taking the constants − 1 , − half dozen , and 11 to the correct-paw side and rewriting the left-hand side every bit the production of a matrix 𝐴 and the solution matrix, 𝑋 = 𝑥 𝑦 𝑧 , we so have − 5 − 2 6 − 7 8 i 1 − iii 3 𝑥 𝑦 𝑧 = 1 6 − 1 1 .

Now, Cramer's dominion tells u.s.a. 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 showtime column) with matrix 𝐵 .

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

Since Δ , the determinant of the unchanged matrix 𝐴 , is common to all three equations, let'due south evaluate this start: Δ = | | | | − five − ii half dozen − seven 8 one 1 − 3 3 | | | | = − 5 | | eight one − 3 three | | + 2 | | − 7 ane one 3 | | + 6 | | − 7 viii 1 − 3 | | = − 5 ( ii four − ( − 3 ) ) + ii ( − ii 1 − ane ) + 6 ( two 1 − 8 ) = − i iii five − 4 4 + 7 8 = − 1 0 1 .

Although this was given in the question, we take at present confirmed that the three planes must intersect at a unmarried indicate, since the determinant Δ is nonzero.

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

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

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

Substituting this value of Δ into Cramer'due south rule, 𝑦 = Δ Δ = 3 0 three − i 0 1 = − three .

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

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

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

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

Key Points

  • Two nonparallel planes in ℝ will intersect over a directly line, which is the one-dimensionally parametrized set of solutions to the equations of both planes.
  • The management vector, ⃑ 𝑑 , of the line of intersection of two planes may exist given past the cross product of the normal vectors of the planes, ⃑ 𝑛 × ⃑ 𝑛 .
  • A line and a nonparallel plane in ℝ will intersect at a single point, which is the unique solution to the equation of the line and the equation of the plane.
  • Iii nonparallel planes volition intersect at a unmarried point if and only if there exists a unique solution to the arrangement of equations of the three planes. When written equally a matrix equation, this equates to the determinant of the coefficient matrix being invertible, that is, Δ ≠ 0 . If the determinant of the coefficient matrix is zero, then the planes do not intersect at a unique point, if at all.

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

Posted by: barrettdetur1973.blogspot.com

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