Explanation:There are several ways to think about this. Here's one... Show
The general equation of a plane in three dimensional (Euclidean) space can be written (non-uniquely) in the form:
Given two planes, we have two linear equations in three variables:
Either these equations will be inconsistent, or they will have an infinite number of solutions. $\begingroup$ In $\Bbb R^3$ two distinct planes either intersect in a line or are parallel, in which case they have empty intersection; they cannot intersect in a single point. In $\Bbb R^n$ for $n>3$, however, two planes can intersect in a point. In $\Bbb R^4$, for instance, let $$P_1=\big\{\langle x,y,0,0\rangle:x,y\in\Bbb R\big\}$$ and $$P_2=\big\{\langle 0,0,x,y\rangle:x,y\in\Bbb R\big\}\;;$$ $P_1$ and $P_2$ are $2$-dimensional subspaces of $\Bbb R^4$, so they are planes, and their intersection $$P_1\cap P_2=\big\{\langle 0,0,0,0\rangle\big\}$$ consists of a single point, the origin in $\Bbb R^4$. Similar examples can easily be constructed in any $\Bbb R^n$ with $n>3$. answered Oct 23, 2012 at 7:27
Brian M. ScottBrian M. Scott 592k53 gold badges717 silver badges1182 bronze badges $\endgroup$ In this explainer, we will learn how to find the points and lines of intersection between lines and planes in space. A plane in 3D space, ℝ, can be described in many different ways. For example, the general equation of a plane is given by 𝑎𝑥+𝑏𝑦
+𝑐𝑧+𝑑=0. This plane has a normal vector ⃑𝑛=(𝑎,𝑏,𝑐), which defines the plane’s orientation in 3D space. This normal vector
is not unique. Any other nonzero scalar multiple of this vector, 𝜆⃑𝑛, is also normal to the plane. The additional constant
𝑑 has no effect on the plane’s orientation but translates the plane 𝑑 units in
the direction of the normal vector ⃑𝑛. For example, for the plane described by the equation
𝑥+2𝑦+3𝑧+10=0, 1𝑥+2𝑦+3𝑧+10=0, a normal vector ⃑𝑛 to the plane is
(1,2,3). Any nonzero scalar multiple of this vector is also a normal
vector to the plane, for example, (−1,−2,−3) or (5,10,15). Any two planes in ℝ with nonparallel normal vectors will intersect over a
straight line. This line is the set of solutions to the simultaneous equations of the planes: 𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑=0,𝑎𝑥+𝑏𝑦+𝑐𝑧+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 former describes the case where the two planes intersect, and the latter describes the case where the two planes are parallel and never intersect. Let’s consider an example of finding the line of intersection between two planes:
First, we need to eliminate one of the three variables. We can eliminate 𝑧 by multiplying equation (1) by 3 and adding it to equation (2) , which gives 3𝑥−12𝑦+9 𝑧−12=0+2𝑥+2𝑦− 9𝑧+7=05𝑥−10𝑦−5=0. We can rearrange for 𝑥 by adding 10𝑦+5 to both sides and dividing by 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 can multiply the second equation by 2 and add it to the first, which gives 𝑥−4𝑦+3𝑧−4= 0+4𝑥+4𝑦−18𝑧+1 4=05𝑥−15𝑧+10=0. We can rearrange for 𝑥 by adding (15𝑧−10) to both sides and dividing by 5, which gives 𝑥=3𝑧−2. Together with equation (3), we now have two expressions for 𝑥, one in terms of 𝑦 and one in terms of 𝑧 : 𝑥=2𝑦+1,𝑥=3𝑧−2. These two equations can be rewritten as one equation with two equalities: 𝑥 =2𝑦+1=3𝑧−2. This is the general equation of a line in 3D space. We cannot reduce the system of equations any further than this, or find values for 𝑥, 𝑦, and 𝑧 that uniquely solve the equations, because we have one more unknown than the number of equations. However, we are free to choose any value for one variable, which has corresponding values to the other two variables that solve the equations. For example, setting 𝑥=1 in the equation above gives 1=2𝑦+1=3𝑧−2. From the first part of the equation, we can rearrange to give 𝑦=0, and from the second part, we can rearrange to give 𝑧=1. Therefore, from setting 𝑥=1, we have 𝑦=0 and 𝑧=1, which gives one point of intersection between the two planes: ( 1,0,1). Likewise, we could set 𝑥= 2, from which we would obtain 𝑦=12 and 𝑧=43, giving another point of intersection between the two planes 2,12,43. We do not, of course, need to choose 𝑥 for the variable to set as a parameter. We could just as freely choose 𝑦 or 𝑧. For example, choosing 𝑦=1 in the main equation above gives 𝑥=2(1)+1=3𝑧−2, which in turn can be rearranged to give 𝑥 =3 and 𝑧=53, so 3,1,53 is another point on the line of intersection between the two planes. These are just some of the infinitely many solutions to the system of equations that form the line of intersection between the two planes. The general form is not the only way 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 Form of the Equation of a Line in 3D SpaceA line in 3D space may be defined by the general set of parametric equations 𝑥=𝑓(𝑡) =𝑥+𝑎𝑡,𝑦=𝑔(𝑡)=𝑦+𝑏𝑡,𝑧=ℎ(𝑡)=𝑧+𝑐𝑡, 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. 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 between Two Planes
Let’s look at an example of constructing a set of parametric equations for a line of intersection given the general equations of two planes. Example 1: Finding the Parametric Equation of the Line of Intersection of Two PlanesFind the parametric equations of the line of intersection between the two planes 𝑥+𝑧=3 and 2𝑥−𝑦 −𝑧=−2.
AnswerOne approach to solving this question is to choose a parametric equation to represent one of our variables. We can do this as we do, in fact, have 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 ways. We can either choose a general parametrization, for example, 𝑥=𝑥+𝑎𝑡, and then fix values for 𝑥 and 𝑎 at a stage of the calculation that is convenient, or we can fix the parametrization at the start of our calculation, for example, 𝑧=𝑡, and adjust our answer at the end as required. We will demonstrate both methods here. Method 1: 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 we would then land on the correct answer. However, we will set 𝑧=𝑡 and then demonstrate that this will 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 first plane, we get 𝑥+𝑡=3, which gives 𝑥=3−𝑡. If we now substitute 𝑥 and 𝑧 into the equation for our second plane, we get 2(3−𝑡)−𝑦−𝑡=−2. If we distribute over the parentheses and simplify, we get 6−2𝑡−𝑡+2=𝑦, which simplifies to 𝑦=8−3𝑡. Therefore, the parametric equations for 𝑥, 𝑦, and 𝑧 are 𝑥=3−𝑡,𝑦=8−3𝑡,𝑧=𝑡.and As 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 point (3,8,0) with direction vector (−1,−3,1) and need 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 and then identify which of the points described also 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 3𝑡 and 𝑡 when compared with the direction vector of our line. The remaining options have a direction vector that is a multiple of −1 of our line and is, therefore, equivalent. We can then discount option A as it passes through the same 𝑥-coordinate but a different 𝑦-coordinate, which leaves options C and D. Option C is described by the same point, (3,8,0), so it must be a solution. Finally, we need to check whether option D is also a solution. We can do this by checking whether (3,8,0) is a point on this particular line: substituting the value 𝑡=2 into each of the parametric equations leads us to the point (3,9,0). Therefore, this is not a valid equation for the line of intersection. Therefore, the answer is option C. Method 2: Using a General Parametrization Recall that the general form for the set of parametric equations for a line in 3D space is given by 𝑥=𝑓(𝑡)=𝑥+𝑎𝑡,𝑦=𝑔(𝑡)=𝑦+𝑏𝑡,𝑧=ℎ(𝑡)=𝑧+𝑐𝑡, 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 set an expression for one variable in terms of the parameter, substitute this expression into the equations of the planes, and then rearrange the resulting equations to find expressions for the other two variables in terms of the parameter. Let 𝑥=𝑥+𝑎𝑡. Substituting this expression into the equations of the planes gives
We now have two 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 can rearrange to give an expression for 𝑧 in terms of 𝑡: 𝑧=3 −𝑥−𝑎𝑡. And substituting this expression for 𝑧 into equation (5) gives 2(𝑥+𝑎𝑡)−𝑦−(3−𝑥−𝑎𝑡)=−2. Distributing over the parentheses and rearranging for 𝑦 gives an expression for 𝑦 in terms of 𝑡: 𝑦=3(𝑥+𝑎𝑡)−1. We can now choose values for 𝑥 and 𝑎 at our convenience to make the equations as simple as possible. We cannot choose 𝑎=0, because the parameter would then be constant and not uniquely define every point on the line, but we can choose any 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 one. This means that we have 𝑥=3,𝑎=1. Substituting these values of 𝑥 and 𝑎 into the expressions for 𝑦 and 𝑧 gives 𝑦=3(3+𝑡)−1=8+3𝑡. And 𝑦=3−3−𝑡=− 𝑡. We then have one possible set of parametric equations for 𝑥, 𝑦, and 𝑧: 𝑥=3+𝑡,𝑦=8 +3𝑡,𝑧=−𝑡,and which matches with answer C. This confirms the answer that we found in method one, option C. A final way of describing the line of intersection between two planes is with a vector equation. Definition: The Vector Form of the Equation of a Line in 3D SpaceA line in 3D space may be defined in vector form by the general equation ⃑𝑟=⃑𝑟+𝑡⃑𝑑, 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 between Two Planes
Let’s look at an example of using the cross product to find the direction vector of the line of intersection between two planes, and then the vector equation of that line. Example 2: Finding the Vector Equation of the Line of Intersection of Two PlanesFind the vector equation of the line of intersection between the two planes 𝑥+3𝑦+2𝑧−6=0 and 2𝑥−𝑦+𝑧+2=0.
AnswerTo find the vector equation of the line of intersection between the two planes, we need to find the position vector, ⃑𝑟, of a point that lies in both planes and then find a nonzero direction vector ⃑𝑑 parallel to the line of intersection. The vector equation of the line is then given by ⃑𝑟=⃑𝑟+𝑡⃑𝑑, where 𝑡 is a scalar. Let’s start with finding the position vector, ⃑𝑟, of a point that lies in both planes. We begin by choosing one 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 zero, it makes sense to set 𝑥=0. Let 𝑥=0. In the equations of the two planes, this gives 3𝑦+2𝑧−6=0, −𝑦+𝑧+2=0. If we do not have 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 𝑦𝑧-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, it will be obvious on replacing the value we have chosen in the equations of the two 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 case here, so we now have two equations for 𝑦 and 𝑧 that can be solved simultaneously. From the equation of the second plane, 𝑦=𝑧+2. Substituting this expression for 𝑦 into the equation for the first plane gives 3(𝑧+2)+2𝑧−6=0. Distributing over the parentheses and rearranging for 𝑧 gives 𝑧=0. From the equation above, 𝑦=𝑧+2, so we have 𝑦=2. 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 two planes. We can do this by taking the cross product (or cross product) of the normal vectors of each plane. We can find normal vectors to the two planes simply by reading off the coefficients of the variables in their equations 1𝑥+3𝑦+2𝑧−6=0,2𝑥−1𝑦+1𝑧+2=0. Therefore, two normal vectors to the planes are ⃑𝑛=(1,3,2) and ⃑𝑛=(2,−1,1) respectively. We can now evaluate the cross product ⃑𝑛×⃑𝑛 by taking the determinant of the matrix: ⃑𝑖⃑ 𝑗⃑𝑘1322−11. Evaluating the determinant, ||||⃑𝑖⃑𝑗⃑𝑘1322−11||||=⃑𝑖||32−11|| −⃑𝑗||1221||+⃑𝑘||132−1||=⃑𝑖(3⋅1−2⋅(−1))−⃑𝑗(1⋅1−2⋅2)+⃑𝑘(1⋅(−1)−3⋅ 2)=⃑𝑖(3+2)−⃑𝑗(1−4)+⃑𝑘(−1− 6)=5⃑𝑖+3⃑𝑗−7⃑𝑘=(5,3,−7). Thus, we have the direction vector for the line of intersection between the two planes: ⃑𝑑=(5,3,−7). Hence, the vector equation of the line of intersection between the two planes is given by ⃑𝑟=⃑𝑟+𝑡⃑𝑑=(0,2,0)+𝑡(5,3,−7). This is option D. Definition: Point of Intersection between a Line and a PlaneA line and a nonparallel plane will intersect at a single point. This point is the unique solution of the equation of the line and the equation of the plane. The equation of the plane, 𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑=0, is one equation, and the equation of the line, 𝑎𝑥+𝑥=𝑏𝑦+𝑦=𝑐𝑧+𝑧, can be rewritten as two distinct equations: 𝑎𝑥+𝑥=𝑏𝑦+𝑦,𝑎𝑥+𝑥=𝑐𝑧+𝑧. This is a system of three distinct equations for three unknowns and therefore will have either no solutions (if the line and plane are parallel and do not intersect), one unique solution (if the line and plane are not coplanar and intersect), or infinitely many solutions (if the line and plane are coplanar). As 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 EquationsFind the point of intersection of the straight line −3𝑥=4𝑦−2=𝑧+1 and the plane −3𝑥+𝑦+ 𝑧=13. AnswerThe point of intersection (𝑥,𝑦,𝑧) between a line and a plane will be given by the unique solution to the system of equations of the straight line and the plane. There are multiple methods of solution. For this example, we will solve the equations algebraically. We begin by rewriting the equation of the line as two distinct equations, both involving 𝑧: −3𝑥=𝑧+1,4𝑦−2=𝑧+1. Rearranging these two equations gives 𝑥 and 𝑦 explicitly in terms of 𝑧: 𝑥=−13(𝑧 +1),𝑦=14(𝑧+3). Substituting these expressions for 𝑥 and 𝑦 into the equation for the plane gives an equation only in 𝑧, which we can solve for 𝑧 : −3−13(𝑧+1)+14(𝑧+3)+𝑧=13. Distributing over the parentheses and simplifying gives 𝑧+1+𝑧4+34+𝑧=139𝑧4=454𝑧=5. Substituting this value for 𝑧 into the equations for 𝑥 and 𝑦, 𝑥=−13(5 +1)=−2.𝑦=14(5+3)=2. Therefore, the point of intersection between the line and the plane is (−2,2,5). 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 PlaneA plane may be defined by a vector equation of the form ⃑𝑛⋅⃑𝑟=𝑐, where ⃑𝑟 is the position vector of a general point on the plane, ⃑𝑛 is a constant vector that is normal to the plane, and 𝑐 is a constant scalar. Also recall that the vector equation of a line in ℝ is given by ⃑𝑟=⃑𝑟+𝑡⃑𝑑, where ⃑𝑟 is the position vector of a point 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 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 plane equal to the general position vector ⃑𝑟=⃑𝑟+𝑡⃑ 𝑑 in the equation of the line, since at the point of intersection (if it exists) the position vectors will be the same. Therefore, we need to find the value of 𝑡 that solves the equation: ⃑𝑛⋅⃑𝑟+𝑡⃑𝑑=𝑐. Let’s look 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 Straight Line and a PlaneFind the coordinates of the point of intersection of the straight line ⃑𝑟=(8,2,−5)+𝑡(−7,−9,13) with the plane (9,4,−5)⋅⃑𝑟=−59. AnswerIf 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=(8−7𝑡,2−9𝑡,−5+13𝑡). At the point of intersection, the position vector ⃑𝑟 will be the same in both equations, so we can substitute the vector ⃑𝑟 from the equation of the line into the equation of the plane. This gives (9,4,−5) ⋅(8−7𝑡,2−9𝑡,− 5+13𝑡)=−59. Expanding the scalar product, 9(8−7𝑡)+4 (2−9𝑡)−5(−5+13𝑡)=−59. Simplifying and solving for 𝑡, 72−63𝑡+ 8−36𝑡+25−65𝑡= −59−164𝑡=−164𝑡=1. This is the value of 𝑡 at the point of intersection between the line and the plane. Substituting this into the equation of the line, ⃑𝑟=(8−7,2−9,−5+13)=(1,−7,8). Therefore, the point of intersection between the line and the plane is (1,−7,8). For three distinct planes in 3D space, there is a much broader range of possible scenarios.
Let’s look at an example of finding the single point of intersection between three planes in scenario 𝑐 above. Example 5: Finding the Point of Intersection of Three PlanesFind the point of intersection of the planes −5𝑥−2𝑦+6𝑧−1=0, −7𝑥+ 8𝑦+𝑧−6=0, and 𝑥−3𝑦+3𝑧+11=0. AnswerIn this example, it is given that there is a single point of intersection between the three planes. Since a point of intersection satisfies the equations of all three planes, there is a unique solution to the system of three equations. Like any system of linear equations, there are multiple methods of solution. Method 1: Geometric Approach One method to find the point of intersection between the three planes is to first find the line of intersection between the first two planes and then find the point of intersection between this line and the third plane. We can do this by finding the parametric equation for the line of intersection between the first two planes, expressing 𝑥, 𝑦, and 𝑧 in terms of a parameter, 𝑡. We can then substitute these expressions for 𝑥, 𝑦, and 𝑧 into the equation for the third plane 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 point of intersection between all three planes. Consider the general equations for the first two planes: −5𝑥−2𝑦+6𝑧−1=0,−7𝑥+8𝑦+𝑧−6=0. We can find the parametric equation for the line of intersection between these two 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 two planes gives
We now need to eliminate one variable from the equations. Multiplying equation (6) by 4 and adding it to equation (7) gives −27𝑥+25𝑡−10=0. Solving for 𝑥, 𝑥=25𝑡−1027. Now, we can substitute this expression for 𝑥 into equation (6) and solve for 𝑦: −525𝑡−1027−2𝑦+6𝑡−1=0𝑦=−5+6𝑡−12𝑦=37𝑡+2354. So, we now have the set 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 can solve for 𝑡, giving the value of 𝑡 at the point of intersection between all three planes. The equation of the third plane is given by 𝑥−3𝑦+3𝑧+11=0. Substituting in the parametric expressions for 𝑥, 𝑦, and 𝑧, 25𝑡−102 7−337𝑡+2354 +3𝑡+11=0. Now, solving for 𝑡, 50𝑡−2054−1 11𝑡+6954+162𝑡54+59454=050𝑡−20−(111𝑡+69)+162𝑡+594=0101𝑡+505=0𝑡=−5. Substituting this value of 𝑡 into the parametric equations for 𝑥, 𝑦, and 𝑧 gives 𝑥=25𝑡−1027=−125−1027=−5,𝑦=37𝑡+2354=−185+2354=−3,𝑧=𝑡=−5. Therefore, the point of intersection between all three planes is (−5,−3,− 5). Method 2: Cramer’s Rule We begin by rewriting the system of equations as a matrix equation of the form 𝐴𝑋=𝐵: −5𝑥−2𝑦+6𝑧−1=0,−7𝑥+8𝑦+𝑧−6=0,𝑥−3𝑦+3𝑧+11=0. Taking the constants −1, −6, and 11 to the right-hand side and rewriting the left-hand side as the product of a matrix 𝐴 and the solution matrix, 𝑋= 𝑥𝑦𝑧, we then have −5−26−7811−33𝑥 𝑦𝑧=16−11. Now, Cramer’s rule tells us that 𝑥=ΔΔ,𝑦=ΔΔ,𝑧=ΔΔ 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’s evaluate this first: Δ=||||−5 −26−7811−33||||=−5||81−33||+2||−7113||+6||−781−3||=−5(24−(−3))+2(−21−1)+6(21−8)=−135−44+78=−101. Although this was given in the question, we have now confirmed that the three planes must intersect at a single point, since the determinant Δ is nonzero. Now, 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 right-hand side: Δ=| |||1−26681−11−33||||=1||81−33||+2||61−113||+6||68−11− 3||=(8⋅3−1⋅(−3))+2(6⋅3−1⋅(−11))+6(6⋅(−3)−8⋅(−11))=27+58+420=505. Substituting this value of Δ into Cramer’s rule, 𝑥=ΔΔ=505−101=−5. We can follow the same procedure for 𝑦 and 𝑧: Δ=||||−516−7611−113||||=−5||61−113||−1||−7113||+6||−761−11||=−5(6⋅3−1⋅(−11))−(−7⋅3−1⋅1)+6(−7⋅(−11)−6⋅1)=−145+22+426=303. Substituting this value of Δ into Cramer’s rule, 𝑦=ΔΔ=303−101=−3. And finally, for 𝑧, Δ=||||−5− 21−7861−3−11||||=−5||86−3−11||+2||−761−11||+1||−781−3||=−5(8⋅(−11)−6⋅(−3))+2(−7⋅(−11)−6⋅1)+(−7⋅(−3)−8⋅1)=350+142+13=505. Substituting this value of Δ into Cramer’s rule, 𝑧=ΔΔ=505−101=−5. So, we have 𝑥 =−5, 𝑦=−3, and 𝑧=−5. This is the unique solution to the equations of the three planes. Therefore, the point of intersection between the three planes is (−5,−3,−5). We conclude our discussion of points and lines of intersection between lines and planes in ℝ by noting some key points. Key Points
Can a line intersect a plane at only one point?If a line and a plane intersect one another, the intersection will be a single point, or a line (if the line lies in the plane). If we want to double-check ourselves, we can plug this coordinate point back into the equation of the plane.
How many planes can intersect at the same exact point?Points that lie in the same plane are said to be coplanar. Planes that intersect do so at a line, and it is possible for three planes to intersect at exactly one point.
Can a line intersect a plane at more than one point?1 Expert Answer. If a line lies within the plane, then it intersects at an infinite number of points.
Is the intersection of two planes always a line?Summary: The intersection of two planes is always a straight line.
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