Lesson Video: Equations of Parallel and Perpendicular Planes | Nagwa Lesson Video: Equations of Parallel and Perpendicular Planes | Nagwa

Lesson Video: Equations of Parallel and Perpendicular Planes Mathematics

In this video, we will learn how to find the equation of a plane that is parallel or perpendicular to another plane given its equation or some properties.

17:12

Video Transcript

In this video, our topic is equations of parallel and perpendicular planes. Weβll see how planes, two-dimensional surfaces in three-dimensional space, can be parallel or perpendicular to one another and how to recognize these conditions mathematically.

As we get started, letβs recall a bit about just how a plane is defined. The two pieces of information we often have about a plane that let us uniquely define it in three-dimensional space are, first, a point that lies on the plane and, second, a vector β weβve called it π§ β that is normal or perpendicular to the plane. In general, given a vector thatβs normal to a plane where that vector has components π, π, and π and a point π zero on the plane with coordinates π₯ zero, π¦ zero, and π§ zero, we can say that our normal vector dotted with the vector that goes to a general point on the plane with coordinates π₯, π¦, π§ is equal to that same normal vector dotted with a vector going to our point π zero.

And we can say that this line here represents the vector form or the vector equation of our plane, while if we multiply out the dot products, then we get this expression here. This is called the scalar form or scalar equation of a plane. And then going a bit further, if we group the terms on the right-hand side together, calling them π, then we arrive at an equation for our plane thatβs sometimes called the general or normal form. So anyway, given a vector thatβs normal to the surface of a plane and a point that is in the plane, weβre able to generate various expressions for the equation of the plane. And it turns out, not only does the normal vector to a plane help us define that planes equation, it also helps us figure out what other mathematical planes this one might be parallel or perpendicular to.

For example, say that we have one plane with a normal vector like this and another plane with this normal vector. If we call these vectors π§ one and π§ two and we assume that these two planes are parallel to each other, then we can say something about the way that normal vector π§ one relates to normal vector π§ two. If the two planes are parallel, then these normal vectors must also be parallel or possibly antiparallel. Either way, we could write the one normal vector in terms of a constant, whether positive or negative, multiplied by the other normal vector.

So just for example, letβs say that our normal vector π§ one has components one, two, three, while our normal vector π§ two has these components. Comparing these vectors, we see that we could multiply the π₯-component of π§ one by negative two to get the π₯-component of π§ two. And likewise, we could multiply the π¦-component of π§ one by that same factor negative two to get the π¦-component of π§ two. And the same is true for the π§-components of these vectors. In this case then, we would say that our constant πΆ is equal to negative two. And because there is such a constant by which we can multiply π§ two so that it equals π§ one, we say that these two planes are parallel.

This expression telling us that the normal vector of one plane equals a constant times the normal vector of a second plane is the condition for two planes being parallel. If two planes have normal vectors that satisfy this condition for any constant value πΆ, then we can safely say that the planes are parallel. But now letβs imagine two planes that are oriented in a different way. Say that our two planes are now perpendicular to one another. Well, since their normal vectors are perpendicular to the surface of each plane, respectively, we can expect that if the planes are perpendicular, then π§ one and π§ two are perpendicular as well. And in general, if two vectors are perpendicular, that means if we take the dot product of those vectors, the result weβll find is zero.

Geometrically, this means that thereβs no overlap between the two vectors weβre dotting together. So, given the normal vectors of two planes, if those normal vectors have a dot product of zero, then the planes must be perpendicular to one another. And these two equations, one for parallel planes and one for perpendicular, give us helpful tests to apply once we know the normal vectors of two planes to see whether they are parallel or perpendicular. The best way to really learn all this is through some practice. So letβs look now at an example exercise.

Given that the plane πΎπ§ plus two π₯ plus three π¦ equals negative four is parallel to the plane πΏπ¦ minus two π₯ minus two π§ equals three, find the values of πΎ and πΏ.

Okay, so in this exercise we have these two planes given by these equations. If we write them in standard form, the first one is two π₯ plus three π¦ plus πΎπ§, where πΎ is some constant value, is all equal to negative four. And the second is negative two π₯ plus πΏπ¦, where πΏ is some unknown constant, minus two times π§ is equal to three. Because these equations are now written in standard form, we can pick out the components of the normal vectors to each plane. The normal vector of the first plane will have an π₯-component of two, a π¦-component of three, and a π§-component of πΎ. Weβll call this vector π§ one and write it out this way. And then for the second plane, its normal vector will have an π₯-component of negative two, a π¦-component of πΏ, and a π§-component of negative two. And weβll call this vector π§ two.

Our problem statement tells us that these two planes are parallel to one another. This fact can point us to recalling the mathematical condition for two planes to be parallel. Two parallel planes with normal vectors π§ one and π§ two have their components related by a constant; weβve called it πΆ. This constant could be positive or negative, but whatever its sign, it means that there is a consistent ratio between the components of these two normal vectors. Knowing this, letβs look at our two normal vectors π§ one and π§ two and see what this constant of proportionality πΆ might be.

Comparing the π₯-components of these vectors, we see the π₯-component of π§ one is two and that of π§ two is negative two. We know that two is equal to negative one times negative two. And in fact, this is the only value by which we can multiply negative two to yield positive two which tells us that πΆ, in this case, is negative one. And this fact is the key for solving for the values of πΎ and πΏ in our normal vector equations because just as we can write this equation out for the π₯-components of our normal vectors, so we can write it out for the π¦- and π§-components.

If we substitute negative one in for πΆ in both of these equations, then we see that three is equal to negative one times πΏ, telling us that πΏ is equal to negative three. And since πΎ is equal to negative one times negative two, that means πΎ is equal to two. So weβve used the fact that our two planes are parallel to solve for the unknown parts of their normal vectors. πΏ is equal to negative three, and πΎ is equal to positive two.

Letβs look now at a second parallel plane example.

Find the equation of the plane passing through the point π, π, π and parallel to the plane π₯ plus π¦ plus π§ equals zero.

All right, so we want to solve for the equation of a plane, and we know that this plane passes through a point with coordinates π, π, π. So if we draw a sketch, letβs say that this is the plane whose equation we want to solve for and we know that somewhere on this plane is this point, π, π, π. Along with this, weβre told that this plane is parallel to another plane whose equation is given here. And note that this planeβs equation is given in such a form that weβre able to use it to solve for the components of the planeβs normal vector. Note that all three of the variables π₯, π¦, and π§ are effectively being multiplied by one. These multiplying factors, these ones, give us components of a vector thatβs normal to this plane.

If we call that vector π§, its π₯-component could be given by one, its π¦-component by one, and its π§-component by the same value. And itβs helpful to know this normal vector because weβre told that the plane whose equation we want to solve for is parallel to the plane whose equation is given here. And therefore, the normal vector of this unknown plane, weβll call this vector π§ sub π, must be parallel or antiparallel to π§. And in fact, we could set π§ sub π equal to π§ in general. And we do this based on our knowledge of the fact that the two planes weβre considering are parallel.

At this point, note that we have a vector thatβs normal to the plane whose equation we want to solve for. And we also have the coordinates of a point on that plane. Taken together, these two pieces of information are enough to let us solve for this planeβs equation. We can recall that given a normal vector to a plane as well as a vector to a random point on that plane, the dot product of those two vectors equals the dot product of the normal vector with a vector to a known point on the plane. To apply this relationship to our scenario, weβll say that our normal vector π§ sub π dotted with a vector that goes to any point on our plane with the general components π₯, π¦, and π§ is equal to our normal vector π§ sub π dotted with the vector to our known point π, π, π.

We can now substitute in the value for our normal vector π§ sub π. That gives us this equation here. And if we then carry out both of these dot product operations, we find that π₯ plus π¦ plus π§ is equal to π plus π plus π. And this result is the equation of the plane that passes through the point π, π, π and is parallel to the plane π₯ plus π¦ plus π§ equals zero.

Now letβs look at an example involving two planes that are perpendicular to one another.

Given that the plane three π₯ minus three π¦ minus three π§ equals one is perpendicular to the plane ππ₯ minus two π¦ minus π§ equals four, find the value of π.

Okay, in this exercise, weβre given equations of two planes and told that these planes are perpendicular to one another. And we can say that if two planes are perpendicular like these are, then it must also be the case that their normal vectors are perpendicular. And this points us to the mathematical condition that is satisfied for two perpendicular planes. The dot product of the normal vectors of these planes must be zero. This suggests that we might solve for the normal vectors of our two given planes and then apply this condition to see if it helps us solve for π. Since both of these equations are given in standard form, we can identify the components of each planeβs normal vector.

For the first plane, its normal vector has an π₯-component of three, a π¦-component of negative three, and a π§-component also of negative three. We can write that like this, calling this normal vector π§ one. For our second plane, its normal vector has an π₯-component of π, a π¦-component of negative two, and a π§-component of negative one. And weβll call this normal vector π§ two. Next, knowing that these two planes are perpendicular, weβll apply this condition. Weβll say that π§ one, this vector here, dotted with π§ two, this vector here, equals zero. If we then carry out this dot product operation, we find that three times π plus six plus three equals zero. And this means that three times the unknown value π is equal to negative nine. Dividing both sides by three, we find that π is equal to negative three. This is the value of π.

Letβs now look at another perpendicular planes example.

Find the general equation of the plane which passes through the point two, eight, one and is perpendicular to the two planes negative six π₯ negative four π¦ plus six π§ equals negative five and five π₯ plus three π¦ minus six π§ equals three.

Okay, so in this exercise, weβre told about three planes, two of which we have the equations for. The third plane whose general equation we want to solve for is perpendicular to these two and it also passes through this given point. Now, in general, we can write the equation for a plane if we have two bits of information about it. First, if we know a vector that is normal to the surface of the plane and, second, if we know a point on the plane. In this scenario, we see that, indeed, we are given a point which lies on this unknown plane. But we donβt yet know the components of a vector normal to it.

However, notice this. We do know that this plane, whatever its equation, is perpendicular to these two planes given by these equations. This implies that the normal vector π§ we want to solve for is also perpendicular to the normal vectors of these two planes, meaning if we can solve for the normal vectors of these planes and then find a vector perpendicular to them both, weβll have an expression for π§. We can begin doing this by recalling that when the equation of a plane is given in whatβs called standard form, like we see it here, then the components of vectors normal to those planes are given by the factors by which we multiply the variables π₯, π¦, and π§.

For example, if we call a vector normal to our first plane π§ one, then this may have components negative six, negative four, six. And if we call a vector normal to our second plane π§ two, this will have components five, three, negative six. And now that we know π§ one and π§ two, recall that π§ is perpendicular to both of these vectors. So hereβs what we can do. We can take the cross product of π§ one and π§ two. And if we recall that the cross product of two vectors β weβve called them π and π, which are three-dimensional β is equal to the determinant of this three-by-three matrix, then we can apply this relationship to our scenario, where the second and third rows of our matrix are populated by the respective components of our vectors π§ one and π§ two.

We see that the π’-component of π§ one is negative six, the π£-component is negative four, and the π€-component is positive six. Likewise, the π’-component of π§ two is five, its π£-component is three, and its π€-component negative six. Weβre now ready to calculate this cross product. And weβll do it component by component, starting with π’. The π’-component of the vector resulting from crossing π§ one and __ two is equal to the determinant of this two-by-two matrix. Thatβs a negative four times negative six or 24 minus six times three or 18. We then move on to the π£-component, which is negative the determinant of this two-by-two matrix. Negative six times negative six is positive 36. And we subtract from that six times five or 30.

And lastly then, we compute the π€-component equal to the determinant of this two-by-two matrix. Negative six times three is negative 18 minus negative four times five or negative 20. Adding these three components together, we get the cross product of π§ one and π§ two. Thatβs equal to six π’ minus six π£ plus two π€ or, written in vector form, six, negative six, two. Now that we know this cross product, letβs recall that we were calculating it because we knew that π§ is perpendicular to π§ one and π§ two. And since this vector resulting from crossing π§ one and π§ two is perpendicular to them both by the rule of the cross product, we can say that itβs equal to the normal vector π§.

Now that we know this and we also know a point on our plane π, we can make quick progress in finding the general equation of the plane. We can write the equation of our plane as the normal vector π§ dotted with a vector to a general point on the plane with coordinates π₯, π¦, and π§ being equal to that same normal vector dotted with the vector to the known point on the plane with coordinates two, eight, one. And since we know the vector form of π§, we could substitute that into this equation. And having done that, if we then calculate both of these dot products, we find that six π₯ minus six π¦ plus two π§ equals 12 minus 48 plus two. This is equal to negative 34.

So if we add 34 to both sides of our equation, we can write that six π₯ minus six π¦ plus two π§ plus 34 equals zero. And noticing that all the terms on the left are divisible by two, we find that three π₯ minus three π¦ plus π§ plus 17 equals zero. And this is the general equation of the plane which passes through the point two, eight, one and is perpendicular to the two planes given.

Letβs finish up now by summarizing a few key points. In this lesson, we saw that two planes are parallel when their normal vectors π§ one and π§ two satisfy this condition, that they are a constant multiple of one another. We also learned that two planes are perpendicular when their normal vectors π§ one and π§ two satisfy the condition that their dot product equals zero. And lastly, we saw that if two planes are parallel or perpendicular, so are their corresponding normal vectors.