Question Video: Finding the Distance between Two Planes | Nagwa Question Video: Finding the Distance between Two Planes | Nagwa

Question Video: Finding the Distance between Two Planes Mathematics • Third Year of Secondary School

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Find the distance between the two planes −𝑥 − 2𝑦 − 2𝑧 = −2 and −2𝑥 − 4𝑦 − 4𝑧 = 3.

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Video Transcript

Find the distance between the two planes negative 𝑥 minus two 𝑦 minus two 𝑧 is equal to negative two and negative two 𝑥 minus four 𝑦 minus four 𝑧 is equal to three.

In this question, we’re asked to find the distance between two given planes. And to do this, the first question we might ask is exactly what it means by the distance between two planes. Whenever we’re asked to find the distance between two planes or two lines or points and planes and points and lines, this always means the perpendicular distance. This is because the perpendicular distance between two of these objects will always be the shortest distance between these two objects. And it’s very useful to know the shortest distance between two objects.

But we don’t know how to find the perpendicular distance between two planes. So instead we’re going to have to get a picture of what’s happening. And in fact, there’s three possibilities. The first possibility would be we’re given two of exactly the same plane. Of course, if we’re given two of the exact same plane, then these planes intersect at every single point because they’re the same plane. So the distance will just be zero.

The second option we could be given is two planes which are not parallel. And just like with two lines that are not parallel, if two planes are not parallel, then they’ll intersect. In fact, they’ll intersect across an entire line. So in this case, the distance will also be equal to zero. One way of showing this would be to solve these equations as simultaneous equations. We could then find the equation of the line of intersection.

The third possibility would be that we’re given two parallel planes. And it’s worth pointing out here when we say that they’re parallel, we also mean that they’re not the same plane since we’ve already covered this in the first option. And this option is a little bit more difficult. How are we going to find the distance between two parallel planes? Because the planes are parallel, we know in particular they’re not going to intersect, so the distance between them must be bigger than zero. If we picked a point on either of our two planes, we could calculate the perpendicular distance between the point and the plane just by using our formula for calculating the distance between a point and a plane.

But then this would give us a new question. How do we know which point to choose? In fact, we can show it doesn’t matter which point we choose. They’ll all be the same distance. Imagine we instead chose a different point on our plane. Once again, we could calculate the perpendicular distance by using our formula. And we know both of these make right angles with our plane because this is a perpendicular distance. We then want to think about the quadrilateral we would get by connecting all four of these vertices together. By using the fact these lines are parallel and by using the fact that they’re perpendicular, we could also show that these must be right angles. Therefore, this is just a rectangle. And we know the opposite sides on a rectangle have to be equal. So it doesn’t matter which point we can choose. We can choose any point.

So to answer this question, we first need to determine which of our three situations are we in. And to do this, it might be easier to write both of our planes in vector form. Let’s start with the first plane of negative 𝑥 minus two 𝑦 minus two 𝑧 is equal to negative two. First, we can find a vector normal to our plane by taking the coefficients of 𝑥, 𝑦, and 𝑧. So we get the normal vector negative one, negative two, negative two. And this is exactly the same as saying that our vector 𝐫 is the vector 𝑥, 𝑦, 𝑧. And now we can see from our equation that the dot product between these two things should be equal to negative two. So this gives us the vector form of our first plane, 𝐫 dot negative one, negative two, negative two should be equal to negative two.

We can then do exactly the same thing for our other plane. Doing exactly the same thing we did before, we get the vector equation of this plane is 𝐫 dot negative two, negative four, negative four is equal to three. So we want to determine which of our three situations we’re in. And the easiest way to do this is to start by checking if our two planes are parallel. And to do this, we need to remember how to check whether two planes are parallel from their vector forms.

We recall that two planes will be parallel if their normal vectors are nonzero scalar multiples of each other. And we can see that this is true in this case. We can see if we set our scalar equal to one-half, then one-half multiplied by the vector negative two, negative four, negative four is equal to the vector negative one, negative two, negative two. So we can eliminate our second option because we know that the planes point in the same direction. However, we do technically need to be careful. We still need to check that they’re not exactly the same plane.

And the easiest way to do this is to multiply our second equation through by our scalar of one-half. If we multiply both sides of this equation by one-half, we get the following equation for our plane. Now remember, one-half multiplied by our normal vector is equal to the normal vector of our other plane. But we notice three multiplied by one-half is not equal to negative two. So these equations don’t represent the same plane. They’re, in fact, parallel. So all we need to do is find the distance between any point on either of our planes and the other plane. And to do this, we’ll start by clearing some space.

Now we need to recall how we find the perpendicular distance between a point and a plane. We recall, the perpendicular distance between the point 𝑥 one, 𝑦 one, 𝑧 one and the plane 𝑎𝑥 plus 𝑏𝑦 plus 𝑐𝑧 plus 𝑑 is equal to zero is given by the distance capital 𝐷 is equal to the absolute value of 𝑎𝑥 one plus 𝑏𝑦 one plus 𝑐𝑧 one plus 𝑑 all divided by the square root of 𝑎 squared plus 𝑏 squared plus 𝑐 squared. In our case, the point 𝑃 can be any point on either of our two planes.

So we just need to find a point on either of our two planes. We can do this by solving either of the equations. We could try 𝑥 is equal to zero, 𝑦 is equal to one, and 𝑧 is equal to zero in our first plane. Substituting these values into our plane, we can see we just get negative two is equal to negative two. So this point lies on our plane. So we’ll set our value of 𝑥 one equal to zero, our value of 𝑦 one equal to one, and our value of 𝑧 one equal to zero. We’re now almost ready to use our formula to find this distance. We just need to find the values of 𝑎, 𝑏, 𝑐, and lowercase 𝑑.

And we can do is directly from the Cartesian equation of our second plane. We see that 𝑎 is equal to negative two, 𝑏 is equal to negative four, and 𝑐 is equal to negative four. However, we do need to be slightly careful because our constant 𝑑 is given on the right-hand side of our equation. However, in our formula, it’s on the left-hand side of our equation. So we need to switch the sign of this. Our value of 𝑑 is going to be equal the negative three.

Now all that’s left to do is substitute our values of 𝑎, 𝑏, 𝑐, lowercase 𝑑, and 𝑥 one, 𝑦 one, 𝑧 one into our formula. We get the distance, capital 𝐷, is equal to the absolute value of negative two times zero plus negative four times one plus negative four multiplied by zero minus three all divided by the square root of negative two squared plus negative four squared plus negative four squared. Calculating the expression in our numerator, we get the absolute value of negative seven. And calculating the expression in our denominator, we get the square root of 36.

And of course, the absolute value of negative seven is seven and the square root of 36 is equal to six. So we just get that this is seven over six. And remember, this represents a distance, so we can say this is seven over six length units. Therefore, we were able to show the distance between the two planes negative 𝑥 minus two 𝑦 minus two 𝑧 is equal to negative two and negative two 𝑥 minus four 𝑦 minus four 𝑧 is equal to three is given by seven over six length units.

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