### Video Transcript

Find the distance between the two planes negative ๐ฅ minus two ๐ฆ minus two ๐ง is equal to negative two and negative two ๐ฅ minus four ๐ฆ minus four ๐ง is equal to three.

In this question, weโre 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 weโre 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 itโs very useful to know the shortest distance between two objects.

But we donโt know how to find the perpendicular distance between two planes. So instead weโre going to have to get a picture of whatโs happening. And in fact, thereโs three possibilities. The first possibility would be weโre given two of exactly the same plane. Of course, if weโre given two of the exact same plane, then these planes intersect at every single point because theyโre 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 theyโll intersect. In fact, theyโll 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 weโre given two parallel planes. And itโs worth pointing out here when we say that theyโre parallel, we also mean that theyโre not the same plane since weโve 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 theyโre 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 doesnโt matter which point we choose. Theyโll 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 theyโre 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 doesnโt 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. Letโs start with the first plane of negative ๐ฅ minus two ๐ฆ minus two ๐ง is equal to negative two. First, we can find a vector normal to our plane by taking the coefficients of ๐ฅ, ๐ฆ, and ๐ง. So we get the normal vector negative one, negative two, negative two. And this is exactly the same as saying that our vector ๐ซ is the vector ๐ฅ, ๐ฆ, ๐ง. 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, ๐ซ 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 ๐ซ dot negative two, negative four, negative four is equal to three. So we want to determine which of our three situations weโre 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 theyโre 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 donโt represent the same plane. Theyโre, 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, weโll 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 ๐ฅ one, ๐ฆ one, ๐ง one and the plane ๐๐ฅ plus ๐๐ฆ plus ๐๐ง plus ๐ is equal to zero is given by the distance capital ๐ท is equal to the absolute value of ๐๐ฅ one plus ๐๐ฆ one plus ๐๐ง one plus ๐ all divided by the square root of ๐ squared plus ๐ squared plus ๐ squared. In our case, the point ๐ 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 ๐ฅ is equal to zero, ๐ฆ is equal to one, and ๐ง 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 weโll set our value of ๐ฅ one equal to zero, our value of ๐ฆ one equal to one, and our value of ๐ง one equal to zero. Weโre now almost ready to use our formula to find this distance. We just need to find the values of ๐, ๐, ๐, and lowercase ๐.

And we can do is directly from the Cartesian equation of our second plane. We see that ๐ is equal to negative two, ๐ is equal to negative four, and ๐ is equal to negative four. However, we do need to be slightly careful because our constant ๐ is given on the right-hand side of our equation. However, in our formula, itโs on the left-hand side of our equation. So we need to switch the sign of this. Our value of ๐ is going to be equal the negative three.

Now all thatโs left to do is substitute our values of ๐, ๐, ๐, lowercase ๐, and ๐ฅ one, ๐ฆ one, ๐ง one into our formula. We get the distance, capital ๐ท, 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 ๐ฅ minus two ๐ฆ minus two ๐ง is equal to negative two and negative two ๐ฅ minus four ๐ฆ minus four ๐ง is equal to three is given by seven over six length units.