Video Transcript
In this lesson, we’ll learn how to find the measure of an angle between two planes or that between a line and a plane. Our planes and lines will exist in three-dimensional space. And as we see from this sketch, finding these angles will have something to do with vectors.
Let’s start out by considering the angle between two planes. If we say that the general equation describing our first plane is 𝑎 one 𝑥 plus 𝑏 one 𝑦 plus 𝑐 one 𝑧 plus 𝑑 one equals zero and that the equation describing our second plane is similar except for the values of these constants 𝑎 two, 𝑏 two, 𝑐 two, and 𝑑 two, then we can solve for the angle in three-dimensional space between these two planes. If we call that angle 𝛼, then the cos of 𝛼, the angle between these two planes, is given by this expression. Now this may look quite involved, but we’ll see in a moment that we can write this in a much simpler way.
For now, though, notice that these two values, 𝑑 one and 𝑑 two, don’t appear anywhere in this equation for the cos of 𝛼. In other words, 𝛼 doesn’t depend on these values. And now notice something else. All of the values that do appear in this equation are multiplied by either an 𝑥-, 𝑦-, or 𝑧-variable in our planes’ equations. From our knowledge of the equation of a plane written in its general form, we know that these quantities represent components of a vector that’s normal to each plane. That is, if we were to draw a vector perpendicular to plane one and we call that 𝐧 one, then this vector could have the components 𝑎 one, 𝑏 one, and 𝑐 one.
The same thing holds true for plane two. A vector normal to its surface, we’ve called it 𝐧 two, could have components 𝑎 two, 𝑏 two, and 𝑐 two. Looking back at our equation for the cosine of the angle between these two planes, we see that it’s written entirely in terms of the components of vectors normal to these planes. In fact, if we look at the numerator of this expression, we see it’s equal to the magnitude of the dot product of 𝐧 one and 𝐧 two. And then in the denominator, this square root and everything underneath it is equal to the magnitude of normal vector 𝐧 one, while this second square root is equal to the magnitude of 𝐧 two.
Therefore, another way we can write the cos of 𝛼 is as the magnitude of the dot product of 𝐧 one and 𝐧 two divided by the magnitude of 𝐧 one times the magnitude of 𝐧 two. This second expression for the cos of 𝛼 shows us that given two planes where we want to solve for the angle between them, if we can somehow find the components of a vector normal to each one, then we’ll be able to solve for the angle between the planes. As a side note, the angle between the normal vectors is the same as the angle between the planes.
Okay, so that’s for two planes. But what if we want to solve for the angle between a plane and a straight line? We see that in this case, it’s not enough to have a vector that’s normal to each of these two objects. And that’s because in the case of a line, a vector normal to it could point in many different directions. We can’t specify a line’s direction by a single normal vector. We can define the axis of a line though by a vector parallel to it.
Let’s say this line runs parallel to a vector we’ll call 𝐩 with components 𝑎 three, 𝑏 three, 𝑐 three. If we now want to solve for the angle between this line and the plane, we’ll call the angle 𝛽, then instead of solving for the cosine of this angle like we did with two planes, we’ll write an expression for the sine of it. Recall that the sine and cosine functions are different by a phase angle of 90 degrees. If there was such a thing as a single vector normal to a line, then that would also be the angular difference between this normal vector and the vector parallel to the line.
In any case, here’s the expression for the sine of the angle between a plane and a straight line. We note that it looks virtually identical to our equation for the angle between two planes. The only real difference, as we’ve seen, is that now we’re calculating the sine rather than the cosine of an angle. Just like we did before, we can express the right-hand side of this equation using vectors. It’s equal to the magnitude of the dot product between a vector normal to our plane and a vector parallel to our line divided by the product of the magnitudes of each of these vectors. And note that if we want to solve for the angle itself, what we’ve called 𝛽, then we would take the inverse or arcsine of this expression.
Knowing all this about finding the angle between planes and straight lines, let’s get a bit of practice through an example.
Find, to the nearest second, the measure of the angle between the planes negative nine 𝑥 minus six 𝑦 plus five 𝑧 equals negative eight and two 𝑥 plus two 𝑦 plus seven 𝑧 equals negative eight.
Okay, so here we have these equations describing two different planes, and we see that they’re almost but not quite written in general form. We recall that in general form, the equation of a plane has zero on one side. For both of these expressions though, if we add positive eight to both sides, then we’ll get these equations in general form. That’s useful to us because now we can more easily pick out the components of vectors that are normal to each one of these planes.
Let’s say that the plane represented by this first equation is plane one. And we’ll call the plane represented by the second equation plane two. When a plane’s equation is given in general form, that means whatever factors we use to multiply 𝑥, 𝑦, and 𝑧 are the components of a vector that is normal to this plane. That is, if we call 𝐧 one a vector normal to plane one, then we know there exists such a vector with components negative nine, negative six, five. Similarly, for plane two, where we can define a normal vector 𝐧 two, this vector will have components two, two, seven. We’ve gone to the effort of solving for vectors normal to each one of our planes because knowing these components, we’re fairly close to solving for the measure of the angle between the planes.
If we call the angle between two planes in general 𝛼, then the cos of 𝛼 is equal to the magnitude of the dot product of vectors normal to each plane divided by the product of the magnitudes of each of these vectors. Knowing 𝐧 one and 𝐧 two for our two planes, we can use this relationship to solve for 𝛼. If we focus first on calculating the magnitude of the dot product of 𝐧 one and 𝐧 two, that equals the magnitude of the dot product of these two vectors. And carrying out this dot product by multiplying together the respective components, we get negative 18 minus 12 plus 35. That’s equal to the magnitude of five or simply five.
Next, we can calculate the denominator of this fraction, the product of the magnitudes of our two normal vectors. That’s equal to the square root of negative nine squared plus negative six squared plus five squared multiplied by the square root of two squared plus two squared plus seven squared. This equals the square root of 81 plus 36 plus 25 times the square root of four plus four plus 49 or the square root of 142 times the square root of 57. This is equal to the square root of 142 times 57 or 8094.
Now that we’ve calculated our numerator and denominator, we can write that the cosine of the angle between our two planes is equal to five divided by the square root of 8094. This means that the angle 𝛼 is the inverse cos of five over the square root of 8094. And if we enter this expression on our calculator and round the result to the nearest second, our result is 86 degrees, 48 minutes, and 51 seconds. This is the angle in degrees, minutes, and seconds between our two planes.
Let’s look now at a second example.
Find, to the nearest degree, the measure of the angle between the planes two times the quantity 𝑥 minus one plus three times the quantity 𝑦 minus four plus four times the quantity 𝑧 plus five equals zero and 𝐫 dot one negative two five equals 16.
Okay, so here we have these equations for two different planes, and we see that they’re given in different forms. This first equation is almost in general form, while the equation for our second plane is given in vector form. In order to solve for the angle between these two planes, we’ll want to identify a vector that is normal to each one.
Let’s recall that when a plane is expressed in vector form, it’s given as a normal vector to the plane dotted with a vector to a general point in the plane as being equal to that normal vector dotted with the vector to a known point in the plane. As a side note, in this equation we’ve used half arrows over letters to represent the vector symbol, while in our problem statement when a letter is written in bold, as this 𝐫 is, that represents a vector. As we work toward our solution, we’ll continue to use this half arrow notation for consistency’s sake.
If we were to match up this second plane equation we’re given with the vector form of a plane’s equation, we could see that this vector one, negative two, five corresponds to the normal vector 𝐧, 𝐫 corresponds to 𝐫 in the equation, and 16 corresponds to the result of this dot product. For our purposes, the important thing is that we now know the components of a vector that is normal to this plane. We’ll call this plane two and so that the vector normal to it we’ll give the name 𝐧 two. As we saw, this vector has components one, negative two, five.
Let’s look now at the first plane equation we were given. And we’ll call this plane plane one. We mentioned that this plane is given to us in almost what is called general form. That form is given this way: 𝑎 times 𝑥 plus 𝑏 times 𝑦 plus 𝑐 times 𝑧 plus 𝑑 equals zero. If we were to take the given form of plane one and multiply through by these factors two, three, and four and if we then collected together all of the values not multiplying a variable, our plane would then be expressed in general form. This helps us because in this form the values that multiply 𝑥, 𝑦, and 𝑧 — 𝑎, 𝑏, and 𝑐 as we’ve written them here — are equal to the 𝑥-, 𝑦-, and 𝑧-components of a vector normal to this plane.
If we call this vector normal to plane one 𝐧 one, we can see then that it will have components two, three, and four. Now that we’ve got these vectors normal to each of our two planes, let’s go about solving for the measure of the angle between the planes. If we call that angle 𝛼, then we can recall that the cos of 𝛼 is equal to the magnitude of the dot product of 𝐧 one and 𝐧 two divided by the product of the magnitude of each one of these vectors. If we write out this expression using the 𝐧 one and 𝐧 two vectors we’ve solved for, it looks like this. Note that in the numerator we’re taking a magnitude of the dot product between 𝐧 one and 𝐧 two. And in the denominator we’re solving for the magnitudes of these two vectors by taking the square root of the sum of the squares of their components.
In the numerator, we calculate this dot product and, in the denominator, square out each of the components so that we get the magnitude of two minus six plus 20 divided by the square root of four plus nine plus 16 times the square root of one plus four plus 25. This equals 16 divided by the square root of 29 times the square root of 30 or 16 divided by the square root of 29 times 30. In order to get the angle 𝛼 itself, what we’ll do is take the inverse cos of this expression. Entering this on our calculator, to the nearest degree, 𝛼 is 57 degrees. This is the measure of the angle between our two planes.
Let’s now look at an example where we calculate the angle between a plane and a line.
Which of the following is the smaller angle between the straight line 𝐫 equals negative seven 𝐢 minus 𝐣 minus nine 𝐤 plus 𝑡 times two 𝐢 plus 𝐣 minus 𝐤 and the plane 𝐫 dot nine 𝐢 minus nine 𝐣 plus two 𝐤 equals 13? (A) The inverse sin of seven times the square root of 249 divided by 498. (B) The inverse cos of seven times the square root of 249 divided by 498. (C) The inverse sin of seven times the square root of 43 over 86. (D) The inverse cos of seven times the square root of 43 over 86.
Okay, in this exercise, we’re given the equation of a line and the equation of a plane. Let’s imagine that this is our line and this is our plane. We want to solve for the smaller angle between these two objects. We imagine this line and plane intersecting in three-dimensional space. And we see that that intersection forms two angles, this one here as well as this one. Our problem statement tells us to solve for the smaller of these two angles.
To do this, we’ll need to know a vector that is normal or perpendicular to our plane as well as a vector that is parallel to our line. Beginning with our plane, let’s say that a vector normal to its surface is called 𝐧. If we look at the form in which the equation of our plane is given, we see it’s presented to us in what’s called vector form. Written this way, a vector to a general point in the plane is dotted with a vector normal to the plane. From this then, we can read off the components of our vector 𝐧, nine, negative nine, two.
Now let’s look to solve for a vector that is parallel to our line. We’ll call this vector 𝐩. If we look at the form in which the equation of our line is given, we can find a vector that’s parallel to this line by looking for the scale factor; in this case it’s called 𝑡. The vector that multiplies the scale factor is parallel to our line. We can say then that the components of 𝐩 are two, one, and negative one. Now that we have our vector normal to our plane and a vector parallel to our line, we can recall the general equation that the sin of the angle 𝛼 between a plane and a line is equal to the magnitude of the dot product of a vector parallel to the line and one normal to the plane all divided by the product of the magnitude of each of these vectors.
To solve for 𝛼 then, we can go about calculating this numerator and denominator and then bringing them together. The magnitude of 𝐩 dot 𝐧 using our 𝐩 and 𝐧 vectors equals the magnitude of two, one, negative one dotted with nine, negative nine, two. This equals the magnitude of 18 minus nine minus two or seven. Knowing this result, now let’s calculate the denominator of our fraction. This equals the square root of the sum of the squares of two, one, and negative one times the square root of the sum of the squares of nine, negative nine, and two. That equals this expression or the square root of six times the square root of 166, which equals the square root of 996.
We can now say that the sin of the angle 𝛼 equals seven over the square root of 996. And if we rationalize our denominator, this equals seven times the square root of 996 over 996. And then we note that 996 is equal to four times 249. Written this way, we can bring the four outside of the square root, where it becomes a two. It turns out, though, that 996 is divisible by two. So if we divide numerator and denominator of this fraction by two, we find it equals seven times the square root of 249 divided by 498. So if the sin of 𝛼 is equal to this expression, then 𝛼 is the inverse sin of seven times the square root of 249 over 498.
Now, if we evaluate this on our calculator, we find a result of about 13 degrees. This confirms to us that, indeed, we’ve solved for the smaller of the two angles of intersection between our line and the plane. Our result matches up with answer option (A). The smaller angle between this line and plane is the inverse sin of seven times the square root of 249 divided by 498.
Let’s finish up our lesson now by summarizing a few key points. In this lesson, we saw that given the equation of two planes in general form, the cosine of the angle between those planes is given by this expression. We saw further that if 𝐧 one and 𝐧 two are normal vectors to each of these respective planes, then the cosine of the angle between the planes also equals the magnitude of the dot product of 𝐧 one and 𝐧 two divided by the product of their magnitudes.
And then if instead of two planes, we have a plane and a straight line, the sine of the angle between them, we’ve called it 𝛽, is equal to this expression or in terms of vectors the magnitude of the dot product of a vector normal to the plane with a vector parallel to the line divided by the product of the magnitudes of these two vectors.
