Question Video: Finding the Measure of the Angle Between Two Planes Mathematics

Find, to the nearest second, the measure of the angle between the planes <βˆ’5, βˆ’9, 4> β‹… 𝐫 = 6 and βˆ’4π‘₯ βˆ’ 𝑦 βˆ’ 6𝑧 = βˆ’3.

02:53

Video Transcript

Find, to the nearest second, the measure of the angle between the planes negative five, negative nine, four dot 𝐫 equals six and negative four π‘₯ minus 𝑦 minus six 𝑧 equals negative three.

All right, so here we have these two planes. Let’s just say they look like this. And we want to solve for the angle between them. The angle between these planes will be the same as the angle between vectors that are normal to each one. And in fact, we’ll work in terms of normal vectors. We’ll call them 𝐧 one and 𝐧 two. In general, if the angle between two planes is πœƒ, then the cos of this angle is equal to the magnitude of the dot product of two vectors 𝐧 one and 𝐧 two that are normal to the planes involved divided by the product of the magnitudes of these vectors.

If we can solve for 𝐧 one and 𝐧 two then, we can use this relationship to ultimately find out πœƒ, the angle between our planes. In our example, our first plane here is given in what’s called vector form. Written this way, on the left-hand side, we have a dot product of a vector to an arbitrary point on our plane and a vector that’s normal to it. In other words, we can say that 𝐧 one is this vector with components negative five, negative nine, four.

Next, we look at the equation of our second plane. And this is given nearly in what’s called general form. A plane written this way is of the form π‘Žπ‘₯ plus 𝑏𝑦 plus 𝑐𝑧 plus 𝑑 equals zero. In the case of the given equation, if we added three to both sides, we would find that negative four π‘₯ minus 𝑦 minus six 𝑧 plus three equals zero. And written in general form, we can recall that the values π‘Ž, 𝑏, and 𝑐 correspond to components of a vector normal to the plane. That is, given a plane written this way, we would say it possesses a normal vector, we can call it 𝐧, with components π‘Ž, 𝑏, and 𝑐.

We see then that in our given equation those values correspond to negative four, negative one, and negative six. We can say then that those are the components of a vector normal to our second plane. Now that we know 𝐧 one and 𝐧 two, we can apply them to this relationship. The cos of the angle between our two planes is equal to this expression. In the numerator, we’re taking the dot product of 𝐧 one and 𝐧 two. And in the denominator, we’re calculating their magnitudes with the square roots of the sum of the squares of their components. And that gives us this expression.

And simplifying further, we get five over the square root of 122 times the square root of 53. Having this expression for the cos of πœƒ means that πœƒ itself is equal to the inverse cos of five over root 122 times root 53. Entering this expression on our calculator and rounding to the nearest second, we get an angle of 86 degrees, 26 minutes, and six seconds. To the nearest second, this is the angle between these two planes.

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