Question Video: Finding the Measure of the Angle between Two Planes | Nagwa Question Video: Finding the Measure of the Angle between Two Planes | Nagwa

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 −9𝑥 − 6𝑦 + 5𝑧 = −8 and 2𝑥 + 2𝑦 + 7𝑧 = −8.

03:13

Video Transcript

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.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy