Question Video: Determining Whether Two Planes Are Parallel or Perpendicular

Determine whether the planes ⟨4, 2, 1⟩ ⋅ 𝐫 = 8 and (−𝑥/5) + (3𝑦/10) + (𝑧/5) = 1 are parallel or perpendicular.

03:50

Video Transcript

Determine whether the planes four, two, one dot 𝐫 equals eight and negative 𝑥 over five plus three 𝑦 over 10 plus 𝑧 over five equals one are parallel or perpendicular.

Looking at the equations of these two planes, the first thing we’ll want to do is identify the components of vectors that are normal to each one. This first plane’s equation, we’ll call it plane one, is given to us in vector form. This means that the vector that we dot with 𝐫 is a normal vector to the plane. If we call that normal vector 𝐧 one, we can say then that it has components four, two, one.

As far as the second plane’s equation, we’ll call this plane two, this is nearly given to us in what’s called general form. To write the equation that way, we would just need a zero to appear on the right instead of a one. We can accomplish this by subtracting one from both sides. But either way, we’re interested in the values by which we multiply 𝑥 and 𝑦 and 𝑧 in this equation. That’s because it’s those values that make up the components of a vector normal to this plane. We’ll call that vector 𝑛 two. And we see it has components negative one-fifth, three-tenths, and one-fifth.

Now we can note that this isn’t the only vector normal or perpendicular to plane number two. In fact, we can multiply this vector by any nonzero constant 𝐶, and it would still be normal to the plane. To make 𝑛 two a bit simpler to work with, let’s let 𝐶 equal positive five. In other words, we’re multiplying all the components of this vector by five. This gives us a resulting vector of negative one, three-halves, one.

Now that we know the components of vectors that are normal to both of our planes, we can start testing whether these planes are parallel or perpendicular. In general, if two planes are parallel, then that means their normal vectors, 𝐧 one and 𝐧 two, are equal to one another to within a constant value. In other words, there exists some constant, we’ve called it 𝐾, by which we can multiply one of the normal vectors so that it equals the other. If the planes are not parallel, then they may be perpendicular. The condition for that is that the dot product of 𝐧 one and 𝐧 two equals zero.

So let’s apply these tests to our two given planes, starting with a test of whether they’re parallel. We can begin to conduct this test by looking at our two normal vectors and specifically analyzing their 𝑥-components. If some constant does exist by which we can multiply 𝐧 two to equal 𝐧 one, then we can solve for the value of that constant by seeing what it must be in order for this equation to be true. If it is true that four equals 𝐾 times negative one, that means 𝐾 must equal negative four. This is the only value 𝐾 can have in order for this equation to hold true.

So now what we do is we look at the 𝑦- and then the 𝑧-components and see if the same 𝐾-value works for them. Let’s try the 𝑦-components first. We want to see if two is equal to 𝐾 times three-halves, where 𝐾 is equal to negative four. Negative four times three over two though equals negative six. And that is not equal to two. So therefore, this 𝐾-value can’t be used consistently across all dimensions to make our normal vectors equal. And that means there is no such constant by which we can multiply these vectors so they are equal. That tells us that our two planes, represented by the vectors 𝐧 one and 𝐧 two, are not parallel.

Let’s consider then whether they’re perpendicular. If they are, the dot product of a vector normal to one plane and one normal to the other should be zero. We see this dot product equals four times negative one plus two times three-halves plus one times one or, simplifying, negative four plus three plus one. This does equal zero, which means that 𝐧 one dot 𝐧 two equal zero for our two planes. And therefore, we can say that the two planes in this example are oriented perpendicularly to one another.

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