WEBVTT
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Find the general equation of the plane which passes through the point three, negative eight, negative seven and contains the 𝑥-axis.
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As we get started, let’s visualize what this plane might look like.
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If we draw in an 𝑥-axis, we’re told that the plane contains this axis, which means that it might look, say, like this.
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But then we realize that it’s also possible to rotate this plane and still have it contain the 𝑥-axis.
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In fact, there are actually infinitely many planes that contain the 𝑥-axis.
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However, only one of those infinitely many contains this given point.
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So whichever of these rotated planes contains the points three, negative eight, negative seven, that’s the plane whose general equation we want to write.
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And we can recall that the general equation of a plane is written in this form.
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𝑎𝑥 plus 𝑏𝑦 plus 𝑐𝑧 plus 𝑑 equals zero.
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Importantly, the factors by which we multiply 𝑥, 𝑦, and 𝑧, respectively, are known to be the components of a vector that is normal or perpendicular to the surface of the plane.
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And in general, if we know a vector that’s normal to the plane’s surface, as well as a point in the plane, we can define it precisely.
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Since we already know a point that lies in our plane, our task now is to solve for a vector normal to it.
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Now, let’s say that out of all the planes that do contain the 𝑥-axis, this is the one that also contains our point three, negative eight, negative seven.
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Let’s call that point 𝑃 zero.
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Now, since our plane also contains the 𝑥-axis, we can name another point in it.
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Since the point zero, zero, zero lies along the 𝑥-axis, it must also lie in our plane.
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The reason we’ve named this second point is that now, since we have two points in the plane, we can connect them by a vector that will itself lie in our plane.
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We can call this vector 𝐫 zero.
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And since it travels from our origin to the known point 𝑃 zero, the components of 𝐫 zero are simply the coordinates of 𝑃 zero.
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Now, we have one vector that we know to lie in our plane.
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If we can find a second vector that also lies in this plane and is not parallel to 𝐫 zero, then we can calculate the cross product of those two vectors and wind up with a vector normal to our plane.
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That’s our objective.
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So now let’s consider another vector that must lie in our plane.
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Well, again, since the 𝑥-axis lies in our plane, so must the point one, zero, zero.
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Drawing a vector from the origin to this new point, we can call the vector 𝐫 one.
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Its components are one, zero, zero.
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And we can see that 𝐫 one and 𝐫 zero are not parallel.
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In other words, it’s impossible to multiply either one of these vectors by a constant and get the other vector.
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This is good because it means if we take the cross product of these two vectors, say 𝐫 zero cross 𝐫 one, then we really will get a vector that’s normal to our plane of interest.
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𝐫 zero cross 𝐫 one is equal to the determinant of this three-by-three matrix.
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Here, we filled in the first row with our 𝐢, 𝐣, and 𝐤 unit vectors and the second and third rows with the corresponding components of our two vectors, 𝐫 zero and 𝐫 one.
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Computing this determinant component by component, we have 𝐢 times the determinant of this matrix, that’s zero, minus 𝐣 times the determinant of this two-by-two matrix, that’s seven, plus 𝐤 times the determinant of this two-by-two matrix, that’s eight.
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Writing it out in simplified form then, this gives us this vector.
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And we’ll call this vector 𝐧 because, as we said, it’s normal to our plane.
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We’re now very close to being able to write the general equation of our plane.
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Now that we have a point in the plane as well as a vector normal to it, we can clear a bit of space and remind ourselves of the vector equation of a plane.
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This equation tells us that if we have a normal vector as well as a point in the plane, then that normal vector dotted with the vector to a general point in the plane equals 𝐧 dot 𝐫 zero, the vector to our known point.
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Applying this to our scenario, we have our normal vector zero, negative seven, eight dotted with a vector to a general point in our plane with components 𝑥, 𝑦, and 𝑧 being equal to our normal vector dotted with a vector to our known point.
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If we now calculate these two dot products, on the left-hand side, we get negative seven plus eight 𝑧.
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And on the right, we get 56 minus 56, or zero.
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And now we have the general equation of our plane.
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It’s negative seven 𝑦 plus eight 𝑧 equals zero.
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And notice that this equation doesn’t depend on the 𝑥-values in our plane.
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This is consistent with the fact that the plane contains the entire 𝑥-axis.