Video Transcript
Determine the general equation of the plane that contains the straight line 𝑥 plus two over seven equals 𝑦 minus six over five equals 𝑧 plus nine over five and that is perpendicular to the plane negative 𝑥 plus 𝑦 minus two 𝑧 equals two.
Okay, so here we’re looking for the general equation of a plane. Let’s say that this is that plane in the plane of our screen. And we know that this plane contains a straight line whose equation we’re given and that it’s perpendicular to another plane whose equation we’re given. Let’s say that this is that perpendicular plane.
In general, to find the equation of a plane, we need two pieces of information. First, we need a vector that is normal or perpendicular to the plane. Note that this vector is not unique. There are actually infinitely many vectors which would be perpendicular to a two-dimensional surface. Anyone of them will work for helping us find the plane’s equation. And second, we need to know a point that lies on the plane.
So let’s think about this with regard to our current scenario. The plane of interest contains a straight line. And the equation of this line is given to us in what’s called symmetric form. It’s called this because all three of these fractions equal to one another are also equal to a scale factor we can call 𝑡. Writing our line’s equation this way sets us up to express it in what’s called parametric form. The idea is we can write three separate equations: one for the 𝑥-, one for the 𝑦-, and one for the 𝑧-coordinates of this line. These three equations come from the fact that these three fractions here are all equal to the same scale factor.
So, for example, the fact that 𝑥 plus two over seven equals 𝑡 implies that 𝑥 itself equals seven times 𝑡 minus two. And then for the 𝑦-equation, since 𝑦 minus six over five equals 𝑡, we can write that 𝑦 equals five 𝑡 plus six. Then lastly, for the 𝑧-equation, we have 𝑧 equals five times 𝑡 minus nine. Our line’s equation is now written in parametric form. And from this form, we can identify first a point that the line passes through and second a vector that’s parallel to the line.
We can see this by writing our line’s equation as a vector, bringing the 𝑥- and 𝑦- and 𝑧-equations together as components. Expressed like this, we say that our line is equal to a vector with components seven, five, five times our scale factor 𝑡 plus a vector from the origin of our coordinate frame to the point with coordinates negative two, six, negative nine.
Since we know then a point on our line and this line lies in our plane of interest, we’ve satisfied our second condition. We now know a point on our plane. We’ll call that point 𝑃 zero. And as we’ve seen, it has coordinates negative two, six, negative nine. Now, this vector here isn’t normal to our plane, but it is parallel to it. It lies in the plane, so we’ll call this vector 𝐏 one. This is all the useful information we’ll get about our plane from this given straight line. But we can now recall that our plane of interest is also perpendicular to the plane with this equation.
If we were to sketch in a vector that’s normal or perpendicular to the second plane, it might look like this. Note that this vector is parallel to our plane of interest. And if we look at the equation of this plane, we can actually pick out components of just such a vector. When our plane is written in this form, the values that multiply 𝑥, 𝑦, and 𝑧 are the components of a vector perpendicular to the plane. These values, we see, are negative one, positive one, and negative two. We note also that a vector that is normal or perpendicular to this plane will be parallel to the one whose equation we’re trying to solve for. We can call this vector then 𝐏 two since it’s parallel to our plane of interest.
So far, we still don’t have a vector that’s normal to this plane. But if we take the cross product of 𝐏 one and 𝐏 two, the vectors parallel to our plane, then we’ll end up with a vector perpendicular to them both. And therefore, that result, the vector that comes from crossing 𝐏 one and 𝐏 two, will be normal or perpendicular to our plane. This cross product equals the determinant of this three-by-three matrix. Note that in the top row, we have the 𝐢 hat, 𝐣 hat, and 𝐤 hat unit vectors and then in the second and third rows the respective components of 𝐏 one and 𝐏 two. Computing this cross product, the 𝐢-component equals the determinant of this two-by-two matrix. Five times negative two minus five times one is negative 10 minus five.
Then the negative of the 𝐣-component equals the determinant of this matrix. Seven times negative two minus five times negative one is negative 14 plus five. And lastly, the 𝐤-component is given by this determinant, which is seven times one minus five times negative one or seven plus five. All together then, these are the components of our cross product, and we can write this simply as a vector with components negative 15, nine, and 12. This, then, is our normal vector, the vector that’s perpendicular to the plane whose equation we want to solve for.
We’ve now satisfied both of our conditions. And to solve for the general equation of this plane, we’ll clear some space. And we can recall that the vector form of a plane’s equation is given by this expression. Here, 𝐧 is a vector normal to the plane, 𝐫 is a vector to an arbitrary point on the plane, while 𝐫 zero is a vector to a known point on the plane. Using the components of the normal vector we’ve solved for as well as the coordinates of the point 𝑃 zero, which lies in our plane, we can write that our vector with components negative 15, nine, 12 is dotted with a general vector with components 𝑥, 𝑦, 𝑧 and that that equals the dot product of our normal vector and a vector to our point 𝑃 zero.
On the left-hand side of this equation, as we calculate this dot product, we get negative 15𝑥 plus nine 𝑦 plus 12𝑧. This equals negative 15 times negative two, that’s 30, plus nine times six, that’s 54, plus 12 times negative nine or negative 108. These three values add up to negative 24. At this point, we’re getting very close to expressing the equation of our plane in general form. All we need to do is rearrange things so that zero appears on the right-hand side of our equation.
We can see that to do that, we’ll add 24 to both sides. That gives us this equation. And notice that everything on the left-hand side of this equation is evenly divisible by negative three. We can divide then both sides of the equation by that value. On the right, zero divided by negative three is still zero. But on the left, we have positive five 𝑥 minus three 𝑦 minus four 𝑧 minus eight. This is the general form of the equation of our plane, five 𝑥 minus three 𝑦 minus four 𝑧 minus eight equals zero.