# Question Video: Finding the General Equation of a Plane Mathematics

Find the general equation of the plane that is perpendicular to the plane −6𝑥 + 3𝑦 + 4𝑧 + 4 = 0 and cuts the 𝑥- and 𝑦-axes at (5, 0, 0) and (0, 1, 0) respectively.

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### Video Transcript

Find the general equation of the plane that is perpendicular to the plane negative six 𝑥 plus three 𝑦 plus four 𝑧 plus four equals zero and cuts the 𝑥- and 𝑦-axes at five, zero, zero and zero, one, zero, respectively.

Let’s say that this shape in orange is our plane. We’re told the coordinates of the points where this plane cuts across the 𝑥- and 𝑦-axes. Knowing this, and also that it’s perpendicular to the plane with this given equation, we want to solve for the general equation of our plane. To do that, we’ll need two pieces of information. First, we’ll need a point that lies on the plane. And second, we’ll need to know a vector that is normal or perpendicular to it. We can see that, thanks to our given information, we’ve already satisfied this first point. In fact, we know the coordinates of two points on our plane.

Our next step then is to figure out the components of a vector that’s normal to it. To start doing that, let’s consider this plane that we’re told is perpendicular to our plane of interest. That perpendicular plane has this general equation. And we know that, written in this form, the values by which 𝑥 and 𝑦 and 𝑧 are multiplied make up the components of a vector that’s normal to this plane.

Now let’s think about this. This vector here is normal or perpendicular to this plane, and this plane is perpendicular to our plane of interest. Therefore, with respect to our plane of interest, this vector with components negative six, three, four is parallel to this plane. We know that because it’s perpendicular or normal to a plane that’s perpendicular to the plane we’re actually interested in. Since this vector is parallel to our plane of interest, let’s call it 𝐩 one. We include this subscript one because, actually, there’s another vector we can identify that’s parallel to our plane. That would be the vector that goes from one of the points on our plane to the other. We can call this vector 𝐩 two and its components are given by the difference between the coordinates of our two known points. This vector then has components five, negative one, and zero.

Now that we know two vectors that are parallel to our plane of interest, we can recall that taking the cross or vector product of two vectors gives a vector that’s perpendicular to both of the vectors that went into that product. This is important for us because it means that whatever results from this cross product will be perpendicular or normal to our plane of interest. And so, we move ahead with this calculation. The cross product of 𝐩 one and 𝐩 two is given by the determinant of this matrix. In the first row, we have our three unit vectors and then in the second and third rows, respectively, the corresponding components of vectors 𝐩 one and 𝐩 two.

The 𝐢-component of this cross product equals the determinant of this two-by-two matrix. Three times zero minus four times negative one is positive four. Then the 𝐣-component is negative the determinant of this matrix. Six times zero minus four times five is negative 20. The 𝐤-component is given by this two-by-two determinant. Negative six times negative one minus three times five is six minus 15. Combining these three components gives us the result of our cross product which we can write as a vector with components four, 20, and negative nine. This then is a vector that is normal to our plane of interest. This means we’ve now satisfied the two conditions necessary for writing the equation of a plane.

So let’s clear some space on screen and start out by recalling the vector form of a plane’s equation. Written this way, we have a vector normal to the plane being dotted with a vector to an arbitrary point on the plane. This is equal to the dot product of that same normal vector and a vector to a known point on the plane surface. As we saw earlier, we have two such points in our case. We could choose to work with either one of these points. But simply because the numbers in this point are smaller than the numbers in this one, we’ll choose this as our point on the plane.

So substituting in our normal vector and our given point, we get this expression. The vector four, 20, negative nine is dotted with a vector to a general point with components 𝑥, 𝑦, 𝑧. This equals the dot product of that same normal vector and a vector to our point 𝐩 zero. Carrying out this dot product on the left, we get four 𝑥 plus 20𝑦 minus nine 𝑧. The same thing on the right gives us four times zero plus 20 times one minus nine times zero or, in other words, 20.

The last step to writing this equation in general form is to subtract 20 from both sides so that we get zero on the right-hand side of this expression. Doing that gives us this equation, which is the general equation of this plane. Four 𝑥 plus 20𝑦 minus nine 𝑧 minus 20 equals zero.