Find the point of intersection of the planes four 𝑥 over seven minus 𝑦 over seven plus two 𝑧 over seven is equal to one, 𝑥 plus three 𝑦 plus five 𝑧 minus 16 is equal to zero, and two 𝑥 plus three 𝑦 minus four 𝑧 minus nine is equal to zero.
We’re asked to find the point of intersection of three planes. And since the planes meet at a single point, this means the point of intersection is a unique solution to the system of three equations, that is, the system formed by the three planar equations. There are a few methods we could use to try and solve this. However, we’re going to use Cramer’s rule, so we’ll need to rewrite all three equations into a form which we can convert into a matrix equation. Labeling our equations one, two, and three, if we multiply both sides of equation one by seven, we have four 𝑥 minus 𝑦 plus two 𝑧 is equal to seven. And now, adding 16 to both sides of equation two gives us 𝑥 plus three 𝑦 plus five 𝑧 is 16. And finally, adding nine to both sides of equation three, we have two 𝑥 plus three 𝑦 minus four 𝑧 is nine.
We can then rewrite this system of equations in matrix form as shown. The elements of the coefficient matrix are the coefficients of 𝑥, 𝑦, and 𝑧 in each of our three equations. This multiplies our variable matrix with elements 𝑥, 𝑦, and 𝑧. And on the right-hand side, we have the constant matrix whose elements are the constants on the right-hand side of our equations.
Now, making some space for our calculations, we’re going to be using Cramer’s rule. And this tells us that 𝑥 is equal to Δ 𝑥 over Δ, 𝑦 is Δ 𝑦 over Δ, and 𝑧 is Δ 𝑧 over Δ is a unique solution to the system of equations 𝐴𝑋 is equal to 𝐵, where Δ is the nonzero determinant of the matrix of coefficients 𝐴. And Δ 𝑥 is the determinant of the matrix formed by replacing the first column of 𝐴 with the matrix 𝐵. That is the matrix of constants. Similarly for Δ 𝑦, we replace the central column by the matrix 𝐵. And similarly for Δ 𝑧, we replace the third column by the matrix 𝐵. It should be stressed that the three planes will intersect at a single point if and only if Δ, that’s the determinant of matrix 𝐴, is nonzero. And that’s equivalent to saying there is a unique solution to the system of equations.
Okay, so let’s begin by finding the determinant of our coefficient matrix 𝐴; that’s Δ. Expanding on our first row, we have four times the determinant of the matrix with elements three, five, three, negative four minus negative one times the determinant of the two-by-two matrix with elements one, five, two, negative four plus two times the determinant of the two-by-two matrix with elements one, three, two, three, noting that we need to be careful with the signs of our cofactors. That’s positive, negative, and positive. And taking our two-by-two determinants, remember that’s 𝑎𝑑 minus 𝑏𝑐, we have four multiplied by three times negative four minus five times three plus one times negative four minus five times two plus two times one times three minus three times two.
Inside our square brackets, we then have negative 12 minus 15, negative four minus 10, and three minus six. This then evaluates to four times negative 27 minus 14 minus six. So the Δ, that’s the determinant of coefficient matrix 𝐴, is negative 128. Since this determinant is nonzero, we’ve confirmed that the three planes intersect at a single point.
So now, making some space and making a note of our value for Δ, we can turn our attention to Δ 𝑥, Δ 𝑦, and Δ 𝑧. And remember, to calculate Δ 𝑥, we replace the elements in the first column by those of the matrix 𝐵. So the first column of our determinant now has elements seven, 16, and nine, and again expanding on the first row and being careful with our signs. Expanding our two-by-two determinants, we have seven multiplied by negative 12 minus 15 plus negative 64 minus 45 plus two times 48 minus 27. And this evaluates to negative 189 minus 109 plus 42. And that’s negative 256. And so we have Δ 𝑥 is negative 256. And making a note of this and making some space, we can now calculate Δ 𝑦.
We recall this time we replace the second column with the elements of the matrix 𝐵. And again, expanding along the first row and calculating our two-by-two determinants, this gives us four times negative 109 minus seven times negative 14 plus two times negative 23. That is negative 436 plus 98 minus 46, which evaluates to negative 384. That is, Δ 𝑦 is negative 384. Now, making some more space, we can finally work out Δ 𝑧, where this time we replace the third column of our determinant with the elements of matrix 𝐵. And so again expanding on the first row, our two-by-two determinants evaluate to negative 21, negative 23, and negative three, and so Δ 𝑧 evaluates to negative 128.
So now making a note of this and making some space, we can finally work out the coordinates of our point of intersection. That’s 𝑥, 𝑦, and 𝑧. Beginning with the point 𝑥, that’s Δ 𝑥 over Δ, we have negative 256 over negative 128, and this evaluates to positive two. Our 𝑥-coordinate is therefore two. Next, for our 𝑦- coordinate, that’s Δ 𝑦 over Δ, which is negative 384 divided by negative 128, and this evaluates to three. And so our 𝑦-coordinate is three. And finally, our third coordinate 𝑧 is Δ 𝑧 over Δ, and that’s negative 128 over negative 128, which of course is one. And so our 𝑧-coordinate is equal to one.
Hence, the point of intersection of the three planes is the point with coordinates 𝑥 is two, 𝑦 is three, and 𝑧 is one.