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Scalar triple product of gradients is zero

Short answer

grad u = (1,1,1), grad v = (2x,2y,2z), grad w = (y+z, z+x, x+y). The determinant is 2|1 1 1; x y z; y+z z+x x+y|. Adding row 2 to row 3 makes row 3 proportional to row 1, so the determinant is zero, showing [grad u, grad v, grad w] = 0.

If u = x+y+z, v = x²+y²+z² and w = xy+yz+zx, show that [grad u, grad v, grad w] = 0.

u = x + y + z, v = x² + y² + z², w = xy + yz + zx.

Gradients

grad u = (1, 1, 1)
grad v = (2x, 2y, 2z)
grad w = (y + z, z + x, x + y)

Scalar triple product (determinant)

                  |  1      1      1    |
[grad u grad v grad w] = |  2x     2y     2z   |
                  | y+z    z+x    x+y  |

Take 2 out of the second row:

= 2 | 1     1     1   |
    | x     y     z   |
    | y+z   z+x   x+y |

Now add the second row to the third row (R₃ → R₃ + R₂):

third row = (y + z + x, z + x + y, x + y + z) = (x + y + z)(1, 1, 1),

which is a multiple of the first row. A determinant with two proportional rows is zero.

Hence [grad u, grad v, grad w] = 0.

This shows u, v, w are functionally dependent: indeed v = u² − 2w.

✓ Answer checked by Claude Opus 5.5 (high effort)

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