(-1,-1,3) x (-1,0,2)

(-1,-1,3) x (-1,0,2)

$$\\\begin{pmatrix} -1&-1&3 \\ \end{pmatrix} &\times \begin{pmatrix} -1& \\ 0&\\ 2& \end{pmatrix} \\\\ =(-1*-1)+(-1*0)+(3*2))\\=1+0+6\\=7$$

Melody has given the dot product - a scalar - of these two vectors...

However....if the cross-product was wanted, we have

(-1,-1,3) x (-1,0,2) = (u1, u2, u3) x (v1, v2, v3)  = (s1, s2, s3)

s1 = [-1*2] - [3*0] = -2 - 0  = -2

s2 = [3*-1] - [-1*2]  =  -3 - (-2)  = -3 + 2 = -1

s3 = [-1*0] - [-1* -1 ] = 0 - 1  = -1

The cross-product produces a vector in three-space that is perpendicular to the given vectors, namely..... (-2, -1, -1)

Here's a picture of the vectors for the cross-product:

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Thanks, Alan.....