Question

Let
A=2x^2i−3yzj+xz^2k
and
ϕ=2z−x^3y
, find
A×▽ϕ
at point (1,-1,1).

Answer 1

$$\nabla \phi=-3x^2yi-x^3j+2k$$

$$A\times \nabla \phi=\begin{vmatrix} i & j & k \\ 2x^2 & -3yz & xz^2 \\ -3x^2y & -x^3 & 2 \end{vmatrix}$$

$$=i\begin{vmatrix} -3yz & xz^2 \\ -x^3 & 2 \end{vmatrix}-j\begin{vmatrix} 2x^2 & xz^2 \\ -3x^2y & 2 \end{vmatrix}+k\begin{vmatrix} 2x^2 & -3yz \\ -3x^2y & -x^3 \end{vmatrix}$$

$$=(-6yz+x^4z^2)i+(-4x^2-3x^3yz^2)j$$

$$+(-2x^5-9x^2y^2z)k$$

$(1,-1,1)$

$$A\times \nabla \phi|_{(1, -1, 1)}=(6+1)i+(-4+3)j+(-2-9)k$$

$$=7i-j-11k$$