Gradient in vector calculus - is this answer correct? $\nabla^2 \| a\times r \| \!\,^2$

Question

Have I done this correctly? I think I have done something wrong but I'm not sure where.

Question: Assume $r=x\textbf i +y\textbf j+z\textbf k$ and $a=a_1\textbf i +a_2\textbf j+a_3\textbf k$ for some constants $a_1,a_2,a_3$. Calculate the following expression.

$\nabla^2 \| a\times r \| \!\,^2$

Answer: $\frac{d^2}{d x^2}$$(\| a\times r \| \!\,)^2 +\frac{d^2}{d y^2}(\| a\times r \| \!\,)^2 +\frac{d^2}{d z^2}(\| a\times r \| \!\,)^2 $

and $(\| a\times r \| \!\,)^2 = (a_1\times x)^2+(a_2\times y)^2 + (a_3\times z)^2$

Therefore, $\nabla^2(\| a\times r \| \!\,)^2 =6$

Answer 1

This is not correct. You should check the definition of the cross product.
$$||a\times r||^2=||(a_2z-a_3y)i+(a_3x-a_1z)j+(a_1y-a_2x)k||^2=(a_2z-a_3y)^2+(a_3x-a_1z)^2+(a_1y-a_2x)^2$$ That is the function whose partial derivatives you will have to take, to find the (scalar) Laplacian. The correct answer is $4(a_1^2+a_2^2+a_3^2)=4||a||^2$.

Answer 2

Since $(a\times r)^2=a^2r^2-(a\cdot r)^2$ and $\nabla^2\phi^2=2\phi\nabla^2\phi+2(\nabla\phi)^2$ for any scalar $\phi$,$$\nabla^2(a\times r)^2=a^2(\partial_x^2+\partial_y^2+\partial_z^2)(x^2+y^2+z^2)-2(a\cdot r)\sum_ia_i\nabla^2x_i-2a^2=4a^2.$$