Way to show curves intersect each other through derivatives and vector question. !->a!=a,then find the value of-: (all a's are vectors and i,j,k are unit vectors )(! is modulus). !a x i!2+!a x j!2+!a x k!2. Can you also suggest something for-:show that the curves xy=a2 and x2 +y2 =2a2 touch each other.

I am not sure whether I understand correctly the notation in the original question, but anyway.

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$\vec a\times\mathbf i=(a_1,a_2,a_3)\times(1,0,0)=(0,a_3,-a_2)$ $\Rightarrow$ $|\vec a\times\mathbf i|^2=a_2^2+a_3^3$
Similarly you get: $|\vec a\times\mathbf j|^2=a_1^2+a_3^3$ and $|\vec a\times\mathbf k|^2=a_1^2+a_2^3$.

Together you get: $|\vec a\times \mathbf i|^2+|\vec a\times \mathbf j|^2+|\vec a\times \mathbf k|^2=2(a_1^2+a_2^2+a_3^2)=2|\vec a|^2$.

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The curve $x^2+y^2=2a^2$ is a circle going through the point $(a,a)$. You can see that the tangent at this point has slope $-1$. (This is clear from the symmetry of $x$ and $y$-coordinate.)

Similarly, $xy=a^2$ is a hyperbole going through the point $(a,a)$ and the tangent at this point has slope $-1$.