Least value of magnitude of a vector
Suppose a and b are vectors such that $a \times b = 2i + j - k$ and $a+b= i - j + k$. The least value of magnitude of vector $a$ is ?
Here $i$, $j$ and $k$ are unit vectors in direction of $x$, $y$ and $z$ axes and $\times$ symbolizes cross product.
Any help would be appreciated.
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$\begingroup$If $a\times b=u$ and $a+b=v$ then $a\times(v−a)=a\times v=u$.
$(a1,a2,a3)\times(1,-1,1)=(a2+a3,a3-a1,-a1-a2)=(2,1,-1)$.
Find one solution, $a_0=(a1,a2,a3)$, to $a\times v=u$. The general solution will be $a_0+kv=(a1+k,a2-k,a3+k)$.
Write $|a_0+kv|^2$ as a function of $k$. It will be a quadratic. Find the minimum value of the quadratic.
$ a \times b = 2i+j-k$ and $a+b = i-j+k $
So $a \times(i-j+k-a)=2i+j-k$, so $a \times (i-j+k)=2i+j-k$
Let $a = xi+yj+zk$
putting on above equation and finding cross product and comparing coefficients we get $x+y=1$,$y+z=2$,$z-x=1$.
The required answer is $(x^2+y^2+z^2)^{1/2}$ using above 3 relations and eliminating $y$ and $z$ we get $(3x^2+2)^{1/2}$. The minimum value of a square of a number is 0(considering real and excluding complex numbers) so the answer is $2^{1/2}$ or $\sqrt{2}$.
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