If $\vec{a}\times \vec{b} = 2\vec{i}+\vec{j}-\vec{k}$ and $\vec{a}+\vec{b}=\vec{i}-\vec{j}+\vec{k}. $ Then Least value of $|\vec{a}|\;,$ is 
If $\vec{a}\times \vec{b} = 2\vec{i}+\vec{j}-\vec{k}$ and $\vec{a}+\vec{b}=\vec{i}-\vec{j}+\vec{k}. $ Then Least value of $|\vec{a}|\;,$ is 

$\bf{My\; Try::}$ Using The formula $$\left|\vec{a}\times \vec{b}\right|^2=|\vec{a}|^2|\vec{b}|^2-|\vec{a}\cdot \vec{b}|^2$$.
So we get $$6=|\vec{a}|^2|\vec{b}|^2-3\Rightarrow |\vec{a}||\vec{b}| = 3.$$
Now How can I calculate it, Help me, Thanks
 A: Hint: use Triangle inequality for $ \vec{b} = \vec i - \vec j + \vec k - \vec a  $ to get $ |\vec b| \leq \sqrt 3 + |\vec a|$. Substitute it on what you got.
A: We can write $\mathbf{a}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}\quad\text{and}\quad\mathbf{b}=(1-a_1)\mathbf{i}+(-1-a_2)\mathbf{j}+(1-a_3)\mathbf{k}$, then after develop the cross product we get
\begin{align*}
\mathbf{a}\times\mathbf{b}&=\left|\begin{matrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\a_1&a_2&a_3\\1-a_1&-1-a_2&1-a_3\end{matrix}\right|\\&=(a_2+a_3)\mathbf{i}+(-a_1+a_3)\mathbf{j}+(-a_1-a_2)\mathbf{k}\\
&=2\mathbf{i}+\mathbf{j}-\mathbf{k}
\end{align*}
We get a system of linear equations, which can be written as
$$\begin{pmatrix}0&1&1&|&2\\-1&0&1&|&1\\-1&-1&0&|&-1\end{pmatrix}\leadsto \begin{pmatrix}1&0&-1&|&-1\\0&1&1&|&2\\0&0&0&|&0\end{pmatrix}$$
Then $a_1=a_3-1,\;a_2=-a_3+2$ and so $$\left\|\mathbf{a}\right\|^2=(a_3-1)^2+(-a_3+2)^2+a_3^2=3a_3^2-6a_3+5=3(a_3-1)^2+2\geq 2$$
Therefore, the least value of $\left\|\mathbf{a}\right\|$ is $\sqrt{2}$.
A: $$2i+j-k = a \times b = a \times (i-j+k - a) = a \times (i-j+k) $$ taking the modulus, we get $$|a|\cdot\sqrt 3\cdot\sin(t) = \sqrt 6 \to |a|=\frac{\sqrt2}{\sin t}\ge \sqrt 2 \text{ since } 0 \le t \le \pi.$$  therefore the least value of $|a|$ is $\sqrt 2.$
