Let $a = i + 2j + k$, $b = i - j + k$, and $c = i + j - k$. $A$ vector in the plane of $a$ and $b$ has a projection of $\frac{1}{\sqrt{3}}$ on $c$. Then, one such vector is:

If $\theta$ is the angle between the unit vectors $\mathbf{a}$ and $\mathbf{b}$,then $\cos \frac{\theta}{2} = $

If $a$ and $b$ are respectively the internal and external bisectors of the angles between the vectors $u = -\hat{i} + 2\hat{j} - 2\hat{k}$ and $v = 3\hat{i} + 4\hat{j}$, and $|a| = \frac{2}{3}\sqrt{6}$, $|b| = \frac{2}{3}\sqrt{3}$, then one of the values of $a - b$ is

Let $\vec{a}=\hat{i}-2\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$ be two vectors. If $\vec{c}$ is a vector such that $\vec{b} \times \vec{c}=\vec{b} \times \vec{a}$ and $\vec{c} \cdot \vec{a}=0$,then $\vec{c} \cdot \vec{b}$ is equal to

If $a, b, c$ are three vectors such that $a \perp (b + c)$, $b \perp (c + a)$ and $c \perp (a + b)$, and $|a| = 1, |b| = 2, |c| = 3$, then $|a + b + c|$ is equal to:

Three vectors $\vec{a}, \vec{b}, \vec{c}$ satisfy the condition $\vec{a}+\vec{b}+\vec{c}=\vec{0}$. If $|\vec{a}|=1, |\vec{b}|=3, |\vec{c}|=4$,then find the value of $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$.