If $ABCDEF$ is a regular hexagon with $\vec{AB} = \vec{a}$ and $\vec{BC} = \vec{b}$,then $\vec{CE}$ equals to

If a vector $\vec{v} = 3\hat{i} + 4\hat{j} - 5\hat{k}$ is rotated about the origin, its magnitude remains constant. If the new vector is $\vec{v}' = (a + 1)\hat{i} - 3\hat{j} + 5\hat{k}$, find the possible values of $a$.

Let $ABC$ be a triangle and $\bar{a}, \bar{b}, \bar{c}$ be the position vectors of $A, B, C$ respectively. If $D$ divides $BC$ in the ratio $2:3$ internally and $E$ divides $CA$ in the ratio $2:1$ internally, then the position vector of the point $P$ which divides $DE$ in the ratio $3:5$ internally is

If $\vec{PO} + \vec{OQ} = \vec{QO} + \vec{OR}$,then

Let $u$ and $v$ be non-collinear vectors in $\mathbb{R}^2$. Let $w$ be the orthogonal projection vector of $u$ on $v$. Consider two statements:
$(i)$ Any vector in $\mathbb{R}^2$ can be written as a linear combination of $u$ and $v$.
(ii) $w$ can be written as a linear combination of $u$ and $v$ as $w = au + bv$,where both $a$ and $b$ are non-zero real numbers.

Three points whose position vectors are $a + b$,$a - b$,and $a + kb$ will be collinear if the value of $k$ is: