If the points $a + b$,$a - b$,and $a + kb$ are collinear,then $k =$

The position vectors of $A$ and $B$ are $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{b} = 3\hat{i} - \hat{j} + 3\hat{k}$. The position vector of the midpoint of the line segment $AB$ is:

Statement $(A):$ If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} + \vec{b} + \vec{c} = 0$,then $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2}$.
Reason $(R): (\vec{x} + \vec{y})^2 = |\vec{x}|^2 + |\vec{y}|^2 + 2(\vec{x} \cdot \vec{y})$.

Let $u, v$ and $w$ be non-coplanar vectors. Then the points corresponding to which of the following vectors are collinear?

The direction cosines of the vector $3\hat{i} - 4\hat{j} + 5\hat{k}$ are

The points with position vectors $10\,i + 3\,j$,$12\,i - 5\,j$,and $a\,i + 11\,j$ are collinear,if $a = $