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

The position vectors of $A$ and $B$ are $(\hat{i}+\hat{j}+\hat{k})$ and $(\frac{1}{3} \hat{j}+\frac{1}{3} \hat{k})$. If $B$ divides the line segment $AC$ in the ratio $2:1$,then the position vector of $C$ is

The direction ratios of the line bisecting the angle between the $x$-axis and the line having direction ratios $(3, -1, 5)$ are

For any two points $M$ and $N$ in the $XY$-plane, let $\overrightarrow{MN}$ denote the vector from $M$ to $N$, and $\overrightarrow{0}$ denote the zero vector. Let $P, Q$ and $R$ be three distinct points in the $XY$-plane. Let $S$ be a point inside the triangle $\triangle PQR$ such that $\overrightarrow{SP} + 5\overrightarrow{SQ} + 6\overrightarrow{SR} = \overrightarrow{0}$. Let $E$ and $F$ be the mid-points of the sides $PR$ and $QR$, respectively. Then the value of $\frac{\text{length of the line segment } EF}{\text{length of the line segment } ES}$ is: (in $.20$)

$P$ is the point of intersection of the diagonals of the parallelogram $ABCD$. If $S$ is any point in space and $\vec{SA} + \vec{SB} + \vec{SC} + \vec{SD} = \lambda \vec{SP}$, then $\lambda$ equals

If $\hat{i}, \hat{j}, \hat{k}$ are unit vectors,then which of the following is true?