If the points whose position vectors are $2 \hat{i}+\hat{j}+\hat{k}$, $6 \hat{i}-\hat{j}+2 \hat{k}$, and $14 \hat{i}-5 \hat{j}+p \hat{k}$ are collinear, then the value of $p$ is:

If $a = (2, 5)$ and $b = (1, 4),$ then the vector parallel to $(a + b)$ is

Let $a$ and $b$ be unit vectors with $\theta$ as the acute angle between them. If $\frac{1}{2}|a-b|=\sin(\lambda \theta)$,then $4 \lambda^2=$

The vectors $\overrightarrow{AB} = 3 \hat{i} + 4 \hat{k}$ and $\overrightarrow{AC} = 5 \hat{i} - 2 \hat{j} + 4 \hat{k}$ are the sides of a triangle $ABC$. The length of the median through $A$ is

Find the unit vector in the direction of the vector $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$.

Let $O$ be the origin, $\vec{OP} = \vec{a}$ and $\vec{OQ} = \vec{b}$. If $R$ is the point on $\vec{OP}$ such that $\vec{OP} = 5\vec{OR}$, and $M$ is the point such that $\vec{OQ} = 5\vec{RM}$, then $\vec{PM}$ is equal to: