The position vectors of three points $A, B$ and $C$ are $(1, 3, x), (3, 5, 8)$ and $(y, -1, -6)$ respectively. If $A, B$ and $C$ are collinear, then $(x, y) =$

If $m_1, m_2, m_3$ and $m_4$ are respectively the magnitudes of the vectors $\overrightarrow{a}_1=2 \hat{i}-\hat{j}+\hat{k}$, $\overrightarrow{a}_2=3 \hat{i}-4 \hat{j}-4 \hat{k}$, $\overrightarrow{a}_3=\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{a}_4=-\hat{i}+3 \hat{j}+\hat{k}$, then the correct order of $m_1, m_2, m_3$ and $m_4$ is

If $\vec{a} = t \vec{b}$ where $t < 0$ is a scalar,then

Let $\vec{a}$ and $\vec{b}$ be non-collinear vectors. If the vectors $(\lambda-1) \vec{a}+2 \vec{b}$ and $3 \vec{a}+\lambda \vec{b}$ are collinear, then the set of all possible values of $\lambda$ is

Let $|\vec{a}| = |\vec{b}| = |\vec{a} - \vec{b}| = 1$,then the angle between $\vec{a}$ and $\vec{b}$ is:

$3\overrightarrow{OD} + \overrightarrow{DA} + \overrightarrow{DB} + \overrightarrow{DC} = $