If the vector $a = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$ and $b$ are collinear and $|b| = 21$, then $b$ is equal to:

If $\vec{a}=\vec{b}+\vec{c},$ then is it true that $|\vec{a}|=|\vec{b}|+|\vec{c}|$ ? Justify your answer.

In a quadrilateral $PQRS$,$M$ and $N$ are mid-points of the sides $PQ$ and $RS$ respectively. If $\vec{PS} + \vec{QR} = t \vec{MN}$,then $t =$

$A$ girl walks $4 \, km$ towards west,then she walks $3 \, km$ in a direction $30^{\circ}$ east of north and stops. Determine the girl's displacement from her initial point of departure.

Let $\alpha, \beta, \gamma$ be distinct real numbers. The points with position vectors $\alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$,$\beta \hat{i} + \gamma \hat{j} + \alpha \hat{k}$,and $\gamma \hat{i} + \alpha \hat{j} + \beta \hat{k}$:

If $M_1, M_2, M_3$ and $M_4$ are respectively the magnitudes of the vectors $\vec{a}_1 = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{a}_2 = -3\hat{i} - 4\hat{j} - 4\hat{k}$,$\vec{a}_3 = -\hat{i} + \hat{j} - \hat{k}$,and $\vec{a}_4 = -\hat{i} + 3\hat{j} + \hat{k}$,then the correct order of $M_1, M_2, M_3$ and $M_4$ is: