The value of $c$ such that for all real $x$, the vectors $\vec{a} = cxi - 6j + 3k$ and $\vec{b} = xi + 2j + 2cxk$ make an obtuse angle is:

Let $ABC$ be a triangle. Let a point $P$ divide $AB$ in the ratio $1:2$ internally and a point $Q$ divide $BC$ in the ratio $1:2$ internally. Let $D$ be the point of intersection of $AQ$ and $CP$. If the area of the triangle $ABC$ is $k$ square units,then the area of the triangle $BCD$ in square units is:

If $P=(0,1,2), Q=(4,-2,1)$ and $O=(0,0,0)$,then $\angle POQ=$

If $\overrightarrow{a}$ and $\overrightarrow{b}$ are unit vectors,then the greatest value of $\sqrt{3}|\overrightarrow{a}+\overrightarrow{b}|+|\overrightarrow{a}-\overrightarrow{b}|$ is

Let $\vec{a} = 6 \hat{i} - 3 \hat{j} - 6 \hat{k}$ and $\vec{d} = \hat{i} + \hat{j} + \hat{k}$. Suppose that $\vec{a} = \vec{b} + \vec{c}$,where $\vec{b}$ is parallel to $\vec{d}$ and $\vec{c}$ is perpendicular to $\vec{d}$. Then $\vec{c}$ is

Suppose $\overrightarrow{a}=\lambda \hat{i}-7 \hat{j}+3 \hat{k}$ and $\overrightarrow{b}=\lambda \hat{i}+\hat{j}+2 \lambda \hat{k}$. If the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is greater than $90^{\circ}$,then $\lambda$ satisfies the inequality: