In a trapezium $ABCD$,$\vec{BC} = \lambda \vec{AD}$ and $\vec{x} = \vec{AC} + \vec{BD}$. If $\vec{x} = p \vec{AD}$,then $p =$

If the angle $\theta$ between the vectors $\overrightarrow{a}=2 x^2 \hat{i}+4 x \hat{j}+\hat{k}$ and $\overrightarrow{b}=7 \hat{i}-2 \hat{j}+x \hat{k}$ is such that $90^{\circ} < \theta < 180^{\circ}$,then $x$ lies in the interval

If $a = 2i + 2j + 3k$,$b = -i + 2j + k$,and $c = 3i + j$,then $a + tb$ is perpendicular to $c$ if $t = $

If $\overline{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}$,$\overline{b}=\hat{i}-2 \hat{j}+\hat{k}$,and $\overline{c}=\hat{i}+\hat{j}-\hat{k}$ are three vectors,and there exists a vector $\overline{r}$ such that $\overline{r} \times \overline{a}=\overline{b}$ and $\overline{r} \cdot \overline{c}=3$,then the value of $|\overline{r}|$ is:

Let $O$ be the origin and let $PQR$ be an arbitrary triangle. The point $S$ is such that $\overline{OP} \cdot \overline{OQ} + \overline{OR} \cdot \overline{OS} = \overline{OR} \cdot \overline{OP} + \overline{OQ} \cdot \overline{OS} = \overline{OQ} \cdot \overline{OR} + \overline{OP} \cdot \overline{OS}$. Then the triangle $PQR$ has $S$ as its

If $l_{1}, m_{1}, n_{1}; l_{2}, m_{2}, n_{2}; l_{3}, m_{3}, n_{3}$ are the direction cosines of three mutually perpendicular lines,prove that the line whose direction cosines are proportional to $l_{1}+l_{2}+l_{3}, m_{1}+m_{2}+m_{3}, n_{1}+n_{2}+n_{3}$ makes equal angles with them.