$A$ hall has a square floor of dimension $10 \, m \times 10 \, m$ and vertical walls. If the angle $GPH$ between the diagonals $AG$ and $BH$ is $\cos^{-1} \frac{1}{5}$,then the height of the hall (in $meters$) is:

For any two non-zero vectors $\vec{a}$ and $\vec{b}$,$(a \vec{b} + b \vec{a}) \cdot (a \vec{b} - b \vec{a})$ is equal to:

Show that the area of the parallelogram whose diagonals are given by $\vec{a}$ and $\vec{b}$ is $\frac{|\vec{a} \times \vec{b}|}{2}$. Also,find the area of the parallelogram whose diagonals are $2 \hat{i}-\hat{j}+\hat{k}$ and $\hat{i}+3 \hat{j}-\hat{k}$.

If $a, b, c$ are the $p^{th}, q^{th}, r^{th}$ terms of an $H.P.$ and $\vec{u} = (q-r)\hat{i} + (r-p)\hat{j} + (p-q)\hat{k}$ and $\vec{v} = \frac{\hat{i}}{a} + \frac{\hat{j}}{b} + \frac{\hat{k}}{c}$,then:

The vector $\vec{a} = (\alpha, 2, \beta)$ lies in the plane of the vectors $\vec{b} = (1, 1, 0)$ and $\vec{c} = (0, 1, 1)$ and bisects the angle between $\vec{b}$ and $\vec{c}$. Then which one of the following gives the possible values of $\alpha$ and $\beta$?

Let a vector $\overrightarrow{a}=\sqrt{2}\hat{i}-\hat{j}+\lambda\hat{k}$,$\lambda>0$,make an obtuse angle with the vector $\overrightarrow{b}=-\lambda^{2}\hat{i}+4\sqrt{2}\hat{j}+4\sqrt{2}\hat{k}$ and an angle $\theta$,$\frac{\pi}{6} < \theta < \frac{\pi}{2}$,with the positive $z$-axis. If the set of all possible values of $\lambda$ is $(\alpha, \beta)-\{\gamma\}$,then $\alpha+\beta+\gamma$ is equal to . . . . . . .