The value of $x$ for which the angle between the vectors $\vec{a} = x\hat{i} - 3\hat{j} - \hat{k}$ and $\vec{b} = 2x\hat{i} + x\hat{j} - \hat{k}$ is acute,and the angle between the vector $\vec{b}$ and the $y$-axis (axis of ordinate) is obtuse,are:

If $\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = 6\hat{i} - 3\hat{j} + 2\hat{k}$,then find $\vec{a} \cdot \vec{b}$.

The angle between the vectors $2 \hat{k} - 3 \hat{j}$ and $\hat{i} - 2 \hat{k}$ is

If $3\vec{a} - 5\vec{b}$ and $2\vec{a} + \vec{b}$ are perpendicular to each other,and $\vec{a} + 4\vec{b}$ and $-\vec{a} + \vec{b}$ are also perpendicular to each other,and $\theta$ is the angle between $\vec{a}$ and $\vec{b}$,then find $\cos \theta$.

If $\vec{a}=\hat{i}+(\tan \theta) \hat{j}+\left(\frac{3}{\sqrt{\sin \frac{\theta}{2}}}\right) \hat{k}$ and $\vec{b}=\tan \theta(\hat{j}-\hat{i})-\left(2 \sqrt{\sin \frac{\theta}{2}}\right) \hat{k}$ are orthogonal vectors and $\vec{c}=(\sin 2 \theta) \hat{i}-2 \hat{j}+2 \hat{k}$ makes an obtuse angle with $X$-axis, then $\theta=$

$A$ parallelogram is constructed with $5\vec{a} + 2\vec{b}$ and $\vec{a} - 3\vec{b}$ as its adjacent sides, where $|\vec{a}| = 2\sqrt{2}$ and $|\vec{b}| = 3$. The angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$. Find the lengths of the diagonals of the parallelogram.