What will be the length of the longer diagonal of the parallelogram constructed on vectors $5a + 2b$ and $a - 3b$,given that $|a| = 2\sqrt{2}$,$|b| = 3$,and the angle between $a$ and $b$ is $\frac{\pi}{4}$?

Two adjacent sides of a parallelogram are represented by $\vec{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\vec{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$. The side $\vec{AD}$ is rotated by an acute angle $\theta$ in the plane of the parallelogram such that $\vec{AD}$ becomes $\vec{AD'}$. If $\vec{AD'}$ is perpendicular to $\vec{AB}$,find $\cos \theta$.

Let $\vec{a}=2 \hat{i}+5 \hat{j}-\hat{k}$,$\vec{b}=2 \hat{i}-2 \hat{j}+2 \hat{k}$ and $\vec{c}$ be three vectors such that $(\vec{c}+\hat{i}) \times (\vec{a}+\vec{b}+\hat{i}) = \vec{a} \times (\vec{c}+\hat{i})$ and $\vec{a} \cdot \vec{c} = -29$. Then $\vec{c} \cdot (-2 \hat{i}+\hat{j}+\hat{k})$ is equal to:

The vectors $\vec{AB} = 3\hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{BC} = \hat{i} - 2\hat{k}$ are the adjacent sides of a parallelogram. The angle between its diagonals is

If $P(6, 10, 10)$,$Q(1, 0, -5)$,$R(6, -10, \lambda)$ are vertices of a triangle right-angled at $Q$,then the value of $\lambda$ is ....

Forces of magnitudes $3$ and $2$ units acting in the directions $5\hat{i} + 3\hat{j} + 4\hat{k}$ and $3\hat{i} + 4\hat{j} - 5\hat{k}$ respectively act on a particle which is displaced from the points $(1, -1, -1)$ to $(3, 3, 1)$. The work done by the forces is equal to