The component of $\hat{i}$ in the direction of the vector $\hat{i}+\hat{j}+2 \hat{k}$ is

If the $p^{th}$,$q^{th}$,and $r^{th}$ terms of a geometric progression are $a$,$b$,and $c$ respectively,then find the angle between the vectors $\vec{u} = (\log a)\hat{i} + (\log b)\hat{j} + (\log c)\hat{k}$ and $\vec{v} = (q - r)\hat{i} + (r - p)\hat{j} + (p - q)\hat{k}$.

Let $\vec{\alpha}=\hat{i}+\hat{j}+\hat{k}$, $\vec{\beta}=\hat{i}-\hat{j}-\hat{k}$ and $\vec{\gamma}=-\hat{i}+\hat{j}-\hat{k}$ be three vectors. $A$ vector $\vec{\delta}$, in the plane of $\vec{\alpha}$ and $\vec{\beta}$, whose projection on $\vec{\gamma}$ is $\frac{1}{\sqrt{3}}$, is given by

Assertion $(A)$: The scalar product of force $\vec{F}$ and displacement $\vec{r}$ is equal to the work done.
Reason $(R)$: The work done is not a scalar quantity.

The set of all $\alpha$,for which the vectors $\vec{a}=\alpha t \hat{i}+6 \hat{j}-3 \hat{k}$ and $\vec{b}=t \hat{i}-2 \hat{j}-2 \alpha t \hat{k}$ are inclined at an obtuse angle for all $t \in R$ is:

If $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ are unit vectors such that $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = 1$ and $\vec{a} \cdot \vec{c} = \frac{1}{2}$,then :-