$a$ and $b$ are unit vectors such that $a+2b$ is also a unit vector. If $\theta$ is the angle between $a$ and $b$,then $\sin \theta + \cos^3 \theta + \tan^5 \theta$ is equal to

$A$ unit vector in the $xy$-plane that makes an angle of $45^{\circ}$ with the vector $(i + j)$ and an angle of $60^{\circ}$ with the vector $(3i - 4j)$ is:

The projection of vector $\vec{a} = 2\hat{i} + 3\hat{j} - 2\hat{k}$ on the vector $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$ is:

Consider the following Assertion $(A)$ and Reason $(R)$:
Assertion $(A)$: The two lines $\bar{r}=\bar{a}+t(\bar{b})$ and $\bar{r}=\bar{b}+s(\bar{a})$ intersect each other.
Reason $(R)$: The shortest distance between the lines $\bar{r}=\bar{p}+t(\bar{q})$ and $\bar{r}=\bar{c}+s(\bar{d})$ is equal to the length of the projection of the vector $(\bar{p}-\bar{c})$ on $(\bar{q} \times \bar{d})$.
The correct answer is:

Let $\theta$ denote the angle between vectors $\vec{a}$ and $\vec{b}$. If $\vec{a}=2 \hat{i}+3 \hat{j}+6 \hat{k}$,$\vec{a} \cdot \vec{b}=4$ and $\theta=\cos ^{-1}\left(\frac{4}{21}\right)$,then $\vec{a}+\vec{b}$ is:

It is given that $a, b, c$ are vectors of lengths $6, 8, 10$ respectively. If $a$ is perpendicular to $(b+c)$, $b$ is perpendicular to $(c+a)$, and $c$ is perpendicular to $(a+b)$, then the length of the vector $a+b+c$ is (in $\sqrt{2}$)