The magnitude of vectors $\vec{a}$ and $\vec{b}$ are $1$ and $2$ respectively,and $\vec{a} \cdot \vec{b} = 1$. Then,the angle between the two vectors $\vec{a}$ and $\vec{b}$ is . . . . . . .

Let two non-collinear vectors $\hat{a}$ and $\hat{b}$ form an acute angle. $A$ point $P$ moves such that at any time $t$,the position vector $\overline{OP}$,where $O$ is the origin,is given by $\hat{a} \sin t + \hat{b} \cos t$. When $P$ is farthest from the origin $O$,let $M$ be the length of $\overline{OP}$ and $\hat{u}$ be the unit vector along $\overline{OP}$. Then:

Let $\vec{a}$ and $\vec{b}$ be two unit vectors and $\theta$ is the angle between them. Then $\vec{a}+\vec{b}$ is a unit vector if $\theta =$

If vectors $\vec{a}$ and $\vec{b}$ have the same magnitude, the angle between them is $60^\circ$, and their scalar product is $1/2$, then $|\vec{a}|$ is:

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 $\frac{\pi}{4}$?

If the vector $a = 3\hat{j} + 4\hat{k}$ is the sum of two vectors $a_1$ and $a_2$, where vector $a_1$ is parallel to $b = \hat{i} + \hat{j}$ and vector $a_2$ is perpendicular to $b$, then $a_1 =$