In a quadrilateral $ABCD$,$\measuredangle A = \frac{2\pi}{3}$ and $\vec{AC}$ is the bisector of angle $A$. If $15|\vec{AC}| = 5|\vec{AD}| = 3|\vec{AB}|$,then the angle between $\vec{AB}$ and $\vec{BC}$ is

The points represented by $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are coplanar and $(\sin A)\vec{a} + (2\sin 2B)\vec{b} + (3\sin 3C)\vec{c} - 4\vec{d} = \vec{0}$. Then the least value of $\frac{21}{8}(\sin^2 A + \sin^2 2B + \sin^2 3C)$ is:

If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors satisfying $|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2=9$,then $|2 \vec{a}+5 \vec{b}+5 \vec{c}|$ is

If $\overrightarrow{a} \cdot \hat{i} = \overrightarrow{a} \cdot (2 \hat{i} + \hat{j}) = \overrightarrow{a} \cdot (\hat{i} + \hat{j} + 3 \hat{k}) = 1$,then $\overrightarrow{a}$ is equal to :

Let $ABCD$ be a parallelogram where $\overrightarrow{AB} = \overrightarrow{a}$,$\overrightarrow{AD} = \overrightarrow{b}$,$|\overrightarrow{a}| = |\overrightarrow{b}| = 2$ and $|\overrightarrow{a} \times \overrightarrow{b}| + \overrightarrow{a} \cdot \overrightarrow{b} = \sqrt{2} |\overrightarrow{a}| |\overrightarrow{b}|$,where $\overrightarrow{a} \cdot \overrightarrow{b} > 0$. Then the area of this parallelogram is (in square units):

If the angle between the vectors $\bar{a}=2 \lambda^2 \hat{i}+4 \lambda \hat{j}+\hat{k}$ and $\bar{b}=7 \hat{i}-2 \hat{j}+\lambda \hat{k}$ is obtuse,then $\lambda \in$