If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}|=|\vec{b}|=\sqrt{14}$ and $\vec{a} \cdot \vec{b}=-7$,then $\frac{|\vec{a} \times \vec{b}|}{|\vec{a} \cdot \vec{b}|}=$

Find a unit vector in the $xy$-plane which makes an angle of $45^{\circ}$ with the vector $i + j$ and an angle of $60^{\circ}$ with the vector $3i - 4j$.

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 :-

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

$\bar{a}, \bar{b}, \bar{c}$ are unit vectors. If $\bar{a}, \bar{b}$ are perpendicular vectors,$(\bar{a}-\bar{c}) \cdot(\bar{b}+\bar{c})=0$ and $\bar{c}=l \bar{a}+m \bar{b}+n(\bar{a} \times \bar{b})$ ($l, m, n$ are scalars),then $n^2=$

Suppose $\overrightarrow{a}=\lambda \hat{i}-7 \hat{j}+3 \hat{k}$ and $\overrightarrow{b}=\lambda \hat{i}+\hat{j}+2 \lambda \hat{k}$. If the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is greater than $90^{\circ}$, then $\lambda$ satisfies the inequality: