If $\bar{a}, \bar{b}, \bar{c}$ are three unit vectors such that $|\bar{a}+\bar{b}+\bar{c}|=1$ and $\bar{b}$ is perpendicular to $\bar{c}$. If $\bar{a}$ makes angles $\alpha$ and $\beta$ with $\bar{b}$ and $\bar{c}$ respectively,then the value of $\cos \alpha+\cos \beta$ is:

In a $\triangle ABC$,let $G$ denote its centroid and let $M, N$ be points in the interiors of the segments $AB, AC$,respectively,such that $M, G, N$ are collinear. If $r$ denotes the ratio of the area of $\triangle AMN$ to the area of $\triangle ABC$,then

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 . . . . . . .

The scalar product of the vector $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ with a unit vector along the sum of the vectors $\vec{b} = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $\vec{c} = \lambda \hat{i} + 2 \hat{j} + 3 \hat{k}$ is equal to $1$. Then,$\lambda =$

If $|\vec{a}|=2, |\vec{b}|=5$ and $|\vec{a} \times \vec{b}|=8$,then $|\vec{a} \cdot \vec{b}|$ is equal to :

In a plane, $\vec{a}$ and $\vec{b}$ are the position vectors of two points $A$ and $B$ respectively. $A$ point $P$ with position vector $\vec{r}$ moves on that plane in such a way that $|\vec{r}-\vec{a}| - |\vec{r}-\vec{b}| = c$ (where $c$ is a real constant). The locus of $P$ is a conic section whose eccentricity is: