Vector Algebra ...

If $$ \sum_{i=1}^{n} \vec{a}_{i}=\vec{0} $$ where $$ \left|\vec{a}_{i}\right|=1 \forall i,  $$ then the value of $$ \sum_{1 \leq i<j \leq n} \vec{a}_{i} \cdot \vec{a}_{j}  $$ is

In parallelogram ABCD, $$|\overline { AB } |=a,|\overline { AD } |=b$$ and $$|\overline { AC } |=c$$ then $$\overline { DB } ,\overline { AB } $$ has the value

$$|\hat{i} \times p|^2 + |\hat{j} \times p|^2 +|\hat{k} \times p|^2$$ is equal to

If the vectors $$\vec{a}$$ , $$\vec{b}$$ , $$\vec{c}$$ satisfying $$\vec{a}$$ + $$\vec{b}$$ + 2$$\vec{c}$$ = 0 . If $$\left | \vec{a} \right |$$ = 1 , $$\left | \vec{b} \right |$$ = 4 , $$\left | \vec{c} \right |$$ = 2 , then $$\vec{a}.\vec{b}$$ + $$\vec{b}.\vec{c}$$ + $$\vec{c}.\vec{a}$$ = 

A stone projected vertically upwards raises 's' feets in 't' seconds where $$_{ S }=112t-{ 16t }^{ 2 }$$ then the maximum height it reached is

The position vector of A is $$2\vec { i } +3\vec { j } +4\vec { k } $$$$\vec { AB } =5\vec { i } +7\vec { j } +6\vec { k } $$, then the position vector of B is

The projection of the join of the point (3, 4, 2), (5, 1, 8) on the line whose d.c.'s are $$\left( \dfrac { 2 }{ 7 } ,-\dfrac { 3 }{ 7 } ,\dfrac { 6 }{ 7 }  \right) $$ is 

Let OAB be a regular triangle with side unity (o being otogin). Also M, N are the points of intersection of AB, M being closer to A and N closer to B. Position vectors of A, B, M and N are $$\vec { a } ,\vec { b } ,\vec { m } $$ and $$\vec { n } $$ respectively. Which of the following hold (s) good ?

Given $$\overline { a } = x \hat { i } + y \hat { j } + 2 \hat { k } , \overline { b } = \hat { i } - \hat { j } + \hat { k } , \overline { c } = \hat { i } + 2 \hat { j } ;( \overline { a } \hat {  } \overline { b } ) = \pi / 2 , \overline { a } .\overline { c } = 4$$ then

If $$\bar { a } ,\bar { b } ,\bar { c } $$ are position vectors of the points A,B,C respectively such that $$9\bar { a } -7\bar { b } -2\bar { c } =\bar { 0 } $$ then point B divides AC in the ratio.....