Vector Algebra ...

The unit vector in the direction of $$\overrightarrow{a}$$ is 

If $$\vec{a}$$ be the position vector whose tip is (5,-3), find the coordinates of a point B such that $$ \vec{AB} = \vec{a},$$ the coordinates of A being (4,-1).

find the coordinate of the tip of the position vector which is equivalent to $$ \vec{AB}$$, where the coordinates of A and B are (-1, 3) and (-2, 1) respectively.

If the position vectors of the points $$A(3,4),B(5, -6)$$ and $$C(4,-1)$$ are $$ \vec{a}, \vec{b}, \vec{c}$$ respectively, compute $$ \vec{a}+2\vec{b}-3\vec{c}. $$

If $$\vec a$$ is parallel to $$\vec b \times \vec c$$, then $$(\vec a \times \vec b) \cdot (\vec a \times \vec c)$$ is equal to

If $$\mid \vec a \mid = 2$$ and $$\mid \vec b \mid = 3$$ and $$\vec a \cdot \vec d = 0$$, then $$(\vec a \times (\vec a \times (\vec a \times (\vec a \times \vec b ))))$$ is equal to

Let $$\mathrm{A}\mathrm{B}\mathrm{C}$$ be a triangle and let $$\mathrm{D},\mathrm{E}$$ be the midpoints of the sides $$\mathrm{A}\mathrm{B},\mathrm{A}\mathrm{C}$$ respectively,then $$\hat { BE } +\hat { DC } =$$

The position vectors of $$A,B,C$$ are $$\overline{i}+\overline{j}+\overline{k},\ 4\overline{i}+\overline{j}+\overline{k},\ 4\overline{i}+5\overline{j}+\overline{k}$$ . Then the position vector of the circumcentre of the triangle $$ABC$$ is

 If $$\mathrm{O}$$ is the circumcentre and $$\mathrm{O}^{'}$$ is the orthocentre of a triangle $$\mathrm{A}\mathrm{B}\mathrm{C}$$ and if $$\mathrm{A}\mathrm{P}$$ is the circumdiameter then
$$\vec{\mathrm{A}\mathrm{O}}+\vec{\mathrm{O}^{'}\mathrm{B}}+\vec{\mathrm{O}^{'}\mathrm{C}}=$$

If the vectors  $$\overline{a}=3\overline{i}+\overline{j}-2\overline{k}$$,$$\overline{b}=-\overline{i}+3\overline{j}+4\overline{k},\ \overline{c}=4\overline{i}-2\overline{j}-6\overline{k}$$  form the sides of the triangle then length of the median bisecting the vector $${c}$$ is