Vector Algebra ...

Let $$a,b,c,d $$ be the position vectors of the points $$\mathrm{A},\mathrm{B},\mathrm{C},\mathrm{D}$$ respectively. The condition for the figure $$ABCD$$ to be a parallelogram is

Let $$2\hat{i}+\hat{k}=\vec{\mathrm{a}},\ 3\hat{j}+4\hat{k}=\vec{{b}}$$, $$8\hat{i}-3\hat{j}$$ $$=\vec{\mathrm{c}}$$. If $$\vec{a}={x}\vec{b}+{y}\vec{{c}}$$, then $$(x,y) $$ is equal to

If $$A(\overline{a})$$ , $$B(\overline{b})$$ and $$C(\overline{c})$$ be the vertices of a triangle $$ABC$$ whose circumcentre is the origin then orthocentre is given by

Let $$\vec{A}= \hat{i}+6\mathrm{i}+6\mathrm{k},\vec{B}=-4\hat{i}+9\hat{i}+6\hat{k},\vec{G}=\displaystyle \dfrac{-5}{3}\hat{i}+\dfrac{22}{3}\hat{j}+\dfrac{22}{3}\hat{k}$$. If $$\mathrm{G}$$ is the centroid then the triangle $$ABC$$ is

If the position vectors of the points $$A, B, C, D$$ are$$(0,2, 1)$$, $$(3,1,1),$$ $$(-5,3,2)$$,$$(2,4,1)$$ respectively and if $$PA+PB+PC+PD=0$$ then the position vector of P is

If $$G$$ is the centroid of the triangle $$ABC$$ then $$\vec{GA}+\vec{GB}+\vec{GC}$$ is equal to 

Let $$\mathrm{A}\mathrm{B}\mathrm{C}$$ be a triangle and let $$\mathrm{S}$$ be its circumcentre and $$\mathrm{O}$$ be its orthocentre. The $$\overline{\mathrm{S}\mathrm{A}}+\overline{\mathrm{S}\mathrm{B}}+\overline{\mathrm{S}\mathrm{C}}= $$

If $$\vec{a} \times \vec{b} = \vec{b} \times \vec{a}$$, then

Taking $$O$$' as origin and the position vectors of $$A, B$$ are $$\vec i+3\vec{j}-2\vec k, 3\vec{i}+\vec{j}-2\vec{k}$$. The vector $$\overrightarrow{OC}$$ is bisecting the angle $$AOB$$ and if $$C$$ is a point on line $$\overrightarrow{AB}$$ then $$C$$ is

If $$\overline{p}$$ is the position vector of the orthocentre and $$\overline{g}$$ is the position vector of the centroid of the triangle $$ABC$$ when circumcenter is the origin and if $$\overline{p}=\lambda\overline{g}$$ then $$\lambda=$$