Vector Algebra ...

Given a parallelogram $$ABCD$$. If $$|\vec{AB}=a, |\vec{AD}|=b$$ and $$|vec{AC}|=c$$, then $$|\vec{DB}|.|\vec{AB}|$$ has the

Let $$\vec a$$ and $$\vec b$$ be two unit vectors such that $$\left| {\vec a + \vec b} \right| = \sqrt 3$$. If $$\vec c = \vec a + 2\vec b + 3(\vec a \times \vec b)$$, then $$2\left| {\vec c} \right|$$ is equal to:

Let $$A(\overline { a } ),B(\overline { b } ), C(\overline { c } )$$ be the vertices of the triangle $$ABC$$ and let $$D,E,F$$ be the mid points of the sides $$BC,CA,AB$$ respectively. If $$P$$ divides the median $$AD$$ in the ratio $$2:1$$ then the position vector of $$P$$ is

If $$|\bar {a}-\bar {b}|=|\bar {a}|=|\bar {b}|$$, where $$\bar a$$ and $$\bar b$$ are non zero vecrors then the angle between $$\bar {a}-\bar {b}$$ and $$\bar b$$ is

$$ABC$$ is an isosceles triangle right angled at $$A$$. Force of magnitude $$2\sqrt{2},5$$ and $$6$$ act along $$\overline { BC } ,\overline { CA } ,\overline { AB } $$ respectively. The magnitude of their resultant force is

The length of the projection of the line segment joining the points $$(5, -1, 4)$$ and $$(4, -1, 3)$$ on the plane , $$ x+ y+ z = 7$$ is : 

$$\vec{a},\vec{b},\vec{c},\vec{d}$$ are the position vectors of four coplanar points A,B,C,D respectively. If no three of them are collinear and $$|\vec{a}-\vec{d}|=|\vec{b}-\vec{d}|=|\vec{c}-\vec{d}|$$ then for triangle ABC, D is

If AD, BE and CF are $$\Delta ABC$$, then $$\\ \vec { AD } +\vec { BE } \vec { +CF } $$

Let $$\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}, b=\hat{i}+2\hat{j}-2\hat{k}$$. Then a unit vector perpendicular to both $$\vec{a}-\vec{b}$$ and $$\vec{a}+\vec{b}$$ is :

If $$\vec a,\vec b, \vec c$$ are three coplanar vectors, then $$v\left[ 2\vec { a } +3\vec { b } ,2\vec { b- } 5\vec { c } ,2\vec { c } +3\vec { a }  \right]$$ is