Vector Algebra ...

If $$ \displaystyle \bar{a}\times \bar{b}=\bar{b}\times \bar{c}=\bar{c}\times \bar{a} $$ then $$ \displaystyle \bar{a}+\bar{b}+\bar{c}=? $$

The vector $$\left ( \bar{a }\times\bar{b } \right )\times \left ( \bar{c }\times\bar{b } \right )$$ is

In a trapezium the vector $$\overline{BC} = \alpha \overline{AD}$$. We will then find that $$\bar{p}= \overline{AC}+\overline{BD}$$ is collinear with $$\overline{AD}$$. if $$\bar{p}= \mu \overline{AD}$$ then

If $$C$$ is the mid point of $$AB$$ and $$P$$ is any point outside $$AB$$, then

Given $$A=ai+bj+ck, \ B=di+3j+4k$$ and $$C=3i+j-2k$$. If the vectors $$A,B$$ and $$C$$ form a triangle such that $$A=B+C$$ and $$area(\Delta ABC)=5\sqrt6$$, then

A unit vector perpendicular to each of the vectors $$\displaystyle 2\hat{i}+4\hat{j}-\hat{k} $$ and $$\displaystyle \hat{i}-2\hat{j}+3\hat{k} $$ forming a right handed system is

If $$ M $$ and $$ N $$ are the midpoints of the diagonals $$AC$$ and $$BD$$, respectively, of a quadrilateral $$ABCD$$, then $$\overrightarrow {AB} + \overrightarrow {AD}  + \overrightarrow {CB} + \overrightarrow {CD} $$ is equal to

Two identical particles are located at $$\overrightarrow{x}$$ and $$ \overrightarrow{y}$$ with reference to the origin of three dimensional co-ordinate system. The position vector of centre of mass of the system is given by

If $$b \neq 0,$$ then every vector $$a$$ can be written in a unique manner as the sum of a vector $$a_{||}$$ parallel to b and a vector $$a_{\perp}$$ perpendicular to $$b$$. If $$a$$ is parallel to $$ b,$$ then $$a_{||} = a $$ and $$ a_{\perp } =0$$. If $$a$$ is perpendicular to $$b$$, then $$a_{||} = 0$$ and $$a_{\perp} = a$$. The vector $$a_{||}$$ is called the projection of $$a$$ on $$b$$ and is denoted by $$proj_{b}a$$. Since $$proj_{b} a$$ is parallel to $$b$$, it is a scalar multiple of the unit vector in the direction of $$b$$, i.e., $$proj _{b} a =\lambda u_{b}$$
The scalar $$\lambda $$ is called the component of $$a$$ in the direction of $$b$$ and is denoted by $$comp _{b} a $$. In fact, $$proj _{b} a = (a \cdot u_{b})u_{b}$$ and $$comp _{b} a = a \cdot u_{b}$$

If the points $$P, \ Q, \ R, \ S$$ have position vectors $$\overrightarrow p, \ \overrightarrow q, \ \overrightarrow r, \ \overrightarrow s$$ such that $$\overrightarrow p - \overrightarrow q = 2(\overrightarrow s - \overrightarrow r)$$, then find out the correct one: