Vector Algebra ...

Let $$A=(-3,4,-8)$$ and $$B=(5,-6,4)$$, then the coordinates of the point in which the $$XY$$- plane or $$XOY$$ plane divides the line segment $$AB$$ is

If $$3\vec{a}+4\vec{b}-7\vec{c}=0$$ then the ratio in which $$C(\vec{c})$$ divides the join of $$A(\vec{a})$$ and $$B(\vec{b})$$ is

The resultant of two concurrent forces $$n\vec{OP}$$ and $$m\vec{OQ}$$ is $$(m+n)\vec{OR}$$. Then $$R$$ divides $$PQ$$ in the ratio

Let $$\vec{A}=\hat{i}+2\hat{j}{+}3\hat{k},\ \vec{B}=4\hat{i}+2\hat{j},\ \vec{C}=2\hat{i}+2\hat{j}{+}2\hat{k}$$. Then the ratio in which $$C$$ divides $$AB$$ is

$$P, Q, R, S$$ have position vectors $$\overline{p},\overline{q},\overline{r},\overline{s}$$ respectively such that $$\overline{p}-\overline{q}=2(\overline{s}-\overline{r})$$, then which of the following is correct

If $$\vec{a}+2\vec{b},2\vec{a}+\vec{b}$$ be the position vectors of the points $$A$$ and $$B$$, then the position vector of the point $$C$$ which divides $$AB$$ internally in the ratio $$2:1$$ is

In $$\Delta ABC$$, $$ P,\ Q,\ R$$ are points on $$BC,\ CA,\ AB$$ respectively, dividing them in the ratio $$1:4$$, $$3:2$$ and $$3 : 7$$. The point $$S$$ divides $$AB$$ in the ratio $$1:3$$. Then $$\displaystyle \dfrac{|\overline{AP}+\overline{BQ}+\overline{CR}|}{|\overline{CS}|}$$ is

If $$\vec a=\hat i+2\hat j$$ and $$\vec b = 3\hat j$$, then $$\vec a\cdot\vec b=$$

If $$\overline{a}$$ and $$\overline{b}$$ are position vectors of $$A$$ and $$B$$ respectively, then the position vector of a point $$C$$ in $$AB$$ produced such that $$\overline{AC}=3\overline{AB}$$ is

If $$\left[ \overrightarrow { a } \overrightarrow { b } \overrightarrow { c }  \right] =1$$ then $$\frac { \overrightarrow { a } .\overrightarrow { b }\times \overrightarrow { c }  }{ \overrightarrow { c }\times \overrightarrow { a } .\overrightarrow { b }  } +\frac { \overrightarrow { b } .\overrightarrow { c }\times \overrightarrow { a }  }{ \overrightarrow { a }\times \overrightarrow { b } .\overrightarrow { c }  } +\frac { \overrightarrow { c } .\overrightarrow { a }\times \overrightarrow { b }  }{ \overrightarrow { b }\times \overrightarrow { c } .\overrightarrow { a }  }$$ is equal to