Question

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

• A. $(0,\frac{5}{2},\frac{5}{4})$
• B. $(\frac{5}{2},\frac{5}{2},\frac{5}{4})$
• C. $(\frac{5}{2},0,\frac{5}{4})$
• D. $(\frac{5}{2},\frac{5}{4},0)$

Answer 1

The correct option is A $(0,\frac{5}{2},\frac{5}{4})$

Taking $P$ common from given condition

$4\overrightarrow{P}-(\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}+\overrightarrow{D})=0$
$\overrightarrow{P}=\frac{\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}+\overrightarrow{D}}{4}$

$\overrightarrow{P}=(0,\frac{10}{4},\frac{5}{4})$