Question

The position vectors of two vertices and the centroid of a triangle are $\overrightarrow{i}+\overrightarrow{j}$, $2\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k}$ and $\overrightarrow{k}$ respectively, then the position vector of the third vertex of the triangle is

• A. $-3\overrightarrow{i}+2\overrightarrow{k}$
• B. $3\overrightarrow{i}-2\overrightarrow{k}$
• C. $\overrightarrow{i}+\frac{2}{3}\overrightarrow{k}$
• D. None of these

Answer 1

The correct option is A $-3\overrightarrow{i}+2\overrightarrow{k}$

Let the third vertex be $C=x\overrightarrow{i}+y\overrightarrow{j}+z\overrightarrow{k}$
and $A=(1,1,0)$, $B=(2,-1,1)$, $G=(0,0,1)$
Hence $G=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3},\frac{z_1+z_2+z_3}{3})$
$(0,0,1)=(\frac{1+2+x}{3},\frac{1-1+y}{3},\frac{0+1+z}{3})$
$(0,0,1)=(\frac{x+3}{3},\frac{y}{3},\frac{z+1}{3})$
Hence $x+3=0$ gives $x=-3$
$\frac{y}{3}=0$ gives $y=0$
And $\frac{z+1}{3}=1$ gives $z=2$
Hence $C=-3\overrightarrow{i}+2\overrightarrow{k}$.