Question

If $\overrightarrow{AB}=3\hat{i}-2\hat{j}+\hat{k}$, $\overrightarrow{BC}=\hat{i}+2\hat{j}-\hat{k}$, $\overrightarrow{CD}=2\hat{i}+\hat{j}+3\hat{k}$, $\overrightarrow{OA}=\hat{i}+\hat{j}+\hat{k}$, then the position vector of $\overrightarrow{OD}$ is

• A. $7\hat{i}-2\hat{j}-3\hat{k}$
• B. $7\hat{i}-2\hat{j}+3\hat{k}$
• C. $7\hat{i}+2\hat{j}-3\hat{k}$
• D. $7\hat{i}+2\hat{j}+4\hat{k}$

Answer 1

The correct option is D $7\hat{i}+2\hat{j}+4\hat{k}$

$OA+AB+BC+CD=OD$

Adding all vectors we get
$OD=7\hat{i}+2\hat{j}+4\hat{k}$

So Position Vector of $OD$ is
$D=7\hat{i}+2\hat{j}+4\hat{k}$