Question

If the position vectors of three consecutive vertex of any parallelogram are respectively $\hat{i}+\hat{j}+\hat{k},\hat{i}+3\hat{j}+5\hat{k},7\hat{i}+9\hat{j}+11\hat{k}$ then the position vector of its fourth vertex is:

• A. $6(\hat{i}+\hat{j}+\hat{k})$
• B. $7(\hat{i}+\hat{j}+\hat{k})$
• C. $2\hat{j}-4\hat{k}$
• D. $\widehat{6i}+8\hat{j}+\widehat{10k}$

Answer 1

Answer: (B)

$7(\hat{i}+\hat{j}+\hat{k})$

Consider the problem

Let the fourth vertex be $D\left(\overrightarrow{d}\right)$ and the remaining three are given as,

$\begin{array}{c}A\left(\overrightarrow{a}\right)=\hat{i}+\hat{j}+\hat{k}\\ B\left(\overrightarrow{b}\right)=\hat{i}+3\hat{j}+5\hat{k}\\ C\left(\overrightarrow{c}\right)=7\hat{i}+9\hat{j}+11\hat{k}\\ D\left(\overrightarrow{d}\right)=\end{array}$

$ABCD$ is a parallelogram,

Then,

$\frac{1}{2}\left(\overrightarrow{b}+\overrightarrow{d}\right)=\frac{1}{2}\left(\overrightarrow{a}+\overrightarrow{c}\right)$

$\begin{array}{c}\overrightarrow{d}=\overrightarrow{a}+\overrightarrow{c}-\overrightarrow{b}\\ =8i+10j+12k-\left(i+3j+5k\right)\\ =7i+7j+7k=7\left(i+j+k\right)\end{array}$

Hence the option $B$ is correct answer.