question_answer

If \[\hat{i}+2\hat{j}+3\hat{k}\]  and \[2\hat{i}-\hat{j}+4\hat{k}\] are the position vectors of the points A and B, then the position vector of the points of trisection of AB are

A)  \[\frac{4}{3}\hat{i}+\hat{j}+\frac{10}{3}\hat{k},\,\frac{5}{3}\hat{i}+\frac{11}{3}\hat{k}\]

B)  \[-\frac{4}{3}\hat{i}-\hat{j}-\frac{10}{3}\hat{k},-\,\frac{5}{3}\hat{i}-\frac{11}{3}\hat{k}\]

C)  \[\frac{4}{3}\hat{i}-\hat{j}-\frac{10}{3}\hat{k},-\,\frac{5}{3}\hat{i}-\frac{11}{3}\hat{k}\]

D)  \[-\frac{4}{3}\hat{i}+\hat{j}-\frac{10}{3}\hat{k},\,\frac{5}{3}\hat{i}-\frac{11}{3}\hat{k}\]

Correct Answer: A

Solution :

Given,  \[\overrightarrow{OA}=\hat{i}+2\hat{j}+3\hat{k}\] and  \[\overrightarrow{OB}=2\hat{i}-\hat{j}+4\hat{k}\] Let point \[C({{x}_{1}},{{y}_{1}},{{z}_{1}})\]divide AB in the ratio \[1:2\] \[\therefore \] \[{{x}_{1}}=\frac{2+2}{1+2}=\frac{4}{3}\] \[{{y}_{1}}=\frac{-1+4}{1+2}=\frac{3}{3}=1\] and \[{{z}_{1}}=\frac{4+6}{1+2}=\frac{10}{3}\] Again let point \[D({{x}_{2}},{{y}_{2}},{{z}_{1}})\] divides AB in the ratio \[2:1\] \[\therefore \] \[{{x}_{2}}=\frac{4+1}{2+1}=\frac{5}{3}\] \[{{y}_{2}}=\frac{-2+2}{2+1}=0\] \[{{z}_{2}}=\frac{8+3}{2+1}=\frac{11}{3}\] So, position vector of the points of trisection of AB are position vector of \[C=\frac{4}{3}\,\,\hat{i}+\hat{j}+\frac{10}{3}\hat{k}\]and position vector of \[D=\frac{5}{3}\hat{i}+\frac{11}{3}\hat{k}\]