Question

If the position vectors of $A,B,C,D$ are $3\hat{i}+2\hat{j}+\hat{k},4\hat{i}+5\hat{j}+5\hat{k},4\hat{i}+2\hat{j}-2\hat{k},6\hat{i}+5\hat{j}-\hat{k}$ respectively then the position vector of the point of intersection of $\overline{AB}$ and $\overline{CD}$ is

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

Answer 1

The correct option is D $2\hat{i}-\hat{j}-3\hat{k}$

$\begin{array}{cc}\overrightarrow{AB}& =\overrightarrow{b}-\overrightarrow{a}\\ & =(4i+5\hat{ȷ}+5\hat{k})-(3\hat{i}+2\hat{ȷ}+\hat{k})\\ & =\hat{ı}+3\hat{ȷ}+4\hat{k},and\end{array}$

$\begin{array}{cc}\overrightarrow{CD}& =\overrightarrow{d}-\overrightarrow{c}\\ & =(6\hat{ı}+5\hat{ȷ}-\hat{k})-(4\hat{ı}+2\hat{ȷ}-2\hat{k})\\ & =\overrightarrow{2}\hat{ı}+3\hat{ȷ}+\hat{k}\\ \text{Let}\overrightarrow{AB}& \text{and}\overrightarrow{CD}\text{intersect at}\overrightarrow{P}.\\ \text{Line}AB=(3i+2j+\hat{k})+\lambda (i+3\hat{j}+4\hat{k})& \text{and}\end{array}$

Line $CD=(4i+2\hat{ı}-2\hat{k})+\mu (2\hat{i}+3\hat{j}+\hat{k})$

W.r.t. line $A_0$,

$P=(3+\lambda ,2+3\lambda ,1+4\lambda )$

and $w\cdot r\cdot t\cdot$ line $CD$,

$P=(4+2\mu ,2+3\mu ,-2+\mu )$

Comparing both points,

$\begin{array}{cc}\Rightarrow & 3+{\lambda }^{\prime }=4+2\mu ,\\ & 2+3\lambda =2+3\mu \\ & 1+4\lambda =-2+\mu \end{array}$

Solving these equs.,

$\therefore \overrightarrow{p}=2\hat{ı}-\hat{ȷ}-3\hat{k}$

$\therefore$ option $D$ is correct.