Question

P,Q,R and S are four points with the position vectors, $3\overrightarrow{i}-4\overrightarrow{j}+5\overrightarrow{k},4\overrightarrow{k},-4\overrightarrow{i}+5\overrightarrow{j}+\overrightarrow{k}\text{and}-3\overrightarrow{i}+4\overrightarrow{j}+3\overrightarrow{k}$ respectively. Then the line PQ meets the line RS at the point

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

Answer 1

The correct option is B

$-3\overrightarrow{i}+4\overrightarrow{j}+3\overrightarrow{k}$

$\text{Equation of PQ}:\frac{x}{3}=\frac{y}{-4}=\frac{z-4}{1}\text{a point A on PQ}=(3\lambda ,-4\lambda ,\lambda +4)\text{Equation of RS}\frac{x+4}{1}=\frac{y-5}{-1}=\frac{z-1}{2}=m\text{a point on RS, B}=(m-4,m+5,2m+1)\text{to intersect A = B}\Rightarrow \lambda =-1,m=1\therefore \text{Point is}(-3,4,3)$