Question

Line passing through the points $2\overline{a}+3\overline{b}-\overline{c}$, $3\overline{a}+4\overline{b}-2\overline{c}$ intersects the line through the points $\overline{a}-2\overline{b}+3\overline{c}$, $\overline{a}-6\overline{b}+6\overline{c}$ at $P$. Position vector of $P=$

• A. $2\overline{a}+\overline{b}$
• B. $\overline{a}+2\overline{b}$
• C. $3\overline{a}+4\overline{b}$
• D. $\overline{a}-2\overline{b}$

Answer 1

The correct option is B $\overline{a}+2\overline{b}$

Direction ratio : $(3\overrightarrow{a}+4\overrightarrow{b}-2\overrightarrow{c})-(2\overrightarrow{a}+3\overrightarrow{b}-\overrightarrow{c})$

$=\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c}$

$\therefore$ Line $1:\overrightarrow{r_1}2\overrightarrow{a}+3\overrightarrow{b}-\overrightarrow{c}+\lambda (\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c})$

Any point on $4=(2+\lambda )\overrightarrow{a},(3+\lambda )\overrightarrow{b},(-1-\lambda )\overrightarrow{c}$

Similarly

Line $2:\overrightarrow{r_2}=\overrightarrow{a}-2\overrightarrow{b}+3\overrightarrow{c}+\mu (-4\overrightarrow{b}+3\overrightarrow{c})$

Any point on $\overrightarrow{r_2}=\overrightarrow{a},(-2-7\mu )\overrightarrow{b},(3+3\mu )\overrightarrow{c}$

$\therefore$ At point $P$ $2+\lambda =1\Rightarrow \lambda =1$ and $3+\lambda =-24\mu$

$\Rightarrow 2+2=-4\mu \Rightarrow \mu =1$

$\therefore$ Position vector of $\overrightarrow{P}=(\overrightarrow{a}+2\overrightarrow{b})$