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 passing through the points $\overline{a}-2\overline{b}+3\overline{c},\overline{a}-6\overline{b}+6\overline{c}$ at $P$. Position vector of $P$ is equal to

• 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 $\overline{a}+2\overline{b}$

Line passing through $2\overrightarrow{a}+3\overrightarrow{b}-\overrightarrow{c}$ & $3\overrightarrow{a}+4\overrightarrow{b}-2\overrightarrow{c}$
So $\overrightarrow{r}=2\overrightarrow{a}+3\overrightarrow{b}-\overrightarrow{c}+\lambda (\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c})$
line passing through $\overrightarrow{a}-2\overrightarrow{b}+3\overrightarrow{c},\overrightarrow{a}-6\overrightarrow{b}+6\overrightarrow{c}$
$so\overrightarrow{r}=\overrightarrow{a}-2\overrightarrow{b}+3\overrightarrow{c}+{\lambda }_1(-4\overrightarrow{b}+3\overrightarrow{c})$
Now equate both line
$\overrightarrow{a}+5\overrightarrow{b}-4\overrightarrow{c}={\lambda }_1(-4\overrightarrow{b}+3\overrightarrow{c})+\lambda (\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c})$
$\lambda =15=-4{\lambda }_1+\lambda$
${\lambda }_1=-1$
So $P=\overrightarrow{a}+2\overrightarrow{b}$