Question

If $\overline{a},\overline{b},\overline{c}$ are non coplanar, then the line passing through the points $2\overline{a}+3\overline{b}-\overline{c}$ and $3\overline{a}+4\overline{b}-2\overline{c}$ intersects the line joining the points $\overline{a}-2\overline{b}+3\overline{c}$ and $\overline{a}-6\overline{b}+6\overline{c}$ at the point whose position vector is

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

Answer 1

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

Equation of line passing through $\overline{m}$ and $\overline{n}$

is given by $\overline{r}=\overline{m}+\lambda (\overline{m}-\overline{n})$

$\Rightarrow \overline{r_1}=2\overline{a}+3\overline{b}-\overline{c}+\lambda (\overline{a}+\overline{b}-\overline{c})=(2+\lambda )\overline{a}+(3+\lambda )\overline{b}+(1+\lambda )\overline{c}$

$\overline{r_2}=\overline{a}-2\overline{b}+3\overline{c}+(0.\overline{a}-4\overline{b}+3\overline{c})=\overline{a}+(-2-4t)\overline{b}+(3+3t)\overline{c}$

At intersection point, coefficients of $\overline{a},\overline{b},\overline{c}$

should be same

$\Rightarrow 2+\lambda =1\Rightarrow \lambda =1;3+\lambda =-2-4t$

$\Rightarrow 2=-2-4t\Rightarrow 4t=-4\Rightarrow t=-1$

$\Rightarrow$ Intersection point $\overline{r}=\overline{a}+2\overline{b}\Rightarrow \left(A\right)$