Question

Point of intersection of the lines
$\overline{r}=(6-6s)\overline{a}+(4s-4)\overline{b}+(4-8s)\overline{c}$ and
$\overline{r}=(2t-1)\overline{a}+(4t-2)\overline{b}-(2t+3)\overline{c}$

• A. $4\overline{c}$
• B. $-4\overline{c}$
• C. $3\overline{c}$
• D. $-2\overline{c}$

Answer 1

The correct option is B $-4\overline{c}$

At point intersection, coefficient of

$\overline{a},\overline{b}$ and $\overline{c}$ would be equal

$\Rightarrow 6-6s=2t-1;4s-4=4t-2$

$\Rightarrow 2t=7-6s\Rightarrow 4(s-t)=2$

$\Rightarrow t=\frac{7}{2}-3s\Rightarrow s-t=\frac{1}{2}$

$\Rightarrow s-\frac{7}{2}+3s=\frac{1}{2}$

$\Rightarrow 4s=4\Rightarrow s=1$

$\Rightarrow t=\frac{7}{2}-3=\frac{1}{2}$

$\Rightarrow \overline{r}=0.\overline{a}+0.\overline{b}+(4-8)\overline{c}$

$\Rightarrow \overline{r}=-4\overline{c}\Rightarrow \left(B\right)$