Question

Test whether the lines
$\overline{p}=\left(\overline{2i}+\overline{4j}+\overline{6k}\right)+\lambda \left(\overline{i}+\overline{4j}+\overline{7k}\right)$and $q=\left(\overline{-i}-\overline{3j}-\overline{5k}\right)+\omega \left(\overline{3i}+\overline{5j}+\overline{7k}\right)$ are co-planer .if so find the equation of the plane containing them.

Answer 1

$\overrightarrow{p}=2\hat{i}+4\hat{j}+6\hat{k}+\lambda (\hat{i}+4\hat{j}+7\hat{k})\overrightarrow{q}=(-\hat{i}-3\hat{j}+(-5\hat{k}))+\omega (3\hat{i}+5\hat{j}+7\hat{k})$

For line $\overrightarrow{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1},\overrightarrow{r}=\overrightarrow{a_2}+\lambda \overrightarrow{b_2}$ to be co-plane

$(\overrightarrow{a_1}-\overrightarrow{a_2})(\overrightarrow{b_1}\times \overrightarrow{b_2})=0$

For line $\overrightarrow{p}\&\overrightarrow{q}$ to be co-planer,

$[(2\hat{i}+4\hat{j}+6\hat{k})-(-\hat{i}-3\hat{j}+(-5\hat{k}))][(\hat{i}+4\hat{j}+7\hat{k})\times (3\hat{i}+5\hat{j}+7\hat{k})]=0\left[(3\hat{i}+7\hat{j}+11\hat{k})\right][(\hat{i}+4\hat{j}+7\hat{k})\times (3\hat{i}+5\hat{j}+7\hat{k})]=0\therefore \left[\begin{array}{ccc}3& 7& 11\\ 1& 4& 7\\ 3& 5& 7\end{array}\right]=0$

$\therefore STP[a_1-a_1\overrightarrow{b_1}\overrightarrow{b_2}]is0$ for the two line hence they are co-planer

Plane through them

$\left[\begin{array}{ccc}x-z& y-4& z-6\\ 1& 4& 7\\ 3& 5& 7\end{array}\right]=0$

$\Rightarrow -7(x-2)+14(y-4)-7(z-6)=0\Rightarrow (x-2)-2(y-4)+(z-6)=0\Rightarrow x-2y+z=0$

is required equation of plane