Question

A plane passing through the point(-1,1,1) is parallel to the vector $2\hat{i}+3\hat{j}-7\hat{k}$ and line $r=(\hat{i}-2\hat{j}-\hat{k})+\lambda (3\hat{i}-8\hat{j}+2\hat{k})$ . Find the vector equation of the plane.

Answer 1

Required plane passing through (-1,1,1)

the parallel to vector is.....

$\begin{array}{c}(2\hat{i}+3\hat{j}-7\hat{k})\\ \overline{r}=(\hat{i}-2\hat{j}-\hat{k})+\lambda (3\hat{i}-8\hat{j}+2\hat{k})\\ so,\\ \overline{u}=(2\hat{i}+3\hat{j}-7\hat{k})\\ \overline{v}=(3\hat{i}-8\hat{j}+2\hat{k})\\ now,\\ u\times v=\left(\begin{array}{ccc}i& j& k\\ 2& 3& -7\\ 3& -8& 2\end{array}\right)=0\\ \Rightarrow i(6-56)-j(+21)+k(-16-9)=0\\ \Rightarrow -50i-25j-25k=0\\ \therefore 2i+j+k=0\\ and,\\ \overline{n}=(2\hat{i}+\hat{j}+\hat{k})\\ equationofrequiredplanewillbe:\\ a(x-x_1)+b(y-y_1)+c(z-z_1)=0\\ 2(i+1)+1(j-1)+1(k-1)=0\\ 2i+2+j-1+k-1=0\\ \therefore 2i+j+k=0\end{array}$