Question

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

Answer 1

Answer:

Equation of the plane 2x+y+z=0

Step-by-step explanation:

To find the equation of the plane,we must find the direction ratio's of the plane,as we know that standard equation of the plane with given point and direction ratio's

$a(x-x_{1})+b(y-y_{1})+c(z-z_{1})=0$

so given point is(-1,1,1)

$a(x+1)+b(y-1)+c(z-1)=0$

Now to find the DR's as the plane is parallel to the vectors

so DR's (2,3,-7)

and (3,-8,2)

So $2a+3b-7c=0\\\\3a-8b+2c=0\\\\ \frac{a}{6-56} =\frac{b}{-21-4}=\frac{z}{-16-9}=\lambda$

$\frac{a}{-50}=\frac{b}{-25} =\frac{c}{-25} =\lambda\\\\\\a=-50\:\lambda\\\\b=-25\:\lambda\\\\c=-25\:\lambda$

$(-50\:\lambda)(x+1)+(-25\:\lambda)(y-1)+(-25\:\lambda)(z-1)=0\\\\-25\lambda(2(x+1)+(y-1)+(z-1))=0\\\\2x+2+y-1+z-1=0\\\\2x+y+z=0$