Question

Find the vector equation of the plane passing through (1,2,3)and perpendicular to the plane r→.(i^+2j^-5k^)+9=0.

Answer 1

The position vector of the point (1,2,3) is $\overrightarrow{r_1}$ =$\hat i+2\hat j^+3\hat k^.$

The direction ratios of the normal to the plane,

$\overrightarrow{r}$.($\hat i + 2\hat {j}-5\hat k$)+9=0, are 1,2,and -5 and the normal vector is $\overrightarrow{N_1}$=$\hat i + 2\hat {j}-5\hat k$

The equation of a line passing through a point and perpendicular to the given plane is given by,

$\overrightarrow{I}$=$\overrightarrow{r}$+λN→, λ∈R

⇒$\overrightarrow{I}$=($\hat i + 2\hat {j}+3\hat k$)+λ($\hat i + 2\hat {j}-5\hat k$).