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(\(\hat i + 2\hat {j}-5\hat k\)).