7.  Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane \overrightarrow{r}.(\widehat{i}+2\widehat{j}-5\widehat{k})+9=0

Answers (1)
D Divya Prakash Singh

Given that the plane is passing through the point A (1,2,3) so, the position vector of the point A is \vec{r_{A}} = \widehat{i}+2\widehat{j}+3\widehat{k}  and perpendicular to the plane \overrightarrow{r}.(\widehat{i}+2\widehat{j}-5\widehat{k})+9=0 whose direction ratios are  1,2,\ and\ -5 and the normal vector is \vec{n} = \widehat{i}+2\widehat{j}-5\widehat{k}

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

\vec{l} = \vec{r} + \lambda\vec{n}, where \lambda \epsilon R

\Rightarrow \vec{l} = (\widehat{i}+2\widehat{j}+3\widehat{k}) + \lambda(\widehat{i}+2\widehat{j}-5\widehat{k}).