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

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 $\lambda \epsilon R$

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

Related Chapters

Preparation Products

Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
Knockout BITSAT 2020

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 1999/-
Knockout NEET May 2022

An exhaustive E-learning program for the complete preparation of NEET..

₹ 34999/- ₹ 24999/-
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-