# 16. If O be the origin and the coordinates of P be (1, 2, – 3), then find the equation of the plane passing through P and perpendicular to OP.

We have the coordinates of the points $O(0,0,0)$  and  $P(1,2,-3)$ respectively.

Therefore, the direction ratios of OP are $(1-0) = 1, (2-0)=2,\ and\ (-3-0)=-3$

And we know that the equation of the plane passing through the point $(x_{1},y_{1},z_{1})$ is

$a(x-x_{1})+b(y-y_{1})+c(z-z_{1})=0$ where a,b,c are the direction ratios of normal.

Here, the direction ratios of normal are $1,2,$ and $-3$ and the point P is $(1,2,-3)$.

Thus, the equation of the required plane is

$1(x-1)+2(y-2)-3(z+3) = 0$

$\implies x+2y -3z-14 = 0$

## 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/-