Question

If $O$ is origin $OP=3$ with direction ratios proportional to $-1,2,-2$ then what are the coordinates of $P$?

Answer 1

Let the coordinates of point $P$ be $(x_1,y_1,z_1)$

Given,

$OP=3$

$\sqrt{(x_1-0{)}^2+(y_1-0{)}^2+(z_1-0{)}^2}=9$

$x_1^2+y_1^2+z_1^2=9$ ----- $\left(1\right)$

Direction ratio of $OP$ are $-1,2,-2$

Therefore,

$x_1-0=-1$

$x_1=-1$

$y_1-0=2$

$y_1=2$

$z_1-0=-2$

$z_1=-2$

Putting the values of $x_1,y_1,z_1$ in $\left(1\right)$ we get,

$(-1{)}^2+(2{)}^2+(-2{)}^2=0$

$9=9$

$x_1,y_1,z_1$ satisfies $\left(1\right)$

Hence, coordinates of point $P$ is $(-1,2,-2)$