Question

If $O$ is the origin $OP = 6$ with $DR's - 2,4, - 4$ then the coordinate of $P$ are
A) $2, - 4,4$
B) $ - 2,4, - 4$
C) $ - \dfrac{1}{3},\dfrac{2}{3}, - \dfrac{2}{3}$
D) None of these

Answer 1

Hint:Let $O$ be the origin and $P$ be any point. If $OP = r$ and $a,b,c$ be the Direction Ratios of$OP$ , then Direction Cosines of $OP$ are $ \pm \dfrac{a}{r}, \pm \dfrac{b}{r}, \pm \dfrac{c}{r}.$
Let $O$ be the origin and $P(x,y,z)$ be any point. Also if $OP = r$ and $l,m,n$ be Direction Cosines of $OP$ then $x = lr,y = mr,z = nr.$ so the coordinates of P are $(lr,mr,nr).$

Complete step-by-step answer:
It is given that $OP = 6$, and also given that their direction ratios as -2, 4,-4.
From the given hint and the given Direction Ratios of $OP$ Also $OP = 6$
The Direction Cosines of $OP$ are $ \pm \left( {\dfrac{{ - 2}}{6}} \right), \pm \left( {\dfrac{4}{6}} \right), \pm \left( {\dfrac{{ - 4}}{6}} \right)$ .
Now let us simplify the direction cosines we get, $ \pm \left( {\dfrac{{ - 1}}{3}} \right), \pm \left( {\dfrac{2}{3}} \right), \pm \left( {\dfrac{{ - 2}}{3}} \right)$
Now let us take positive signs in the Direction Cosines of $OP$.
Hence we get the following values $\dfrac{{ - 1}}{3},\dfrac{2}{3},\dfrac{{ - 2}}{3}.$
From $ \pm \left( {\dfrac{{ - 1}}{3}} \right), \pm \left( {\dfrac{2}{3}} \right), \pm \left( {\dfrac{{ - 2}}{3}} \right)$ this value let us consider the negative sign in the Direction Cosines of $OP$ $\dfrac{1}{3},\dfrac{{ - 2}}{3},\dfrac{2}{3}.$
Let the coordinates of P be $(x,y,z).$
Now let us take the following values $\dfrac{{ - 1}}{3},\dfrac{2}{3},\dfrac{{ - 2}}{3}.$
Hence we get, $l = \dfrac{{ - 1}}{3},\quad m = \dfrac{2}{3},\quad n = \dfrac{{ - 2}}{3}$ and $r = 6$ we get,
From the given hint we can find the value of x, y, z $x = \dfrac{{ - 1}}{3} \times 6 = - 2,\;\;y = \dfrac{2}{3} \times 6 = 4,\;\;z = \dfrac{{ - 2}}{3} \times 6 = - 4.$
Here x=-2, y=4 and z=-4.
So the coordinates of $P$ are $\left( { - 2,4, - 4} \right).$
Now let us consider the following values $\dfrac{1}{3},\dfrac{{ - 2}}{3},\dfrac{2}{3}.$
Hence we get, $l = \dfrac{1}{3},\quad m = \dfrac{{ - 2}}{3},\quad n = \dfrac{2}{3}$ and $r = 6$ we get,
Let us solve using the hint for x,y,z $x = \dfrac{1}{3} \times 6 = 2,\;\;y = \dfrac{{ - 2}}{3} \times 6 = - 4,\;\;z = \dfrac{2}{3} \times 6 = 4.$
Here x=2, y=-4 and z=4.
In this case the coordinates of $P$ are $\left( {2, - 4,4} \right).$
Hence,
We have found that the coordinates of $P$ are either $A)\;\;2,\; - 4,\;4$ or $B)\;\; - 2,\;4,\; - 4.$

Note:In the process of finding Direction Cosines from Direction Ratios the same sign must be taken throughout. Sign should not be changed.