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Question: If \(O\) is the origin \(OP = 6\) with \(DR's - 2,4, - 4\) then the coordinate of \(P\) are A) \(2...

If OO is the origin OP=6OP = 6 with DR′s−2,4,−4DR's - 2,4, - 4 then the coordinate of PP are
A) 2,−4,42, - 4,4
B) −2,4,−4 - 2,4, - 4
C) −13,23,−23 - \dfrac{1}{3},\dfrac{2}{3}, - \dfrac{2}{3}
D) None of these

Explanation

Solution

Let OO be the origin and PP be any point. If OP=rOP = r and a,b,ca,b,c be the Direction Ratios ofOPOP , then Direction Cosines of OPOP are ±ar,±br,±cr. \pm \dfrac{a}{r}, \pm \dfrac{b}{r}, \pm \dfrac{c}{r}.
Let OO be the origin and P(x,y,z)P(x,y,z) be any point. Also if OP=rOP = r and l,m,nl,m,n be Direction Cosines of OPOP then x=lr,y=mr,z=nr.x = lr,y = mr,z = nr. so the coordinates of P are (lr,mr,nr).(lr,mr,nr).

Complete step-by-step answer:
It is given that OP=6OP = 6, and also given that their direction ratios as -2, 4,-4.
From the given hint and the given Direction Ratios of OPOP Also OP=6OP = 6
The Direction Cosines of OPOP are ±(−26),±(46),±(−46) \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, ±(−13),±(23),±(−23) \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 OPOP.
Hence we get the following values −13,23,−23.\dfrac{{ - 1}}{3},\dfrac{2}{3},\dfrac{{ - 2}}{3}.
From ±(−13),±(23),±(−23) \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 OPOP 13,−23,23.\dfrac{1}{3},\dfrac{{ - 2}}{3},\dfrac{2}{3}.
Let the coordinates of P be (x,y,z).(x,y,z).
Now let us take the following values −13,23,−23.\dfrac{{ - 1}}{3},\dfrac{2}{3},\dfrac{{ - 2}}{3}.
Hence we get, l=−13,m=23,n=−23l = \dfrac{{ - 1}}{3},\quad m = \dfrac{2}{3},\quad n = \dfrac{{ - 2}}{3} and r=6r = 6 we get,
From the given hint we can find the value of x, y, z x=−13×6=−2,    y=23×6=4,    z=−23×6=−4.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 PP are (−2,4,−4).\left( { - 2,4, - 4} \right).
Now let us consider the following values 13,−23,23.\dfrac{1}{3},\dfrac{{ - 2}}{3},\dfrac{2}{3}.
Hence we get, l=13,m=−23,n=23l = \dfrac{1}{3},\quad m = \dfrac{{ - 2}}{3},\quad n = \dfrac{2}{3} and r=6r = 6 we get,
Let us solve using the hint for x,y,z x=13×6=2,    y=−23×6=−4,    z=23×6=4.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 PP are (2,−4,4).\left( {2, - 4,4} \right).
Hence,
We have found that the coordinates of PP are either A)    2,  −4,  4A)\;\;2,\; - 4,\;4 or B)    −2,  4,  −4.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.