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Question: If \[\text{O}=\left( 0,0,0 \right),\text{ OP = 5}\] and the DRs of \[\text{OP}\] is \[\text{1,2,2}\]...

If O=(0,0,0), OP = 5\text{O}=\left( 0,0,0 \right),\text{ OP = 5} and the DRs of OP\text{OP} is 1,2,2\text{1,2,2} then Px+Py+Pz=?{{P}_{x}}+{{P}_{y}}+{{P}_{z}}=?
(a) 25
(b) 259\dfrac{25}{9}
(c) 253\dfrac{25}{3}
(d) (53,103,103)\left( \dfrac{5}{3},\dfrac{10}{3},\dfrac{10}{3} \right)

Explanation

Solution

In this type of question we have to use the concept of direction ratios. We know that, if O is the origin and P is any point, and a, b, c are the Direction Ratios of OP, then Direction Cosines of OP are given by, ±ar,±br,±cr\pm \dfrac{a}{r},\pm \dfrac{b}{r},\pm \dfrac{c}{r} where r=a2+b2+c2r=\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}. Also if OP = k and l, m, n are the Direction Cosines of OP then we know that, Px=lk,Py=mk,Pz=nk{{P}_{x}}=lk,{{P}_{y}}=mk,{{P}_{z}}=nk.

Complete step by step answer:
Now, we have to find the value of Px+Py+Pz{{P}_{x}}+{{P}_{y}}+{{P}_{z}} if O=(0,0,0), OP = 5\text{O}=\left( 0,0,0 \right),\text{ OP = 5} and the DRS of OP\text{OP} is 1,2,2\text{1,2,2}
As we know that, if a, b, c are direction ratios of OP then Direction Cosines of OP can be given by, ±ar,±br,±cr\pm \dfrac{a}{r},\pm \dfrac{b}{r},\pm \dfrac{c}{r} where r=a2+b2+c2r=\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}.
Here, OP = 5\text{OP = 5} and the Direction ratios of OP are 1,2,2\text{1,2,2}, hence the Direction Cosines of OP are ±13,±23,±23\pm \dfrac{1}{3},\pm \dfrac{2}{3},\pm \dfrac{2}{3} since r=12+22+22=9=3r=\sqrt{{{1}^{2}}+{{2}^{2}}+{{2}^{2}}}=\sqrt{9}=3.
Now let us take positive signs in the Direction Cosines of OP.
Hence, we get the values, 13,23,23\dfrac{1}{3},\dfrac{2}{3},\dfrac{2}{3}
l=13,m=23,n=23\Rightarrow l=\dfrac{1}{3},m=\dfrac{2}{3},n=\dfrac{2}{3}
Now, as if OP = k and l, m, n are the Direction Cosines of OP then we know that, Px=lk,Py=mk,Pz=nk{{P}_{x}}=lk,{{P}_{y}}=mk,{{P}_{z}}=nk.
Here, OP = 5 and l=13,m=23,n=23l=\dfrac{1}{3},m=\dfrac{2}{3},n=\dfrac{2}{3}. Hence, we can calculate, Px{{P}_{x}}, Py{{P}_{y}} and Pz{{P}_{z}} as follows:

& \Rightarrow {{P}_{x}}=lk=\dfrac{1}{3}\times 5=\dfrac{5}{3} \\\ & \Rightarrow {{P}_{y}}=mk=\dfrac{2}{3}\times 5=\dfrac{10}{3} \\\ & \Rightarrow {{P}_{z}}=nk=\dfrac{2}{3}\times 5=\dfrac{10}{3} \\\ \end{aligned}$$ Thus we can write our final answer as, $$\Rightarrow {{P}_{x}}+{{P}_{y}}+{{P}_{z}}=\dfrac{5}{3}+\dfrac{10}{3}+\dfrac{10}{3}=\dfrac{25}{3}$$ **So, the correct answer is “Option c”.** **Note:** In this type of question students have to remember to calculate Direction Cosines from the given Direction Ratios and then have to perform calculation for the coordinates of P. Also students have to take care in the calculation they have to maintain the same sign throughout, signs should not be changed.