Question

The coordinates of the point P are $\left( 3,12,4 \right)$ with respect to origin O, then the direction cosines of OP are
(A) $3,12,4$
(B) $\dfrac{1}{4},\dfrac{1}{3},\dfrac{1}{2}$
(C) $\dfrac{3}{\sqrt{13}},\dfrac{1}{\sqrt{13}},\dfrac{2}{\sqrt{13}}$
(D) $\dfrac{3}{13},\dfrac{12}{13},\dfrac{4}{13}$

Answer 1

First of all, convert the coordinate of the point P into vector form that is $3\widehat{i}+12\widehat{j}+4\widehat{k}$ . We know the property that the direction ratio of a point is its vector form. Now, get the magnitude of the direction ratio vector. Using the magnitude of the direction ratio vector, get the unit vector in the direction of the direction ratio vector. We know the property that direction cosine is the unit vector in the direction of the direction ratio of a vector. Now, solve it further and get the direction cosine.

Complete step by step answer:
According to the question, it is given that
The coordinate of the point P = $\left( 3,12,4 \right)$ ………………………………………….(1)
The x coordinate of the point P = 3
The vector form for the x coordinate of the point P = $3\widehat{i}$ ……………………………………….(2)
The y coordinate of the point P = 12
The vector form for the y coordinate of the point P = $12\widehat{j}$ ……………………………………………(3)
The z coordinate of the point P = 4
The vector form for the z coordinate of the point P = $4\widehat{k}$ …………………………………………(4)
From equation (2), equation (3), and equation (4), we have
The vector form of the point P = $3\widehat{i}+12\widehat{j}+4\widehat{k}$ ……………………………………………………..(5)
We know the property that the direction ratio of a point is its vector form. …………………………………………(6)
Now, from equation (5) and equation (6), we get
The direction ratio vector of the point P = $3\widehat{i}+12\widehat{j}+4\widehat{k}$ ……………………………………….(7)
Magnitude of the direction ratio vector = $\sqrt{{{3}^{2}}+{{12}^{2}}+{{4}^{2}}}=\sqrt{169}=13$ ……………………………….(8)
We know the property that direction cosine is the unit vector in the direction of the direction ratio of a vector ……………………………………..(9)
We also know that when a vector is divided by its magnitude then we get a unit vector in the direction of the original vector.
From equation (7) and equation (8), we have the direction ratio vector and its magnitude.
Now, using equation (9) and on dividing the direction ratio vector by its magnitude, we get
The direction cosine vector = $\dfrac{3}{13}\widehat{i}+\dfrac{12}{13}\widehat{j}+\dfrac{4}{13}\widehat{k}$ …………………………………..(10)
Therefore, the direction cosine of the point P is $\dfrac{3}{13},\dfrac{12}{13},\dfrac{4}{13}$ .

So, the correct answer is “Option D”.

Note: In this question, one might make a silly mistake and take the direction ratio vector $3\widehat{i}+12\widehat{j}+4\widehat{k}$ equal to the direction cosine vector. This is wrong because we know that the magnitude of the direction cosine vector is always equal to 1 and the magnitude of the direction ratio is equal to 13. So, this is not our direction ratio. To get the direction cosine, we have to divide the direction ratio vector by its magnitude.