Question

The direction ratios of a vector are 2,-3, 4. Find its direction cosines.

Answer 1

Hint: Here in this question the direction ratios are given. By using the formula $l=\cos \alpha ={\displaystyle \frac{x}{\sqrt{x^2+y^2+z^2}}}$ , $m=\cos \beta ={\displaystyle \frac{y}{\sqrt{x^2+y^2+z^2}}}$ and $n=\cos \gamma ={\displaystyle \frac{z}{\sqrt{x^2+y^2+z^2}}}$ we can determine the direction cosines of the vector where x, y, and z are the direction ratios of the vector.

Complete step-by-step answer:
 Suppose if we have a vector in a plane in the form of $\overrightarrow{OP}=x\hat{i}+y\hat{j}+z\hat{k}$ , where $\hat{i}$ , $\hat{j}$ and $\hat{k}$ are the unit vectors of x, y and z axis respectively. The numbers which are proportional to the direction cosines are direction ratios. The direction cosine is the cosines of the angles between the vector and the three coordinate axes.
We have formula for direction cosines and it is given as
 $l=\cos \alpha ={\displaystyle \frac{x}{\sqrt{x^2+y^2+z^2}}}$ , $m=\cos \beta ={\displaystyle \frac{y}{\sqrt{x^2+y^2+z^2}}}$ and $n=\cos \gamma ={\displaystyle \frac{z}{\sqrt{x^2+y^2+z^2}}}$
Now we will find the direction cosine along the x axis
 $l=\cos \alpha ={\displaystyle \frac{x}{\sqrt{x^2+y^2+z^2}}}$ , where $x=2,y=-3$ and $z=4$
by substituting all the values in l we get
 $$\begin{gathered} l=\cos \alpha ={\displaystyle \frac{x}{\sqrt{x^2+y^2+z^2}}} \\ \Rightarrow l=\cos \alpha ={\displaystyle \frac{2}{\sqrt{2^2+{(-3)}^2+4^2}}} \end{gathered}$$
on simplification we have
 $$\begin{gathered} \Rightarrow l=\cos \alpha ={\displaystyle \frac{2}{\sqrt{4+9+16}}} \\ \Rightarrow l={\displaystyle \frac{2}{\sqrt{29}}} \end{gathered}$$
Now we will find the direction cosine along the y axis
 $m=\cos \beta ={\displaystyle \frac{y}{\sqrt{x^2+y^2+z^2}}}$ , where $x=2,y=-3$ and $z=4$
By substituting all the values in m we get
 $$\begin{gathered} m=\cos \beta ={\displaystyle \frac{y}{\sqrt{x^2+y^2+z^2}}} \\ \Rightarrow m=\cos \beta ={\displaystyle \frac{-3}{\sqrt{2^2+{(-3)}^2+4^2}}} \end{gathered}$$
On simplification we have
 $$\begin{gathered} \Rightarrow m=\cos \beta ={\displaystyle \frac{-3}{\sqrt{4+9+16}}} \\ \Rightarrow m={\displaystyle \frac{-3}{\sqrt{29}}} \end{gathered}$$
Now we will find the direction cosine along the z axis
 $n=\cos \gamma ={\displaystyle \frac{z}{\sqrt{x^2+y^2+z^2}}}$ , where $x=2,y=-3$ and $z=4$
By substituting all the values in n we get
 $$\begin{gathered} n=\cos \gamma ={\displaystyle \frac{z}{\sqrt{x^2+y^2+z^2}}} \\ \Rightarrow n=\cos \gamma ={\displaystyle \frac{4}{\sqrt{2^2+{(-3)}^2+4^2}}} \end{gathered}$$
On simplification we have
 $$\begin{gathered} \Rightarrow n=\cos \gamma ={\displaystyle \frac{4}{\sqrt{4+9+16}}} \\ \Rightarrow n={\displaystyle \frac{4}{\sqrt{29}}} \end{gathered}$$
Hence we obtain the direction cosines from the direction ratios and that are
 $l={\displaystyle \frac{2}{\sqrt{29}}}$ , $m={\displaystyle \frac{-3}{\sqrt{29}}}$ and $n={\displaystyle \frac{4}{\sqrt{29}}}$
Therefore the direction cosines are $\left({\displaystyle \frac{2}{\sqrt{29}}},{\displaystyle \frac{-3}{\sqrt{29}}},{\displaystyle \frac{4}{\sqrt{29}}}\right)$
So, the correct answer is “ $\left({\displaystyle \frac{2}{\sqrt{29}}},{\displaystyle \frac{-3}{\sqrt{29}}},{\displaystyle \frac{4}{\sqrt{29}}}\right)$ ”.

Note: The direction ratios are proportional direction cosines of the vector. If we know the values of direction ratios of a vector then by applying the formula we can obtain the result. The direction ratios are the coefficient of the vector.