If a directed line L, passing through the origin, makes angles $\alpha, \beta$, and $\gamma$ with the $x, y$ and $z$ axes respectively called direction angles, then the cosines of these angles, namely, cos($\alpha$), cos($\beta$) cos($\gamma$), are called the direction cosines of the directed line $L$. In real life, we use vectors for tracking objects like airplanes and drones.
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In this article, we will cover the concept of Direction Cosines & Direction Ratios Of A Line. This topic lies under the broader category of 3-dimensional geometry, which is an important chapter in Class 12 Mathematics. It is an important topic for boards and the competitive level as well. The exams include: JEE Main, SRM Joint Engineering Entrance, BITSAT, WBJEE, and BCECE. A total of 14 questions have been asked on this topic in JEE Main from 2013-2023 including one in 2013, two in 2014, one in 2016, one in 2017, three in 2018, three in 2019, and two in 2021.
If the point $R(x, y, z)$ divides the line joining the points $P\left(x_1, y_1, z_1\right)$ and $Q\left(x_2, y_2, z_2\right)$ internally in the ratio $m: n$, then the coordinate of $R(x, y, z)$ is given by :
$\mathrm{R}(\mathrm{x}, \mathrm{y} . \mathrm{z})=\left(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y_1}{m+n}, \frac{m z_2+n z_1}{m+n}\right)$
If the point $R$ divides $P Q$ externally in the ratio $m$ : $n$, then its coordinates are obtained by replacing $n$ with $(-n)$, so that the coordinates of point R will be
$\mathrm{R}(\mathrm{x}, \mathrm{y}, \mathrm{z})=\left(\frac{m x_2-n x_1}{m-n}, \frac{m y_2-n y_1}{m-n}, \frac{m z_2-n z_1}{m-n}\right)$
NOTE:
The coordinates of the midpoint ( $\mathrm{m}: \mathrm{n}=1: 1$ ) of the line joining the points $P\left(x_1, y_1, z_1\right)$ and $Q\left(x_2, y_2\right.$, $z_2$ ) are as:
$\mathrm{R}(\mathrm{x}, \mathrm{y}, \mathrm{z})=\left(\frac{x_2+x_1}{2}, \frac{y_2+y_1}{2}, \frac{z_2+z_1}{2}\right)$
The coordinates of the point $R$ which divides $PQ$ in the ratio $k : 1$ are obtained by taking $\mathrm{k}=\mathrm{m} / \mathrm{n}$ which are as given below:
$\mathrm{R}(\mathrm{x}, \mathrm{y}, \mathrm{z})=\left(\frac{k x_2+x_1}{1+k}, \frac{k y_2+y_1}{1+k}, \frac{k z_2+z_1}{1+k}\right)$
The coordinates of the centroid of a triangle $R$ with vertices $A\left(x_1, y_1, z_1\right) B\left(x_2, y_2, z_2\right)$, and $C\left(x_3, y_3, z_3\right)$ are
$\mathrm{R}(\mathrm{x}, \mathrm{y}, \mathrm{z})=\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$
Example 1: Let the position vectors of two points P and Q be $3 \hat{i}-\hat{j}+2 \hat{k}$ and $\hat{i}+2 \hat{j}-4 \hat{k}$, respectively. Let $R$ and $S$ be two points such that the direction ratios of lines $PR$ and $QS$ are ( $4,-1,2)$ and ( $-2,1,-2$ ), respectively. Let lines $PR$ and $QS$ intersect at T. If the vector $\overrightarrow{T A}$ is perpendicular to both $\overrightarrow{P R}$ and $\overrightarrow{Q S}$ and the length of the vector $\overrightarrow{T A}$ is $\sqrt{5}$ units, then the modulus of a position vector of $A$ is :
[JEE MAINS 2021]
Solution
$
\begin{aligned}
& \mathrm{P}(3,-1,2) \\
& \mathrm{Q}(1,2,-4) \\
& \overrightarrow{\mathrm{PR}} \| 4 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}} \\
& \overrightarrow{\mathrm{QS}} \|-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}
\end{aligned}
$
dr's of normal to the plane containing P, T, and Q will be proportional to :
$
\therefore \quad \frac{\ell}{0}=\frac{\mathrm{m}}{4}=\frac{\mathrm{n}}{2}
$
For point, $T: \overrightarrow{\mathrm{PT}}=\frac{\mathrm{x}-3}{4}=\frac{\mathrm{y}+1}{-1}=\frac{\mathrm{z}-2}{2}=\lambda$
$
\overrightarrow{\mathrm{QT}}=\frac{\mathrm{x}-1}{-2}=\frac{\mathrm{y}-1}{1}=\frac{\mathrm{z}+4}{-2}=\mu
$
T: $(4 \lambda+3,-\lambda-1,2 \lambda+2) \cong(2 \mu+1, \mu+2,-2 \mu-4)$
$4 \lambda+3=-2 \mu+1 \quad \Rightarrow 2 \lambda+\mu=-1$
$\lambda+\mu=-3 \quad \Rightarrow \quad \lambda=2$
\& $\quad \mu=-5 \quad \lambda+\mu=-3 \quad \Rightarrow \quad \lambda=2$
So point $T:(11,-3,6)$
$
\begin{aligned}
& \overrightarrow{\mathrm{OA}}=(11 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}) \pm\left(\frac{2 \hat{\mathrm{j}}+\hat{k}}{\sqrt{5}}\right) \sqrt{5} \\
& \overrightarrow{\mathrm{OA}}=(11 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}) \pm(2 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \\
& \overrightarrow{\mathrm{OA}}=11 \hat{\mathrm{i}}-\hat{\mathrm{j}}+7 \hat{\mathrm{k}}
\end{aligned}
$
Or,
$
\begin{aligned}
& 9 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+5 \hat{\mathrm{k}} \\
& |\overrightarrow{\mathrm{OA}}|=\sqrt{121+1+49}=\sqrt{171}
\end{aligned}
$
or
$
\sqrt{81+25+25}=\sqrt{131}
$
Hence, the answer is $\sqrt{171}$
Example 2: If a unit vector $\vec{a}$ makes angles $\pi / 3$ with $\hat{i}, \pi / 4$ with $\hat{j}$ and $\theta \epsilon(0, \pi)$ with $\hat{k}$ then a value of $\theta$ is :
[JEE MAINS 2019]
Solution:
Direction Cosines - If $\alpha, \beta, \gamma$ are the angles which a vector makes with positive $x$ -axis, $y$ -axis, and $z$ -axis respectively then $\cos \alpha, \cos \beta, \cos \gamma$ are known as direction cosines, generally denoted by $(l, m, n)$.
$
\begin{aligned}
& l=\cos \alpha, m=\cos \beta, n=\cos \gamma \\
& l^2+m^2+n^2=1 \\
& \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1
\end{aligned}
$
$\begin{aligned} & \cos ^2 \frac{\pi}{3}+\cos ^2 \frac{\pi}{4}+\cos ^2 \theta=1 \\ & \left(\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2+\cos ^2 \theta=1 \\ & \frac{1}{4}+\frac{1}{2}+\cos ^2 \theta=1 \\ & \cos ^2 \theta=\frac{1}{4} \\ & \cos \theta= \pm \frac{1}{2} \\ & \therefore \frac{\pi}{3} \text { or } \pi-\frac{\pi}{3}=\frac{2 \pi}{3}\end{aligned}$
Example 3: Let $A(3,0,-1), B(2,10,6)$ and $C(1,2,1)$ be the vertices of a triangle and $M$ be the midpoint of $A C$. If $G$ divides $B M$ in the ratio, of 2:1, then $\cos (\angle G O A)$ ( O being the origin) is equal to :
[JEE MAINS 2019]
Solution:
Centroid of the triangle -
$
\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)
$
The angle between two lines (Vector form ) - Let the two lines be $\vec{r}=\vec{a}+\lambda \vec{b}$ and $\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$. The angle between two lines will be equal to the angle between their parallel vectors $\vec{b}$ and $\overrightarrow{b_1}$.
$
\cos \Theta=\frac{\vec{b} \cdot \overrightarrow{b_1}}{|\vec{b}|\left|\overrightarrow{b_1}\right|}
$
Given that $M$ is the midpoint of $A C$ and $G$ divides $B M$ in the ratio $2:1$
$\therefore G$ is the centroid of $A B C$
$
\begin{aligned}
& G=\left(\frac{3+2+1}{3}, \frac{0+10+2}{3}, \frac{-1+6+1}{3}\right)=(2,4,2) \\
& 0 A=3 \hat{i}+\hat{j}-\hat{k} \\
& 0 G=2 \hat{i}+4 \hat{j}+2 \hat{k} \\
& \therefore \cos (G O A)=\frac{\overrightarrow{O G} \cdot \overrightarrow{O A}}{|\overrightarrow{O G}||\overrightarrow{O A}|}=\frac{6-2}{2 \sqrt{10} \sqrt{6}} \\
& \therefore \cos (G O A)=\frac{1}{\sqrt{15}}
\end{aligned}
$
Hence, the answer is $\frac{1}{\sqrt{15}}$
Example 4: An angle between the lines whose direction cosines are given by the equations, $l+3 m+5 n=0$ and $5 l m-2 m n+6 n l=0$, is :
[JEE MAINS 2018]
Solution:
Direction Cosines - If $\alpha, \beta, \gamma$ are the angles which a vector makes with positive $X$-axis, $Y$-axis, and $Z$-axis respectively then $\cos \alpha, \cos \beta, \cos \gamma$ are known as direction cosines, generally denoted by $(l,m,n).$
$
\begin{aligned}
& l=\cos \alpha, m=\cos \beta, n=\cos \gamma \\
& l^2+m^2+n^2=1 \\
& \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1
\end{aligned}
$
$\begin{aligned}
&\begin{aligned}
& l+3 m+5 n=0 \Rightarrow l=-(3 m+5 n) \cdots \cdots(i) \\
& 5 l m-2 m n+6 n l=0 \cdots \cdots(i i)
\end{aligned}\\
&\text { Substitute (i) in (ii) }\\
&\begin{aligned}
& -(5 m+6 n)(3 m+5 n)-2 m n=0 \\
& -15 m^2-30 n^2-43 m n-2 m n=0 \\
& m^2+2 n^2+3 m n=0 \\
& m^2+2 m n+m n+2 n^2=0 \\
& m(m+2 n)+n(m+2 n)=0 \\
& m=-n \text { and } m=-2 n
\end{aligned}
\end{aligned}$
$\begin{aligned}
&\begin{aligned}
& l^2+m^2+n^2=1 \\
& l^2+m^2+n^2=1 \\
& 6 n^2=1 \Rightarrow n= \pm \frac{1}{\sqrt{6}} \\
& 6 n^2=1 \Rightarrow n= \pm \frac{1}{\sqrt{6}} \\
& \text { DCs are }+\frac{2}{\sqrt{6}} \hat{i}+\frac{1}{\sqrt{6}} \hat{j}-\frac{1}{\sqrt{6}} \hat{k} ; \frac{1}{\sqrt{6}} \hat{i}-\frac{2}{\sqrt{6}} \hat{j}+\frac{1}{\sqrt{6}} \hat{k} \\
& \text { Angle } \Rightarrow \cos \theta=\frac{2}{6}-\frac{2}{6}-\frac{1}{6}=\frac{-1}{6} \quad \theta=\cos ^{-1}\left(\frac{1}{6}\right) \\
& \text { Hence, the answer is } \cos ^{-1}\left(\frac{1}{6}\right)
\end{aligned}
\end{aligned}$
Example 5: If the position vectors of the vertices A, B, and C of a ΔABC are respectively and then the position vector of the point, where the bisector of ∠A meets BC is [JEE MAINS 2018]
Solution: Let the angular bisector of A meet side BC at
P(x,y,z)
By Angular Bisector theorem, AB: AC=BP: PC
BP:PC = :2:1=m:n
B (2,3,4) C(2,5,7)
Put values
Hence, the answer is
Example 5: If the position vectors of the vertices $A, B$, and $C$ of a $\triangle A B C$ are respectively $4 \hat{i}+7 \hat{j}+8 \hat{k}, 2 \hat{i}+3 \hat{j}+4 \hat{k}$ and $2 \hat{i}+5 \hat{j}+7 \hat{k}$ then the position vector of the point, where the bisector of $\angle A$ meets $B C$ is
[JEE MAINS 2018]
Solution:
Let the angular bisector of $A$ meet side $B C$ at
$
P(x, y, z)
$
By Angular Bisector theorem, $\mathrm{AB}: \mathrm{AC}=\mathrm{BP}: \mathrm{PC}$
$
B P: P C=: 2: 1=m: n
$
$\mathrm{B} \equiv(2,3,4) \quad \mathrm{C} \equiv(2,5,7)$
Put values
$
P(x, y, z)=\left(\frac{6}{3}, \frac{13}{3}, \frac{18}{3}\right)
$
Hence, the answer is $\frac{1}{3}(6 \hat{i}+13 \hat{j}+18 \hat{k})$
Summary
Direction cosines are essential mathematical representations for describing the orientation of vectors in three-dimensional space. By defining the cosines of angles between vectors and coordinate axes, it offers a standardized method to specify direction independently of magnitude. Understanding of Direction cosines helps us to analyze and solve varieties of Problems.
If a directed line $L$, passing through the origin, makes angles $\alpha, \beta$, and $\gamma_{\text {with }}$ the $\mathrm{x}-\mathrm{y}$ - and $z$-axes, respectively, called direction angles, then the cosines of these angles, namely, $\cos (\alpha), \cos \left({ }^\beta\right) \cos (\gamma)$, are called the direction cosines of the directed line $L$.
Direction Ratios are any set of three numbers that are proportional to the Direction cosines. If $\mathrm{l}, \mathrm{m}, \mathrm{n}$ are DCs of a vector then $\lambda l, \lambda m, \lambda n$ are DRs of this vector, where a can take any real value.
A vector has only one set of direction cosines, but infinite sets of direction ratios.
A given line in space can be extended in two opposite directions, and so it has two sets of direction cosines. In order to have a unique set of direction cosines for a given line in space, we must take the given line as a directed line.
The point $R(x, y, z)$ divides the line joining the points $P\left(x_1, y_1, z_1\right)$ and $Q\left(x_2, y_2, z_2\right)$ internally in the ratio $m: n$, then the coordinate of $R(x, y, z)$ is given by
$
\mathrm{R}(\mathrm{x}, \mathrm{y} . \mathrm{z})=\left(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y_1}{m+n}, \frac{m z_2+n z_1}{m+n}\right)
$
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