Section Formula in 3D

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.

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.

What is Section Formula?

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:

1. 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:

$\mathrm{R}(\mathrm{x}, \mathrm{y}, \mathrm{z})=\left(\frac{x_2+x_1}{2}, \frac{y_2+y_1}{2}, \frac{z_2+z_1}{2}\right)$

2. 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:

$\mathrm{R}(\mathrm{x}, \mathrm{y}, \mathrm{z})=\left(\frac{k x_2+x_1}{1+k}, \frac{k y_2+y_1}{1+k}, \frac{k z_2+z_1}{1+k}\right)$

3. 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

$\mathrm{R}(\mathrm{x}, \mathrm{y}, \mathrm{z})=\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$

Solved Examples

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.