26.6 - Section Formula

Exercise 26.2In the previous section, we completed a discussion on components of a vector. In this  section, we will see vector joining two points. Later in this section, we will see section formula also.

Vector joining two points

This can be explained in 6 steps:
1. In fig.26.22 below, P₁ and P₂ are any two points in space. We want $\small{\vec{P_1 P_2}}$ in component form.

Fig.26.22

2. Consider the three vectors $\small{\vec{OP_1},~\vec{OP_2}~\vec{P_1 P_2}}$.
They form the sides of a triangle.
3. Applying triangle law, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{OP_1}~+~\vec{P_1 P_2}}    & {~=~}    &{\vec{OP_2}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\vec{OP_1}~+~\vec{P_1 P_2}~+~\left(-\vec{OP_1} \right)}    & {~=~}    &{\vec{OP_2}~+~\left(-\vec{OP_1} \right)}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\vec{P_1 P_2}}    & {~=~}    &{\vec{OP_2}~-~\vec{OP_1}}
\\ \end{array}}$
◼ Remarks:
2 (magenta color): Here we add  the −ve of $\small{\vec{OP_1}}$ on both sides
4. We have the component form of $\small{\vec{OP_1}~\text{and}~\vec{OP_2}}$:
$\small{\vec{OP_1}=x_1\hat{i}+y_1\hat{j}+z_1\hat{k}}$
$\small{\vec{OP_2}=x_2\hat{i}+y_2\hat{j}+z_2\hat{k}}$
5. Therefore:
$\small{\vec{P_1 P_2}=\left(x_2 - x_1 \right)\hat{i}~+~\left(y_2 - y_1 \right)\hat{j}~+~\left(z_2 - z_1 \right)\hat{k}}$
6. We can write the magnitude also:
$\small{\left| \vec{P_1 P_2}\right| = \sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left(z_2 - z_1 \right)^2}}$

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Now we will see a solved example.

Solved example 26.28
Find the vector joining the points P(2,3,0) and Q(−1,−2,−4) directed from P to Q
Solution:
• We want the vector directed from P to Q. So P is the initial point and Q is the terminal point.
• Then we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{PQ}}    & {~=~}    &{(-1-2)\hat{i}+(-2-3)\hat{j}+(-4-0)\hat{k}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\vec{PQ}}    & {~=~}    &{-3\hat{i}-5\hat{j}-4\hat{k}}
\\ \end{array}}$

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Section formula

• P and Q are two points in space.
    ♦ $\small{\vec{OP}}$ is the position vector of P.
    ♦ $\small{\vec{OQ}}$ is the position vector of Q
• We know that, a line of infinite length can be drawn connecting P and Q. Consider a point R on this line. We want the position vector of R
• Two cases can arise in this situation.
Case I: R is within the line segment PQ
This can be analyzed in 6 steps:
1. In fig.26.23 below, point R is within PQ such that:
    ♦ Length PR = $\small{m\left|\vec{PQ} \right|}$
    ♦ Length QR = $\small{n\left|\vec{PQ} \right|}$
• $\small{m~\text{and}~n}$ are +ve scalars

Fig.26.23

2. We can write:
$\small{\frac{\left|\vec{PR} \right|}{\left|\vec{RQ} \right|} = \frac{m\left|\vec{PQ} \right|}{n\left|\vec{PQ} \right|} = \frac{m}{n}}$
• That means, R divides PQ internally in the ratio m:n
3. In the above step, all quantities are scalars because, we took the ratio of magnitudes. Let us try to bring vectors also into the equation.
• In the fig.26.23 above, $\small{\vec{PR}~\text{and}~\vec{RQ}}$ have the same direction. So their corresponding unit vectors will be equal.
We get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{PR}}    & {~=~}    &{\hat{RQ}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{\vec{PR}}{\left|\vec{PR} \right|}}    & {~=~}    &{\frac{\vec{RQ}}{\left|\vec{RQ} \right|}}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{\vec{PR}}{mn\left|\vec{PR} \right|}}    & {~=~}    &{\frac{\vec{RQ}}{mn\left|\vec{RQ} \right|}}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\frac{\vec{PR}}{m\left(n\left|\vec{PR} \right| \right)}}    & {~=~}    &{\frac{\vec{RQ}}{mn\left|\vec{RQ} \right|}}
\\ {~\color{magenta}    5    }    &{{\Rightarrow}}    &{\frac{\vec{PR}}{m\left(m\left|\vec{RQ} \right| \right)}}    & {~=~}    &{\frac{\vec{RQ}}{mn\left|\vec{RQ} \right|}}
\\ {~\color{magenta}    6    }    &{{\Rightarrow}}    &{\frac{\vec{PR}}{m}}    & {~=~}    &{\frac{\vec{RQ}}{n}}
\\ {~\color{magenta}    7    }    &{{\Rightarrow}}    &{n\,\vec{PR}}    & {~=~}    &{m\,\vec{RQ}}
\\ \end{array}}$

◼ Remarks:
• 3 (magenta color): Here we divide both sides by mn
• 5 (magenta color): Here we use the result
$\small{n\left|\vec{PR} \right| = m\left|\vec{RQ} \right|}$, which can be obtained from (2)

4. From triangle ORP, we get:
$\small{\vec{PR} = \vec{OR} - \vec{OP}}$
5. From triangle ORQ, we get:
$\small{\vec{RQ} = \vec{OQ} - \vec{OR}}$
6. From (3), we have: $\small{n\,\vec{PR} = m\,\vec{RQ}}$
• Substituting from (4) and (5), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{n\,\vec{PR}}    & {~=~}    &{m\,\vec{RQ}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{n\left(\vec{OR} - \vec{OP} \right)}    & {~=~}    &{m\left(\vec{OQ} - \vec{OR} \right)}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{n\,\vec{OR} - n\,\vec{OP}}    & {~=~}    &{m\,\vec{OQ} - m\,\vec{OR}}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{(m+n)\vec{OR}}    & {~=~}    &{m\,\vec{OQ}+n\,\vec{OP}}
\\ {~\color{magenta}    5    }    &{{\Rightarrow}}    &{\vec{OR}}    & {~=~}    &{\frac{m\,\vec{OQ}~+~n\,\vec{OP}}{m+n}}
\\ \end{array}}$

Case II: R is outside the line segment PQ, on the extension of PQ
This can be analyzed in 6 steps:
1. In fig.26.24 below, point R is outside PQ such that:
    ♦ Length PR = $\small{m\left|\vec{PQ} \right|}$
    ♦ Length QR = $\small{n\left|\vec{PQ} \right|}$
• $\small{m~\text{and}~n}$ are +ve scalars

Fig.26.24

2. We can write:
$\small{\frac{\left|\vec{PR} \right|}{\left|\vec{RQ} \right|} = \frac{m\left|\vec{PQ} \right|}{n\left|\vec{PQ} \right|} = \frac{m}{n}}$
• That means, R divides PQ externally in the ratio m:n
3. In the above step, all quantities are scalars because, we took the ratio of magnitudes. Let us try to bring vectors also into the equation.
• In the fig.26.24 above, $\small{\vec{PR}~\text{and}~\vec{RQ}}$ have opposite directions. So their corresponding unit vectors will differ by sign only.
We get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{PR}}    & {~=~}    &{-\hat{RQ}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{\vec{PR}}{\left|\vec{PR} \right|}}    & {~=~}    &{\frac{-\vec{RQ}}{\left|\vec{RQ} \right|}}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{\vec{PR}}{mn\left|\vec{PR} \right|}}    & {~=~}    &{\frac{-\vec{RQ}}{mn\left|\vec{RQ} \right|}}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\frac{\vec{PR}}{m\left(n\left|\vec{PR} \right| \right)}}    & {~=~}    &{\frac{-\vec{RQ}}{mn\left|\vec{RQ} \right|}}
\\ {~\color{magenta}    5    }    &{{\Rightarrow}}    &{\frac{\vec{PR}}{m\left(m\left|\vec{RQ} \right| \right)}}    & {~=~}    &{\frac{-\vec{RQ}}{mn\left|\vec{RQ} \right|}}
\\ {~\color{magenta}    6    }    &{{\Rightarrow}}    &{\frac{\vec{PR}}{m}}    & {~=~}    &{\frac{\vec{-RQ}}{n}}
\\ {~\color{magenta}    7    }    &{{\Rightarrow}}    &{n\,\vec{PR}}    & {~=~}    &{-m\,\vec{RQ}}
\\ \end{array}}$

◼ Remarks:
• 3 (magenta color): Here we divide both sides by mn
• 5 (magenta color): Here we use the result
$\small{n\left|\vec{PR} \right| = m\left|\vec{RQ} \right|}$, which can be obtained from (2)

4. From triangle ORP, we get:
$\small{\vec{PR} = \vec{OR} - \vec{OP}}$
5. From triangle ORQ, we get:
$\small{\vec{RQ} = \vec{OQ} - \vec{OR}}$
6. From (3), we have: $\small{n\,\vec{PR} = -m\,\vec{RQ}}$
• Substituting from (4) and (5), we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{n\,\vec{PR}}    & {~=~}    &{-m\,\vec{RQ}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{n\left(\vec{OR} - \vec{OP} \right)}    & {~=~}    &{-m\left(\vec{OQ} - \vec{OR} \right)}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{n\,\vec{OR} - n\,\vec{OP}}    & {~=~}    &{-m\,\vec{OQ} + m\,\vec{OR}}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{(m-n)\vec{OR}}    & {~=~}    &{m\,\vec{OQ}-n\,\vec{OP}}
\\ {~\color{magenta}    5    }    &{{\Rightarrow}}    &{\vec{OR}}    & {~=~}    &{\frac{m\,\vec{OQ}~-~n\,\vec{OP}}{m-n}}
\\ \end{array}}$

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Now we will see a special case. It can be written in 2 steps:
1. Let R be the midpoint of PQ. Then we can apply case I because, R will be between P and Q
• So we have: $\small{\vec{OR} = \frac{m\,\vec{OQ}~+~n\,\vec{OP}}{m+n}}$
2. Since R is the midpoint, we can write: m = n = 1
• Substituting these values of m and n in (1), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{OR}}    & {~=~}    &{\frac{(1)\,\vec{OQ}~+~(1)\,\vec{OP}}{1+1}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\vec{OR}}    & {~=~}    &{\frac{\vec{OQ}~+~\vec{OP}}{2}}
\\ \end{array}}$

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Now we will see some solved examples.

Solved example 26.29
Consider two points P and Q with position vectors $\small{\vec{OP} = 3\vec{a}-2\vec{b}}$ and $\small{\vec{OQ} = \vec{a}+\vec{b}}$. Find the position vector of a point R which divides the line joining P and Q in the ratio 2:1, (I) internally and (ii) externally.
Solution:
Part (i):
• For internal division, we have the formula:
$\small{\vec{OR} = \frac{m\,\vec{OQ}~+~n\,\vec{OP}}{m+n}}$
• Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{OR}}    & {~=~}    &{\frac{(2)\left(\vec{a}+\vec{b} \right)~+~(1)\left(3\vec{a}-2\vec{b} \right)}{2+1}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{2\vec{a} + 2\vec{b}+3\vec{a}-2\vec{b}}{3}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{5\vec{a}}{3}}
\\ \end{array}}$

Part (ii):
• For external division, we have the formula:
$\small{\vec{OR} = \frac{m\,\vec{OQ}~-~n\,\vec{OP}}{m-n}}$
• Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{OR}}    & {~=~}    &{\frac{(2)\left(\vec{a}+\vec{b} \right)~-~(1)\left(3\vec{a}-2\vec{b} \right)}{2-1}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{2\vec{a} + 2\vec{b}-3\vec{a}+2\vec{b}}{1}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{4\vec{b}-\vec{a}}
\\ \end{array}}$

Solved example 26.30
Show that the points
$\small{A\left(2\hat{i}-\hat{j}+\hat{k} \right)}$
$\small{B\left(\hat{i}-3\hat{j}-5\hat{k} \right)}$
$\small{C\left(3\hat{i}-4\hat{j}-4\hat{k} \right)}$
are the vertices of a right angled triangle
Solution:
1. Let us write the vectors connecting the points
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{AB}}    & {~=~}    &{\vec{OB} - \vec{OA}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\hat{i}-3\hat{j}-5\hat{k}~-~\left[2\hat{i}-\hat{j}+\hat{k} \right]}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{-\hat{i}-2\hat{j}-6\hat{k}}
\\ \end{array}}$

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{BC}}    & {~=~}    &{\vec{OC} - \vec{OB}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{3\hat{i}-4\hat{j}-4\hat{k}~-~\left[\hat{i}-3\hat{j}-5\hat{k} \right]}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{2\hat{i}-\hat{j}+\hat{k}}
\\ \end{array}}$

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{CA}}    & {~=~}    &{\vec{OA} - \vec{OC}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{2\hat{i}-\hat{j}+\hat{k}~-~\left[3\hat{i}-4\hat{j}-4\hat{k} \right]}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{-\hat{i}+3\hat{j}+5\hat{k}}
\\ \end{array}}$

2. Now we can write the squares of the magnitudes of the above vectors:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{AB}}    & {~=~}    &{-\hat{i}-2\hat{j}-6\hat{k}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left|\vec{AB} \right|^2}    & {~=~}    &{(-1)^2 + (-2)^2 + (-6)^2}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{41}
\\ \end{array}}$

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{BC}}    & {~=~}    &{2\hat{i}-\hat{j}+\hat{k}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left|\vec{BC} \right|^2}    & {~=~}    &{(2)^2 + (-1)^2 + (1)^2}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{6}
\\ \end{array}}$

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{CA}}    & {~=~}    &{-\hat{i}+3\hat{j}+5\hat{k}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left|\vec{CA} \right|^2}    & {~=~}    &{(-1)^2 + (3)^2 + (5)^2}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{35}
\\ \end{array}}$

3. We see that:
$\small{\left|\vec{AB} \right|^2 = \left|\vec{BC} \right|^2 + \left|\vec{CA} \right|^2}$

4. Applying Pythagoras theorem, we can say that:
    ♦ AB is the hypotenuse
    ♦ BC and CA form base and altitude
• So the three points are the vertices of a right angled triangle.

Solved example 26.31
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are:
$\small{\vec{OP}=\hat{i}+2\hat{j}-\hat{k}}$ and $\small{\vec{OQ} = -\hat{i}+\hat{j}+\hat{k}}$, respectively in the ratio 2:1
(i) internally (ii) externally.
Solution:
Part (i):
• For internal division, we have the formula:
$\small{\vec{OR} = \frac{m\,\vec{OQ}~+~n\,\vec{OP}}{m+n}}$
• Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{OR}}    & {~=~}    &{\frac{(2)\left(-\hat{i}+\hat{j}+\hat{k} \right)~+~(1)\left(\hat{i}+2\hat{j}-\hat{k} \right)}{2+1}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{-2\hat{i}+2\hat{j}+2\hat{k}~+~\left(\hat{i}+2\hat{j}-\hat{k} \right)}{3}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{-\hat{i}+4\hat{j}+\hat{k}}{3}}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\left(\frac{-1}{3} \right)\hat{i}+\left(\frac{4}{3} \right)\hat{j}+\left(\frac{1}{3} \right)\hat{k}}
\\ \end{array}}$

Part (ii):
• For external division, we have the formula:
$\small{\vec{OR} = \frac{m\,\vec{OQ}~-~n\,\vec{OP}}{m-n}}$
• Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{OR}}    & {~=~}    &{\frac{(2)\left(-\hat{i}+\hat{j}+\hat{k} \right)~-~(1)\left(\hat{i}+2\hat{j}-\hat{k} \right)}{2-1}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{-2\hat{i}+2\hat{j}+2\hat{k} ~-~\left(\hat{i}+2\hat{j}-\hat{k} \right)}{1}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{-3\hat{i}+3\hat{k}}
\\ \end{array}}$

Solved example 26.32
Show that the points A, B and C with position vectors
$\small{3\hat{i}-4\hat{j}-4\hat{k}}$
$\small{2\hat{i}-\hat{j}+\hat{k}}$
$\small{\hat{i}-3\hat{j}-5\hat{k}}$,
respectively form the vertices of a right angled triangle
Solution:
1. Let us write the vectors connecting the points
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{AB}}    & {~=~}    &{\vec{OB} - \vec{OA}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{2\hat{i}-\hat{j}+\hat{k}~-~\left[3\hat{i}-4\hat{j}-4\hat{k} \right]}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{-\hat{i}+3\hat{j}+5\hat{k}}
\\ \end{array}}$

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{BC}}    & {~=~}    &{\vec{OC} - \vec{OB}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\hat{i}-3\hat{j}-5\hat{k}~-~\left[2\hat{i}-\hat{j}+\hat{k} \right]}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{-\hat{i}-2\hat{j}-6\hat{k}}
\\ \end{array}}$

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{CA}}    & {~=~}    &{\vec{OA} - \vec{OC}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{3\hat{i}-4\hat{j}-4\hat{k}~-~\left[\hat{i}-3\hat{j}-5\hat{k} \right]}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{2\hat{i}-\hat{j}+\hat{k}}
\\ \end{array}}$

2. Now we can write the squares of the magnitudes of the above vectors:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{AB}}    & {~=~}    &{-\hat{i}+3\hat{j}+5\hat{k}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left|\vec{AB} \right|^2}    & {~=~}    &{(-1)^2 + 3^2 + 5^2}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{35}
\\ \end{array}}$

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{BC}}    & {~=~}    &{-\hat{i}-2\hat{j}-6\hat{k}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left|\vec{BC} \right|^2}    & {~=~}    &{(-1)^2 + (-2)^2 + (-6)^2}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{41}
\\ \end{array}}$

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{CA}}    & {~=~}    &{2\hat{i}-\hat{j}+\hat{k}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left|\vec{CA} \right|^2}    & {~=~}    &{(2)^2 + (-1)^2 + (1)^2}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{6}
\\ \end{array}}$

3. We see that:
$\small{\left|\vec{BC} \right|^2 = \left|\vec{AB} \right|^2 + \left|\vec{CA} \right|^2}$

4. Applying Pythagoras theorem, we can say that:
    ♦ BC is the hypotenuse
    ♦ AB and CA form base and altitude
• So the three points are the vertices of a right angled triangle.

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The link below gives a few more solved examples:

Exercise 26.2

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After completing the above exercise, the reader may attempt the two problems given below:

Solved example 26.33
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are:
$\small{\left(2\vec{a}+\vec{b} \right)}$ and $\small{\left(\vec{a}-3\vec{b} \right)}$, externally in the ratio 1:2. Also show that P is the midpoint of the line segment RQ
Solution:
Part (i):
1. Given that, P is to be the midpoint of RQ. So we draw the rough sketch in such a way that, P is some where between R and Q. It is shown in fig.26.25 below:

Fig.26.25

2. OP and OQ are the original vectors. They are shown in magenta color. Point R divides QP externally into two parts: QR and PR

3. For this problem, we make the following changes:
(i) We consider $\small{\vec{QP}}$ instead of the usual $\small{\vec{PQ}}$
(ii) In the usual case, 'm' is related to the end P of $\small{\vec{PQ}}$. In the present case, 'm' is related to end Q of $\small{\vec{QP}}$
• The segment related to end Q is QR. In this problem, for external division, we consider QR:PR. Segment QR is larger than segment PR. So we can write:
QR:PR = m:n = 2:1
(iii) In the usual case, the formula that we use is:
$\small{\vec{OR} = \frac{m\,\vec{OQ}~-~n\,\vec{OP}}{m-n}}$
• So for the present case, we must change the formula to:
$\small{\vec{OR} = \frac{m\,\vec{OP}~-~n\,\vec{OQ}}{m-n}}$

4. Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{OR}}    & {~=~}    &{\frac{m\,\vec{OP}~-~n\,\vec{OQ}}{m-n}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{(2)\left(2\vec{a}+\vec{b} \right)~-~(1)\left(\vec{a}-3\vec{b} \right)}{2-1}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{4\vec{a}+2\vec{b} ~-~\left(\vec{a}-3\vec{b} \right)}{2-1}}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{3\vec{a}+5\vec{b}}
\\ \end{array}}$

Part (ii):
If P is the midpoint of RQ, we can say that, P divides RQ in the ratio 1:1.
1. We can treat $\small{\vec{OQ}~\text{and}~\vec{OR}}$ as the original vectors. Then we can find the position vector $\small{\vec{OS}}$ of the "assumed midpoint" S of QR
2. The original formula is:
$\small{\vec{OR} = \frac{\vec{OQ}~+~\vec{OP}}{2}}$
• In this original case, R is the midpoint of PQ
3. For the present case, we must change the formula to:
$\small{\vec{OS} = \frac{\vec{OR}~+~\vec{OQ}}{2}}$
4. Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{OS}}    & {~=~}    &{\frac{\vec{OR}~+~\vec{OQ}}{2}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{\left(3\vec{a}+5\vec{b} \right)~+~\left(\vec{a}-3\vec{b} \right)}{2}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{4\vec{a}+2\vec{b}}{2}}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{2\vec{a}+\vec{b}}
\\ \end{array}}$
5. We see that: $\small{\vec{OP}=\vec{OS}}$
• That means, points P and S are the same.
• That means, P is the midpoint of QR

Solved example 26.34
Show that the points A(1,−2,−8), B(5,0,−2) and C(11,3,7) are collinear, and find the ratio in which B divides AC
Solution:
Part (i):
1. Fig.26.26(i) below shows the rough sketch

Fig.26.26

2. First we write $\small{\vec{AB}}$
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{AB}}    & {~=~}    &{(5-1)\hat{i}+(0-(-2))\hat{j}+(-2-(-8))\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{4\hat{i}+2\hat{j}+6\hat{k}}
\\ \end{array}}$

• Now we write the unit vector:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{AB}}    & {~=~}    &{\frac{\vec{AB}}{\left|\vec{AB} \right|}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{4\hat{i}+2\hat{j}+6\hat{k}}{\sqrt{16+4+36}}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{4\hat{i}+2\hat{j}+6\hat{k}}{\sqrt{56}}~=~\frac{4\hat{i}+2\hat{j}+6\hat{k}}{\sqrt{4(14)}}}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\frac{4\hat{i}+2\hat{j}+6\hat{k}}{2\sqrt{14}}}
\\ {~\color{magenta}    5    }    &{}    &{}    & {~=~}    &{\frac{2\hat{i}+\hat{j}+3\hat{k}}{\sqrt{14}}}
\\ \end{array}}$

3. Next we write $\small{\vec{BC}}$
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{BC}}    & {~=~}    &{(11-5)\hat{i}+(3-0)\hat{j}+(7-(-2))\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{6\hat{i}+3\hat{j}+9\hat{k}}
\\ \end{array}}$

• Now we write the unit vector:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{BC}}    & {~=~}    &{\frac{\vec{BC}}{\left|\vec{BC} \right|}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{6\hat{i}+3\hat{j}+9\hat{k}}{\sqrt{36+9+81}}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{6\hat{i}+3\hat{j}+9\hat{k}}{\sqrt{126}}~=~\frac{6\hat{i}+3\hat{j}+9\hat{k}}{\sqrt{9(14)}}}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\frac{6\hat{i}+3\hat{j}+9\hat{k}}{3\sqrt{14}}}
\\ {~\color{magenta}    5    }    &{}    &{}    & {~=~}    &{\frac{2\hat{i}+\hat{j}+3\hat{k}}{\sqrt{14}}}
\\ \end{array}}$

4. We see that: $\small{\hat{AB}=\hat{BC}}$
• So $\small{\vec{AB}~\text{and}~\vec{BC}}$ are parallel.
• But B is a common point. So the three points are collinear.

Part (ii):
1. Fig.26.26(ii) above shows the rough sketch. From part (i), we know that A, B and C are collinear. We are asked to find the ratio $\small{\left|\vec{CB} \right|:\left|\vec{BA} \right|}$. We will denote this ratio as m:n

2. When CA is divided in this way, we get:
    ♦ $\small{\left|\vec{CB} \right|=m\left|\vec{CA} \right|}$
    ♦ $\small{\left|\vec{BA} \right|=n\left|\vec{CA} \right|}$
• So we want: $\small{\left|\vec{CB} \right|,~\left|\vec{BA} \right|~\text{and}~\left|\vec{CA} \right|}$

3. From the coordinates of A and C, we get:
$\small{\vec{CA} = -10\hat{i}-5\hat{j}-15\hat{k}}$
• Therefore, $\small{\left|\vec{CA} \right|=\sqrt{350}=5\sqrt{14}}$

4. From Part (i), we have: $\small{\vec{CB} = -6\hat{i}-3\hat{j}-9\hat{k}}$
• Therefore, $\small{\left|\vec{CB} \right|=\sqrt{126}=3\sqrt{14}}$ 

5. From Part (i), we have: $\small{\vec{BA} = -4\hat{i}-2\hat{j}-6\hat{k}}$
• Therefore, $\small{\left|\vec{BA} \right|=\sqrt{56}=2\sqrt{14}}$

6. So from (2), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{CB} \right|}    & {~=~}    &{m\left|\vec{CA} \right|}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{3\sqrt{14}}    & {~=~}    &{m\left(5\sqrt{14} \right)}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{m}    & {~=~}    &{\frac{3}{5}}
\\ \end{array}}$

7. Also from (2), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{BA} \right|}    & {~=~}    &{n\left|\vec{CA} \right|}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{2\sqrt{14}}    & {~=~}    &{n\left(5\sqrt{14} \right)}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{n}    & {~=~}    &{\frac{2}{5}}
\\ \end{array}}$

8. Thus the ratio m:n is $\small{\frac{3}{5}:\frac{2}{5}}$
• This is same as 3:2

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In the next section, we will see scalar product.

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Chapter 12.6 - Miscellaneous Examples

In the previous section, we saw some solved examples on section formula. In this section, we will see some miscellaneous examples.

Solved example 12.11
Show that the points A(1,2,3), B(-1,-2,-1), C(2,3,2) and D(4,7,6) are the vertices of a parallelogram ABCD but it is not a rectangle.
Solution:
1. A rough sketch is shown in fig.12.19 below:

Fig.12.19

2. Using the distance formula, we can find the lengths of sides.
   ♦ AB = √36 = 6
   ♦ BC = √43
   ♦ CD = √36 = 6
   ♦ DA = √43
3. We see that, both pairs of opposite sides (AB, CD) and (BC, DA) are equal. So it is a parallelogram.
4. A rectangle is a parallelogram in which both diagonals are of the same length.
• If a parallelogram is not to be a rectangle, it's diagonals should not be of the same length.
5. Using distance formula, let us write the lengths of the diagonals:
   ♦ AC = √3
   ♦ BD = √155
• We see that, diagonals are not equal. So it a parallelogram which is not a rectangle.

Solved example 12.12
Find the equation of the set of points P such that it's distances from the points A(3,4,-5) and B(-2,1,4) are equal.
Solution:
1. Let P(x,y,z) be equidistant from A and B
2. Let us use the distance formula.
• Square of the distance PA can be obtained as follows:
PA² = (x-3)² + (y-4)² + (z+5)²
= x² -6x +9 +y² -8y +16 +z² +10z +25
• Square of the distance PB can be obtained as follows:
PB² = (x+2)² + (y-1)² + (z-4)²
= x² +4x +4 +y² -2y +1 +z² -8z +16
3. Since the distances are equal, the square of the distances will also be equal. We can write:
x² -6x +9 +y² -8y +16 +z² +10z +25 = x² +4x +4 +y² -2y +1 +z² -8z +16
⇒ -6x +9 -8y +16 +10z +25 = +4x +4 -2y +1 -8z +16
⇒ -6x -4x +9 -8y +2y +16 +10z +8z +25 = +4  +1 +16
⇒ -10x +9 -6y +16 +18z +25 = +4  +1 +16
⇒ -10x -6y +18z +50 = 21
⇒ -10x -6y +18z +29 = 0
⇒ 10x +6y -18z -29 = 0

Solved example 12.13
The centroid of a triangle ABC is at G(1,1,1). If the coordinates of A and B are (3,-5,7) and (-1,7,-6) respectively, find the coordinates of C.
Solution:
1. Let the coordinates of C be (x,y,z)
Then we can draw a rough sketch as shown below:

Fig.12.20

2. We know that, the coordinates of the centroid of any triangle can be obtained using the expression:
$\left( \frac{x_1 + x_2 + x_3}{3},~ \frac{y_1 + y_2 + y_3}{3},~ \frac{z_1 + z_2 + z_3}{3}\right)$  

3. Substituting the known values, we get:
• $ \frac{3 -1 + x}{3}~=~1$
⇒ 2+x = 3
⇒ x = 1 
• $ \frac{-5 +7 + y}{3}~=~1$
⇒ 2+y = 3
⇒ y = 1 
• $ \frac{7 -6 + z}{3}~=~1$
⇒ 1+x = 3
⇒ z = 2

4. So the coordinates of C are: (1,1,2)

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The link below gives a few more solved examples:

Miscellaneous Exercise

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We have completed a disscusion on three dimensional geometry. In the next chapter, we will see limits and derivatives.

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Chapter 12.5 - Solved Examples on Section Formula

In the previous section, we saw section formula. We also saw a solved example. In this section, we will see a few more solved examples.

Solved example 12.8
Using the section formula, prove that the three points A(-4,6,10), B(2,4,6) and C(14,0,-2) are collinear.
Solution:
1. Let us assume that, the third point C is collinear with A and B. Then two possibilities arise:
(i) C lies between A and B
(ii) C lies in the extension of AB
2. If the case is 1(i), then C will be dividing AB internally in some ratio k:1
3. If the case is 1(ii), then C will be dividing AB externally in some ratio k:1
4. For a point which divides AB (internally or externally) in the ratio k:1, the coordinates will be:
$\left(\frac{k x_2 + x_1}{1+k},~\frac{k y_2 + y_1}{1+k},~\frac{k z_2 + z_1}{1+k} \right)$
⇒ $\left(\frac{k × 2 + -4}{1+k},~\frac{k × 4 + 6}{1+k},~\frac{k × 6 + 10}{1+k} \right)$
⇒ $\left(\frac{2k - 4}{1+k},~\frac{4k + 6}{1+k},~\frac{6k + 10}{1+k} \right)$
5. Let us check whether the coordinates of C satisfy the conditions in (4).
The check can be done in 4 steps:
(i) Equating the x coordinates, we get:

$\begin{array}{ll}
{}&{\frac{2k - 4}{1+k}}
& {~=~}& {14}
&{} \\

{\Rightarrow}&{2k-4}
& {~=~}& {14 +14k }
&{} \\

{\Rightarrow}&{-12k}
& {~=~}& {18}
&{} \\

{\Rightarrow}&{-2k}
& {~=~}& {3}
&{} \\

{\Rightarrow}&{k}
& {~=~}& {- \frac{3}{2}}
&{} \\

\end{array}$

(ii) If k = -3/2, then the y-coordinate can be calculated from the result in (4). We get:

$\begin{array}{ll}
{}&{\rm{y-coordinate}}
& {~=~}& {\frac{4k + 6}{1+k}}
&{} \\

{}&{}
& {~=~}& {\frac{4(-3/2) + 6}{1+(-3/2)}}
&{} \\

{}&{}
& {~=~}& {\frac{-6+6}{1+(-3/2)}}
&{} \\

{}&{}
& {~=~}& {0}
&{} \\

\end{array}$

• Note that, the y-coordinate of C is also '0'.

(iii) If k = -3/2, then the z-coordinate can be calculated from the result in (4). We get:

$\begin{array}{ll}
{}&{\rm{z-coordinate}}
& {~=~}& {\frac{6k + 10}{1+k}}
&{} \\

{}&{}
& {~=~}& {\frac{6(-3/2) + 10}{1+(-3/2)}}
&{} \\

{}&{}
& {~=~}& {\frac{-9+10}{-(1/2)}}
&{} \\

{}&{}
& {~=~}& {\frac{1}{-(1/2)}}
&{} \\

{}&{}
& {~=~}& {-2}
&{} \\

\end{array}$

• Note that, the z-coordinate of C is also '-2'.

(iv) So the coordinates of C satisfy the conditions in (4).

6. That means, the point C indeed divides AB internally or externally in the ratio k:1
Since k (= -3/2) is a -ve value, it is an external division.
7. Point C can divide AB (internally or externally) only if A, B and C is collinear.

Solved example 12.9
Find the coordinates of the centroid of the triangle PQR whose vertices are P(x₁,y₁,z₁), Q(x₂,y₂,z₂), R(x₃,y₃,z₃).
Solution:
1. Let A be the midpoint of QR. Then the coordinates of A will be:
$\left(\frac{x_2 + x_3}{2},~\frac{y_2 + y_3}{2},~\frac{z_2 + z_3}{2} \right)$
2. Let G be the centroid.
3. We have the coordinates of P and A.
• The centroid G will divide PA internally in the ratio 2:1
• So the coordinates of G will be:
$\left(\frac{m x_2 + n x_1}{m+n},~\frac{m y_2 + n y_1}{m+n},~\frac{m z_2 + n z_1}{m+n} \right)$
⇒ $\left(\frac{2 × \frac{x_2 + x_3}{2} + 1 × x_1}{2+1},~\frac{2 × \frac{y_2 + y_3}{2} + 1 × y_1}{2+1},~\frac{2 × \frac{z_2 + z_3}{2} + 1 × z_1}{2+1} \right)$
⇒ $\left(\frac{x_2 + x_3 + 1 × x_1}{2+1},~\frac{y_2 + y_3 + 1 × y_1}{2+1},~\frac{z_2 + z_3 + 1 × z_1}{2+1} \right)$
⇒ $\left(\frac{x_1 + x_2 + x_3}{3},~\frac{y_1 + y_2 + y_3}{3},~\frac{z_1 + z_2 + z_3}{3} \right)$

Solved example 12.10
Find the ratio in which the line segment joining the points A(4,8,10) and B(6,10,-8) is divided by the YZ-plane. Also find the coordinates of the point at which the division is done.
Solution:
1. Let the YZ-plane divide AB in the ratio k:1
2. Let the point of division be C.
3. We can write the coordinates of C as follows:
$\left(\frac{k x_2 + x_1}{1+k},~\frac{k y_2 + y_1}{1+k},~\frac{k z_2 + z_1}{1+k} \right)$
⇒ $\left(\frac{k × 6 + 4}{1+k},~\frac{k × 10 + 8}{1+k},~\frac{k × -8 + 10}{1+k} \right)$
⇒ $\left(\frac{6k + 4}{1+k},~\frac{10k + 8}{1+k},~\frac{-8k + 10}{1+k} \right)$
4. Consider any point on the YZ-plane. The x-coordinate of that point will be zero.
So the x-coordinate written in (3) is zero. We get:

$\begin{array}{ll}
{}&{\frac{6k + 4}{1+k}}
& {~=~}& {0}
&{} \\

{\Rightarrow}&{6k + 4}
& {~=~}& {0}
&{} \\

{\Rightarrow}&{6k}
& {~=~}& {-4}
&{} \\

{\Rightarrow}&{3k}
& {~=~}& {-2}
&{} \\

{\Rightarrow}&{k}
& {~=~}& {- \frac{2}{3}}
&{} \\

\end{array}$

5. Thus we get the value of k.
• We have seen that, k:1 is just another form of m:n. Both represent the same ratio.
So we can write:

$\begin{array}{ll}
{}&{k:1}
& {~=~}& {\frac{k}{1}}
&{} \\

{\Rightarrow}&{\frac{k}{1}}
& {~=~}& {\frac{m}{n}}
&{} \\

{\Rightarrow}&{\frac{-(2/3)}{1}}
& {~=~}& {\frac{m}{n}}
&{} \\

{\Rightarrow}&{-\frac{2}{3}}
& {~=~}& {\frac{m}{n}}
&{} \\

\end{array}$

• That means, the YZ-plane divides AB in the ratio m:n where m/n = -(2/3)

6. Now we can calculate the y-coordinate of C.
• Substituting the value of k in the expression for y-coordinate in (3), we get:

$\begin{array}{ll}
{}&{\rm{y-coordinate}}
& {~=~}& {\frac{10k + 8}{1+k}}
&{} \\

{}&{}
& {~=~}& {\frac{10(-2/3) + 8}{1+(-2/3)}}
&{} \\

{}&{}
& {~=~}& {\frac{-20+24}{3-2}}
&{} \\

{}&{}
& {~=~}& {4}
&{} \\

\end{array}$

7. Finally we can calculate the z-coordinate of C.
• Substituting the value of k in the expression for z-coordinate in (3), we get:

$\begin{array}{ll}
{}&{\rm{z-coordinate}}
& {~=~}& {\frac{-8k + 10}{1+k}}
&{} \\

{}&{}
& {~=~}& {\frac{-8(-2/3) + 10}{1+(-2/3)}}
&{} \\

{}&{}
& {~=~}& {\frac{16+30}{3-2}}
&{} \\

{}&{}
& {~=~}& {46}
&{} \\

\end{array}$

8. So the division is done at C(0,4,46)
• Fig.12.18 below shows the actual plot.

Fig.12.18

We can write 3 points:
(i) We see that:
• The YZ-plane does not divide AB
• But the YZ plane divides the extension of AB. So it is an external division. The division is done at C(0,4,46)
(ii) We have the coordinates of all three points:
A(4,8,10), B(6,10,-8) and C(0,4,46)
    ♦ Using the distance formula, BC = √2988
    ♦ Using the distance formula, AC = √1328
(iii) So we get:
$\frac{AC}{BC}~=~\frac{\sqrt{1328}}{\sqrt{2988}}~=~\sqrt{\frac{1328}{2988}}~=~\sqrt{\frac{4}{9}}~=~\frac{2}{3}$

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The link below gives a few more solved examples:

Exercise 12.3

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In the next section, we will see some miscellaneous examples.

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Chapter 12.4 - Section Formula

In the previous section, we completed a discussion on distance formula. In this section, we will see section formula.

• In our earlier classes, we have seen the application of section formula in two-dimensional problems.
• Let us recall the important points related to the section formula. It can be written in 2 steps:
1. In fig.12.13 (a) below, the line segment PQ lies in the XY-plane.    
    ♦ Point R lies between P and Q.
    ♦ Point R divides the segment PQ internally in the ratio m:n
    ♦ Coordinates of P and Q are (x₁,y₁) and (x₂,y₂) respectively.
• Then the coordinates of R will be: $\left(\frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n} \right)$.

Fig.12.13

2. In fig.12.13 (b) above, the line segment PQ lies in the XY-plane.    
    ♦ Point R does not lie between P and Q. It is on the extension of the line PQ.
    ♦ Point R divides the segment PQ externally in the ratio m:n
    ♦ Coordinates of P and Q are (x₁,y₁) and (x₂,y₂) respectively.
• Then the coordinates of R will be:$\left(\frac{m x_2 - n x_1}{m-n}, \frac{m y_2 - n y_1}{m-n} \right)$.

---

• Now we will see the section formula when P(x₁,y₁,z₁) and Q(x₂,y₂,z₂) are two points in three-dimensional space.
    ♦ R (x,y,z) divides PQ in the ratio m:n.
    ♦ We need to derive expressions for x, y and z.
• First, we will derive an expression for z. It can be derived in 8 steps:

1. The first step is to drop perpendiculars from P, Q and R, on to the XY-plane.
• In fig.12.14 below, the XY-plane is shown in blue color.
    ♦ L is the foot of the perpendicular from P.
    ♦ M is the foot of the perpendicular from Q.
    ♦ N is the foot of the perpendicular from R.

Fig.12.14

2. Draw the line LM.
    ♦ Point N will lie on LM.
3. Through R, draw a line parallel to LM.
    ♦ Let this line meet PL (actually, it is the ‘extension of PL’) at S.
    ♦ Let this line meet QM at T.
4. The points P, Q, R, S and T are on the same plane. So we can show them separately as in fig.12.15 below:

Fig.12.15

• Consider the two triangles ΔPRS and ΔQRT.
• Let us analyze the angles of the two triangles. The analysis can be written in 5 steps:
(i) PS and QT are vertical lines because, they are drawn perpendicular to the XY-plane.
(ii) ST is a horizontal line because, it is drawn parallel to the XY-plane.
(iii) So we get: ∠PSR = ∠QTR = 90^o
(iv) Angle at R is common to both triangles.
(v) So we have two angles same in both triangles. Consequently, the third angles (∠SPR and ∠TQR) will also be the same.
5. Since angles are the same in the two triangles, they are similar triangles.
• Since they are similar triangles, we can take ratio of sides:

(i) $\frac{\text{Side opposite ∠R in ΔPRS}}{\text{Side opposite ∠R in ΔQRT}}~=~\frac{\text{SP}}{\text{TQ}}$

(ii) $\frac{\text{Side opposite ∠PSR in ΔPRS}}{\text{Side opposite ∠QTR in ΔQRT}}~=~\frac{\text{PR}}{\text{QR}}~=~\frac{m}{n}$

(iii) $\frac{\text{Side opposite ∠SPR in ΔPRS}}{\text{Side opposite ∠TQR in ΔQRT}}~=~\frac{\text{SR}}{\text{TR}}$

6. For similar triangles, the three ratios will be equal. So we get:

$\frac{\text{SP}}{\text{TQ}}~=~\frac{m}{n}~=~\frac{\text{SR}}{\text{TR}}$ 

• We need only the first two ratios. We can write:

$\frac{m}{n}~=~\frac{\text{SP}}{\text{TQ}}$

7. Now we can write the lengths in terms of 'z'. It can be done in two steps:
(i) From fig.12.14, we get:
SP = SL – PL
• S is at the same level as R.
So SL = RN = z
• Also, PL = height of P = z₁
• Therefore, SP = z-z₁
(ii) Again from Fig.12.14, we get:
TQ = QM – TM
• T is at the same level as R
So TM = RN = z
• Also, QM = height of Q = z₂
• Therefore, TQ = z₂ – z
8. Substituting for SP and TQ in (6), we get:

$\begin{array}{ll}
{}&{\frac{m}{n}}
& {~=~}& {\frac{z - z_1}{z_2 - z}}
&{} \\

{\Rightarrow}&{m(z_2 - z)}
& {~=~}& {n(z - z_1)}
&{} \\

{\Rightarrow}&{m z_2 ~-~m z}
& {~=~}& {n z ~-~n z_1}
&{} \\

{\Rightarrow}&{m z_2 ~+~n z_1}
& {~=~}& {n z ~+~ m z}
&{} \\

{\Rightarrow}&{(m~+~n)z}
& {~=~}& {m z_2 ~+~n z_1}
&{} \\

{\Rightarrow}&{z}
& {~=~}& {\frac{m z_2 + n z_1}{m+n}}
&{} \\

\end{array}$

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• So we obtained the z-coordinate of R. We did this by dropping perpendiculars from P and Q, onto the XY-plane.
• If we drop perpendiculars from P and Q, onto the XZ-plane, we will get the y-coordinate of R.
• If we drop perpendiculars from P and Q, onto the YZ-plane, we will get the x-coordinate of R.
• The perpendiculars onto the XZ-plane is shown in fig.12.16 below:

Fig.12.16

• In the above fig.12.16, we see the two similar triangles ΔPRS and ΔQRT.
• Based on the above fig., the reader may write all the steps for the y-coordinate in his/her own notebooks.
   ♦ The result will be: $y~=~\frac{m y_2 + n y_1}{m+n}$
• Similarly, the reader may draw the diagrams and write the steps for x-coordinate also.
   ♦ The result will be: $x~=~\frac{m x_2 + n x_1}{m+n}$

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So we can write a summary:

Case 1:
   ♦ P(x₁,y₁,z₁) and Q(x₂,y₂,z₂) are two points in space.
   ♦ Point R divides the line segment PQ internally in the ratio m:n
   ♦ Then the coordinates of R will be: $\left(\frac{m x_2 + n x_1}{m+n},~\frac{m y_2 + n y_1}{m+n},~\frac{m z_2 + n z_1}{m+n} \right)$

Case 2:
   ♦ P(x₁,y₁,z₁) and Q(x₂,y₂,z₂) are two points in space.
   ♦ Point R divides the line segment PQ externally in the ratio m:n
   ♦ Then the coordinates of R will be: $\left(\frac{m x_2 - n x_1}{m-n},~\frac{m y_2 - n y_1}{m-n},~\frac{m z_2 - n z_1}{m-n} \right)$

Case 3:
   ♦ Consider case 1.
   ♦ If R is the midpoint of PQ, then m:n will be 1:1
   ♦ So the coordinates of R will be: $\left(\frac{1 × x_2 + 1 × x_1}{1+1},~\frac{1 × y_2 + 1 × y_1}{1+1},~\frac{1 × z_2 + 1 × z_1}{1+1} \right)$
   ♦ That means: If R is the midpoint, then it's coordinates will be: $\left(\frac{x_2 + x_1}{2},~\frac{y_2 + y_1}{2},~\frac{z_2 + z_1}{2} \right)$

Case 4:
   ♦ Any ratio m:n can be written as k:1
   ♦ For that, we divide both m and n by n
   ♦ $\frac{m}{n}~=~\frac{m/n}{n/n}~=~\frac{m/n}{1}~=~\frac{k}{1}$
         ✰ So we get: $k~=~\frac{m}{n}$
   ♦ So the coordinates of R will be: $\left(\frac{k × x_2 + 1 × x_1}{k+1},~\frac{k × y_2 + 1 × y_1}{k+1},~\frac{k × z_2 + 1 × z_1}{k+1} \right)$
   ♦ That means, coordinates of R are: $\left(\frac{k x_2 + x_1}{1+k},~\frac{k y_2 + y_1}{1+k},~\frac{k z_2 + z_1}{1+k} \right)$
• This case 4 is helpful in some special problems where, we need to calculate only one unknown value 'k' instead of two unknown values 'm' and 'n'.

---

Now we will see a solved example:

Solved example 12.7
Find the coordinates of point R which divides the line segment joining P(1,-2,3) and Q(3,4,-5) in the ratio 2:3 (i) internally, (ii) externally.
Solution:
Part (i): Dividing internally
1. The ratio m:n is 2:3
2. Coordinates of R will be: $\left(\frac{m x_2 + n x_1}{m+n},~\frac{m y_2 + n y_1}{m+n},~\frac{m z_2 + n z_1}{m+n} \right)$
3. Substituting the values, we get:
$\left(\frac{2 × 3 + 3 × 1}{2+3},~\frac{2 × 4 + 3 × -2}{2+3},~\frac{2 × -5 + 3 × 3}{2+3} \right)$
⇒ $\left(\frac{6 + 3}{5},~\frac{8 -6}{5},~\frac{-10 + 9}{5} \right)$
⇒ $\left(\frac{9}{5},~\frac{2}{5},~\frac{-1}{5} \right)$

Part (ii): Dividing externally
1. The ratio m:n is 2:3
2. Coordinates of R' will be: $\left(\frac{m x_2 - n x_1}{m-n},~\frac{m y_2 - n y_1}{m-n},~\frac{m z_2 - n z_1}{m-n} \right)$
3. Substituting the values, we get:
$\left(\frac{2 × 3 - 3 × 1}{2-3},~\frac{2 × 4 - 3 × -2}{2-3},~\frac{2 × -5 - 3 × 3}{2-3} \right)$
⇒ $\left(\frac{6 - 3}{-1},~\frac{8 +6}{-1},~\frac{-10 - 9}{-1} \right)$
⇒ $\left(\frac{3}{-1},~\frac{14}{-1},~\frac{-19}{-1} \right)$
⇒ (-3, -14, 19)

Check:
1. Fig.12.17 below shows the actual plot:

Fig.12.17

2. We have the coordinates of all four points:
P(1,-2,3), Q(3,4,-5), R(9/5, 2/5, -1/5) and R'(-3, -14, 19)
3. First we will check the internal division:
• Using the distance formula, we can find the lengths:
    ♦ PR = 37/9 units
    ♦ QR = 49/8 units
• Thus the ratio PR/QR = $\frac{37/9}{49/8}~=~\frac{2}{3}$
4. Next we will check the external division:
• Using the distance formula, we can find the lengths:
    ♦ PR' = 102/5 units
    ♦ QR' = 153/5 units
• Thus the ratio PR'/QR' = $\frac{102/5}{153/5}~=~\frac{2}{3}$ 

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In the next section, we will see a few more solved examples. 

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Chapter 10 - Straight Lines

In the previous section, we completed a discussion on sequences and series. In this chapter, we will see Straight lines.

In our earlier classes, we have discussed problems related to coordinate geometry. Let us recall some of those topics:

A. Significance of coordinates
This can be written in 3 steps:
1. In fig.10.1 below, points P and Q lie on the XY plane.

Fig.10.1

2. The coordinates of the point P are (4,1.5)
• Then we can write:
    ♦ P is at a distance of 4 units from the y-axis.
        ✰ This distance is measured along the +ve x-axis.
    ♦ P is at a distance of 1.5 units from the x-axis.
        ✰ This distance is measured along the +ve y-axis.
3. The coordinates of the point Q are (-2.5,-2)
• Then we can write:
    ♦ Q is at a distance of 2.5 units from the y-axis.
        ✰ This distance is measured along the -ve x-axis.
    ♦ Q is at a distance of 2 units from the x-axis.
        ✰ This distance is measured along the -ve y-axis

B. Distance between two points
This can be written in 3 steps:
1. In fig.10.2 below, points P (x₁,y₁) and Q (x₂,y₂) lie on the XY-plane.

Fig.10.2

2. Then the distance between the two points can be obtained using the formula:
$\rm{PQ~=~\sqrt{\left(x_2 - x_1 \right)^2~+~\left(y_2 - y_1 \right)^2}}$
3. For example, if (x₁,y₁) is (-2,2) and (x₂,y₂) is (3,-1), then:

$\begin{array}{ll}
{PQ}&{}={}
&{\sqrt{\left(3 - (-2) \right)^2~+~\left(-1 - 2 \right)^2}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\sqrt{\left(5 \right)^2~+~\left(-3 \right)^2}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\sqrt{25~+~9}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\sqrt{34}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{5.83~\text{units}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
\end{array}$

◼ We have seen this topic in our earlier classes (Details here) 

C. Division of a line in the ratio m:n
This can be written in 4 steps:
1. In fig.10.3 below, PQ is a line with end points P (x₁,y₁) and Q (x₂,y₂)

Fig.10.3

2. Point A divides the line in the ratio m:n
That is:
$\text{Distance PA = Distance PQ}~ × ~\frac{m}{m+n}$

$\text{Distance QA = Distance PQ}~ × ~\frac{n}{m+n}$
3. Then the coordinates of A are:
$\frac{m\, x_2 ~+~n\, x_1}{m+n},~\frac{m\, y_2 ~+~n\, y_1}{m+n}$       
◼ Note:
• For this formula to be valid,
    ♦ The "segment represented by the first component m", must be adjacent to the first point (x₁, y₁).
4. For example, let:
    ♦ P (x₁,y₁) be P (-2,2)
    ♦ Q (x₂,y₂) be Q (3,-1)
    ♦ Point A divides PQ in the ratio 3:10
• Then:
$\begin{array}{ll}
{\text{Coordinates of A}}&{}={}
&{\left(\frac{3 × 3~+~10 × -2}{3+10}\right.~,}& {\left. \frac{3 × -1~+~10 × 2}{3+10} \right)}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\left(\frac{9~-~20}{13}\right.~,}& {\left. \frac{-3~+~20}{13} \right)}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\left(\frac{-11}{13}\right.~,}& {\left. \frac{17}{13} \right)}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\left(-0.85 \right.~,}& {\left. 1.3 \right)}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

• In this example, the ratio m:n is 3:10. The first component of the ratio is 3. The "segment represented by the first component 3", is adjacent to the first point P (-2,2).

◼ We have seen this topic in our earlier classes (Details here)

D. Division of a line in the ratio m:m
This can be written in 3 steps:
1. In this case, the point A divides the line PQ into two equal segments (remember that, m:m is same as 1:1).
• That means, A is the midpoint of PQ
2. We can write:
$\begin{array}{ll}
{\text{Coordinates of A}}&{}={}
&{\left(\frac{m\, x_2 ~+~n\, x_1}{m+n}\right.~,}& {\left. \frac{m\, y_2 ~+~n\, y_1}{m+n} \right)}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\left(\frac{m\, x_2 ~+~m\, x_1}{m+m}\right.~,}& {\left. \frac{m\, y_2 ~+~m\, y_1}{m+m} \right)}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\left(\frac{m(x_2 ~+~x_1)}{2m}\right.~,}& {\left. \frac{m(y_2 ~+~y_1}{2m} \right)}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\left(\frac{x_1 ~+~x_2}{2}\right.~,}& {\left. \frac{y_1 ~+~y_2}{2} \right)}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
\end{array}$

3. This result can be used as a formula:
Coordinates of the midpoint of the line PQ with P (x₁,y₁) and Q (x₂,y₂) are:
$\left(\frac{x_1~+~x_2}{2},\frac{y_1~+~y_2}{2} \right)$

E. Area of a Triangle
This can be written in 4 steps:
1. In fig.10.4 below, PQR is a triangle with vertices P (x₁,y₁), Q (x₂,y₂) and R (x₃,y₃)

Fig.10.4

2. Area of the triangle PQR will be equal to:
$\frac{1}{2}\left|x_1(y_2 - y_3)~+~x_2(y_3 - y_1)~+~x_3(y_1 - y_2) \right|$
• The calculations may give a -ve result. But we can ignore the -ve sign. This is indicated by the || lines.
3. Let us see an example:
If the vertices of a triangle are P (-1,5,2), Q (-2.5,-2) and R (4,1.5), then the area of that triangle will be:

$\begin{array}{ll}
{\text{Area}}&{}={}
&{\frac{1}{2}~ × ~\left |-1.5(-2 - 1.5)~+~-2.5(1.5 - 2)~+~4(2 - -2)\right |}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{1}{2}~ × ~\left |-1.5(- 3.5)~+~-2.5(-0.5)~+~4(4)\right |}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{1}{2}~ × ~\left |5.25~+~1.25~+~16\right |}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{1}{2}~ × ~\left |22.5\right |}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{11.25~\text{square units}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
\end{array}$

4. Suppose that we are given three points P, Q and R.
• Using their coordinates, we find the area formed by the three points.
• If that area works out to zero, we can confirm that, P, Q and R lie on a line.
    ♦ In other words, P, Q and R are collinear.

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• We can draw geometric figures like triangles, rectangles, parallelograms etc., on the XY-plane. All we need is the "coordinates of the vertices" of those geometric figures.
• So it is clear that, coordinate geometry is a combination of coordinates and geometry.
• A line can be drawn between any two points on the XY-plane.
• A line can be defined by the coordinates of any two points on that line.
• By combining different lines, we can draw various geometric figures. So line is a basic concept in coordinate geometry.
• In the next section, we will see more details about lines.

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