In 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:
$\small{\left| \vec{P_1 P_2}\right| = \sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left(z_2 - z_1 \right)^2}}$
Now we will see a solved example.
Solved example 26.24
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:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{PQ}} & {~=~} &{(-1-2)\hat{i}+(-2-3)\hat{j}+(-4-0)\hat{k}}
\\ {~\color{magenta} 2 } &{{\Rightarrow}} &{\vec{PQ}} & {~=~} &{-3\hat{i}-5\hat{j}-4\hat{k}}
\\ \end{array}}$
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:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{n\,\vec{PR}} & {~=~} &{-m\,\vec{RQ}}
\\ {~\color{magenta} 2 } &{{\Rightarrow}} &{n\left(\vec{OR} - \vec{OP} \right)} & {~=~} &{-m\left(\vec{OQ} - \vec{OR} \right)}
\\ {~\color{magenta} 3 } &{{\Rightarrow}} &{n\,\vec{OR} - n\,\vec{OP}} & {~=~} &{-m\,\vec{OQ} + m\,\vec{OR}}
\\ {~\color{magenta} 4 } &{{\Rightarrow}} &{(m-n)\vec{OR}} & {~=~} &{m\,\vec{OQ}-n\,\vec{OP}}
\\ {~\color{magenta} 5 } &{{\Rightarrow}} &{\vec{OR}} & {~=~} &{\frac{m\,\vec{OQ}~-~n\,\vec{OP}}{m-n}}
\\ \end{array}}$
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:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{OR}} & {~=~} &{\frac{(1)\,\vec{OQ}~+~(1)\,\vec{OP}}{1+1}}
\\ {~\color{magenta} 2 } &{{\Rightarrow}} &{\vec{OR}} & {~=~} &{\frac{\vec{OQ}~+~\vec{OP}}{2}}
\\ \end{array}}$
Now we will see some solved examples.
Solved example 26.25
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 26.26
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.
The link below gives a few more solved examples:
Exercise 26.2
In the next section, we will see scalar product.
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