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:
$\small{\left| \vec{P_1 P_2}\right| = \sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left(z_2 - z_1 \right)^2}}$
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:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{PQ}} & {~=~} &{(-1-2)\hat{i}+(-2-3)\hat{j}+(-4-0)\hat{k}}
\\ {~\color{magenta} 2 } &{{\Rightarrow}} &{\vec{PQ}} & {~=~} &{-3\hat{i}-5\hat{j}-4\hat{k}}
\\ \end{array}}$
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 26.29
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:
$\small{\left|\vec{BC} \right|^2 = \left|\vec{AB} \right|^2 + \left|\vec{CA} \right|^2}$
♦ BC is the hypotenuse
♦ AB 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
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:
$\small{\begin{array}{ll}
{~\color{magenta} 1 } &{{}} &{\left|\vec{BA}
\right|} & {~=~} &{n\left|\vec{CA} \right|}
\\ {~\color{magenta} 2 } &{\Rightarrow} &{2\sqrt{14}} & {~=~} &{n\left(5\sqrt{14} \right)}
\\ {~\color{magenta} 3 } &{\Rightarrow} &{n} & {~=~} &{\frac{2}{5}}
\\ \end{array}}$
• This is same as 3:2
In the next section, we will see scalar product.
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