26.8 - Cauchy-Schwartz Inequality

In the previous section, we completed a discussion on the scalar product of two vectors. We saw some solved examples also. In this section, we will see a few more solved examples. Later in this section, we will see the Cauchy-Schwartz inequality.

Solved example 26.43
Evaluate the product $\small{\left(3\vec{a}-5\vec{b} \right).\left(2\vec{a}+7\vec{b} \right)}$.
Solution:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(3\vec{a}-5\vec{b} \right).\left(2\vec{a}+7\vec{b} \right)}    & {~=~}    &{3\vec{a}\left(2\vec{a}+7\vec{b} \right)-5\vec{b}\left(2\vec{a}+7\vec{b} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{6\left|\vec{a} \right|^2+21\vec{a}.\vec{b}-10\vec{b}.\vec{a}-35\left|\vec{b} \right|^2}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{6\left|\vec{a} \right|^2+11\vec{a}.\vec{b}-35\left|\vec{b} \right|^2}
\\ \end{array}}$

Solved example 26.44
Find the magnitude of two vectors $\small{\vec{a}~\text{and}~\vec{b}}$, having the same magnitude and such that the angle between them is $\small{60^o}$ and their scalar product is $\small{\frac{1}{2}}$.
Solution:
1. Based on the given information, we can write:
   ♦ $\small{\left|\vec{a} \right|~=~\left|\vec{b} \right|}$
   ♦ $\small{\theta~=~60^o~=~\frac{\pi}{3}}$
   ♦ $\small{\vec{a}.\vec{b}~=~\frac{1}{2}}$

2. So we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}.\vec{b}}    & {~=~}    &{\frac{1}{2}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left|\vec{a} \right|\left|\vec{b} \right|\cos\theta}    & {~=~}    &{\frac{1}{2}}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\left|\vec{a} \right|^2 \cos\left(\frac{\pi}{3} \right)}    & {~=~}    &{\frac{1}{2}}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{\left|\vec{a} \right|^2 \left(\frac{1}{2} \right)}    & {~=~}    &{\frac{1}{2}}
\\ {~\color{magenta}    5    }    &{\Rightarrow}    &{\left|\vec{a} \right|^2}    & {~=~}    &{1}
\\ {~\color{magenta}    6    }    &{\Rightarrow}    &{\left|\vec{a} \right|}    & {~=~}    &{1~=~\,\left|\vec{b} \right|}
\\ \end{array}}$

◼ Remarks:
• 6 (magenta color): Here we discard the −ve root because, length cannot be −ve.

Solved example 26.45
Find $\small{\left|\vec{x} \right|}$ if for a unit vector $\small{\vec{a}}$, $\small{\left(\vec{x}-\vec{a} \right).\left(\vec{x}+\vec{a} \right)}$ = 12.
Solution:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{x}-\vec{a} \right).\left(\vec{x}+\vec{a} \right)}    & {~=~}    &{\vec{x}\left(\vec{x}+\vec{a} \right)-\vec{a}\left(\vec{x}+\vec{a} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left|\vec{x} \right|^2+\vec{x}.\vec{a}-\vec{a}.\vec{x}-\left|\vec{a} \right|^2}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left|\vec{x} \right|^2-\left|\vec{a} \right|^2}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\left|\vec{x} \right|^2-(1)^2}
\\ {~\color{magenta}    5    }    &{\Rightarrow}    &{12}    & {~=~}    &{\left|\vec{x} \right|^2-1}
\\ {~\color{magenta}    6    }    &{\Rightarrow}    &{\left|\vec{x} \right|^2}    & {~=~}    &{13}
\\ {~\color{magenta}    7    }    &{\Rightarrow}    &{\left|\vec{x} \right|}    & {~=~}    &{\sqrt{13}}
\\ \end{array}}$

Solved example 26.46
If $\small{\vec{a}=2\hat{i}+2\hat{j}+3\hat{k}}$, $\small{\vec{b}=-\hat{i}+2\hat{j}+\hat{k}}$ and $\small{\vec{c}=3\hat{i}+\hat{j}}$ are such that $\small{\left(\vec{a}+\lambda \vec{b} \right)}$ isperpendicular to $\small{\vec{c}}$, then find the value of $\small{\lambda}$.
Solution:
1.First we write the sum:
$\small{\left(\vec{a}+\lambda \vec{b} \right)}$
= $\small{\vec{b}=\left(2-\lambda \right)\hat{i}+\left(2+2\lambda \right)\hat{j}+\left(3+\lambda \right)\hat{k}}$ 

2. The resultant vector obtained above is perpendicular to $\small{\vec{c}}$. So their dot product will be zero. We can write:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left[\left(2-\lambda \right)\hat{i}+\left(2+2\lambda \right)\hat{j}+\left(3+\lambda \right)\hat{k} \right].\left[\vec{c} \right]}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left[\left(2-\lambda \right)\hat{i}+\left(2+2\lambda \right)\hat{j}+\left(3+\lambda \right)\hat{k} \right].\left[3\hat{i}+\hat{j} \right]}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\left(2-\lambda \right)(3)+\left(2+2\lambda \right)(1)+\left(3+\lambda \right)(0)}    & {~=~}    &{0}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{6-3\lambda+2+2\lambda +0}    & {~=~}    &{0}
\\ {~\color{magenta}    5    }    &{\Rightarrow}    &{8-\lambda}    & {~=~}    &{0}
\\ {~\color{magenta}    6    }    &{\Rightarrow}    &{\lambda}    & {~=~}    &{8}
\\ \end{array}}$

Solved example 26.47
Show that $\small{\left|\vec{a} \right|\vec{b}+\left|\vec{b} \right|\vec{a}}$, is perpendicular to $\small{\left|\vec{a} \right|\vec{b}-\left|\vec{b} \right|\vec{a}}$, for any two nonzero vectors $\small{\vec{a}~\text{and}~\vec{b}}$.
Solution:
1. First we write the scalar product:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\left|\vec{a} \right|\vec{b}+\left|\vec{b} \right|\vec{a} \right).\left(\left|\vec{a} \right|\vec{b}-\left|\vec{b} \right|\vec{a} \right)}    & {~=~}    &{\left(\left|\vec{a} \right|\vec{b}+\left|\vec{b} \right|\vec{a} \right).\left|\vec{a} \right|\vec{b}~-~\left(\left|\vec{a} \right|\vec{b}+\left|\vec{b} \right|\vec{a} \right).\left|\vec{b} \right|\vec{a}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left|\vec{a} \right|^2\,\left|\vec{b} \right|^2+\left(\left|\vec{a} \right|\left|\vec{b} \right| \right)\vec{a}.\vec{b}-\left(\left|\vec{a} \right|\left|\vec{b} \right| \right)\vec{b}.\vec{a}-\left|\vec{b} \right|^2\,\left|\vec{a} \right|^2}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left|\vec{a} \right|^2\,\left|\vec{b} \right|^2+\left(\left|\vec{a} \right|\left|\vec{b} \right| \right)\vec{a}.\vec{b}-\left(\left|\vec{a} \right|\left|\vec{b} \right| \right)\vec{a}.\vec{b}-\left|\vec{b} \right|^2\,\left|\vec{a} \right|^2}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{0}
\\ \end{array}}$

2. Since the scalar product is zero, we can write:
The vectors are perpendicular to each other for any two nonzero vectors $\small{\vec{a}~\text{and}~\vec{b}}$ 

Solved example 26.48
If $\small{\vec{a}.\vec{a}=0}$ and $\small{\vec{a}.\vec{b}=0}$, then what can be concluded about $\small{\vec{b}}$?
Solution:
1. We know that:
• If $\small{\vec{p}.\vec{q}=0}$, then either
   ♦ $\small{\vec{p}}$ is a zero vector
   ♦ or $\small{\vec{q}}$ is a zero vector.
2. Given first information is that:  $\small{\vec{a}.\vec{a}=0}$
• It is clear that, $\small{\vec{a}}$ is a zero vector.
3. Now we consider the second information:
$\small{\vec{a}.\vec{b}=0}$
• We found out that, $\small{\vec{a}}$ is zero vector. So $\small{\vec{b}}$ can be a zero vector or any nonzero vector.

Solved example 26.49
If $\small{\vec{a},~\vec{b},~\vec{c}}$ are unit vectors such that $\small{\vec{a}+\vec{b}+\vec{c}=\vec{0}}$, find the value of $\small{\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}}$
Solution:
1. Given that: $\small{\vec{a}+\vec{b}+\vec{c}=\vec{0}}$
Multiplying both sides by $\small{\vec{a}}$, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}.\left[\vec{a}+\vec{b}+\vec{c} \right]}    & {~=~}    &{\vec{a}.\vec{0}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\vec{a}.\vec{a}+\vec{a}.\vec{b}+\vec{a}.\vec{c}}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{1+\vec{a}.\vec{b}+\vec{a}.\vec{c}}    & {~=~}    &{0}
\\ \end{array}}$

2. Multiplying both sides by $\small{\vec{b}}$, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{b}.\left[\vec{a}+\vec{b}+\vec{c} \right]}    & {~=~}    &{\vec{b}.\vec{0}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\vec{b}.\vec{a}+\vec{b}.\vec{b}+\vec{b}.\vec{c}}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\vec{b}.\vec{a}+1+\vec{b}.\vec{c}}    & {~=~}    &{0}
\\ \end{array}}$

3. Multiplying both sides by $\small{\vec{c}}$, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{c}.\left[\vec{a}+\vec{b}+\vec{c} \right]}    & {~=~}    &{\vec{c}.\vec{0}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\vec{c}.\vec{a}+\vec{c}.\vec{b}+\vec{c}.\vec{c}}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\vec{c}.\vec{a}+\vec{c}.\vec{b}+1}    & {~=~}    &{0}
\\ \end{array}}$

4. We have three results:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{1+\vec{a}.\vec{b}+\vec{a}.\vec{c}}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{}    &{\vec{b}.\vec{a}+1+\vec{b}.\vec{c}}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{}    &{\vec{c}.\vec{a}+\vec{c}.\vec{b}+1}    & {~=~}    &{0}
\\ \end{array}}$

5. Adding the L.H.S together, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{3 + 2\vec{a}.\vec{b} + 2\vec{b}.\vec{c} + 2\vec{c}.\vec{a}}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{3 + 2\left(\vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} \right)}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{ 2\left(\vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} \right)}    & {~=~}    &{-3}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{ \vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} }    & {~=~}    &{\frac{-3}{2}}
\\ \end{array}}$

---

Cauchy-Schwartz Inequality 

This can be explained in 5 steps:
1. Let $\small{\vec{a}~\text{and}~\vec{b}}$ be two nonzero vectors. And let $\small{\theta}$ be the angle between them.

2. We can write the dot product and make some rearrangements as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}.\vec{b}}    & {~=~}    &{\left|\vec{a} \right|\left|\vec{b} \right|\cos\theta}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\frac{\vec{a}.\vec{b}}{\left|\vec{a} \right|\left|\vec{b} \right|}}    & {~=~}    &{\cos\theta}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\frac{\left|\left(\vec{a}.\vec{b} \right) \right|}{\left|\vec{a} \right|\left|\vec{b} \right|}}    & {~=~}    &{\left|\cos\theta \right|}
\\ \end{array}}$

◼ Remarks:
• 3 (magenta color): The denominator of the L.H.S is the product of two lengths. So the product will be +ve. We do not need to take the absolute value of that product

3. In the above result, the absolute value of $\small{\cos \theta}$ will be always less than or equal to 1. So we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{\left|\left(\vec{a}.\vec{b} \right) \right|}{\left|\vec{a} \right|\left|\vec{b} \right|}}    & {~\le~}    &{1}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left|\left(\vec{a}.\vec{b} \right) \right|}    & {~\le~}    &{\left|\vec{a} \right|\left|\vec{b} \right|}
\\ \end{array}}$

4. So we can write:
   ♦ Absolute value of the dot product
   ♦ will be less than or equal to
   ♦ the product of the individual absolute values

• This is known as the Cauchy-Schwartz inequality

5. Note that, we used nonzero vectors to prove the triangle inequality. What if either $\small{\vec{a}~\text{or}~\vec{b}}$ is a zero vector?

The answer can be written in 2 steps:
(i) We have the inequality: $\small{\left|\vec{a}.\vec{b} \right| ~\le~\left|\vec{a} \right|\,\left|\vec{b} \right|}$
(ii) Suppose that, $\small{\vec{b}=\vec{0}}$. Then we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\left|\vec{a}.\vec{b} \right|}    & {~\le~}    &{\left|\vec{a} \right|\,\left|\vec{b} \right|}
\\ {~\color{magenta}    2    }    &{\Rightarrow}&{\left|\vec{a}.\vec{0} \right|}    & {~\le~}    &{\left|\vec{a} \right|\,\left|\vec{0} \right|}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{0}    & {~\le~}    &{\left|\vec{a} \right|(0)}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{0}    & {~\le~}    &{0}
\\ \end{array}}$

This is true.

---

Triangle Inequality 

This can be explained in 12 steps:
1. Let $\small{\vec{a}~\text{and}~\vec{b}}$ be two nonzero vectors. And let their sum be $\small{\vec{c}}$.

2. We have:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{c}.\vec{c}}    & {~=~}    &{\left|\vec{c} \right|^2}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left(\vec{a}+\vec{b} \right).\left(\vec{a}+\vec{b} \right)}    & {~=~}    &{\left|\left(\vec{a}+\vec{b} \right) \right|^2}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\left|\left(\vec{a}+\vec{b} \right) \right|^2}    & {~=~}    &{\left(\vec{a}+\vec{b} \right).\left(\vec{a}+\vec{b} \right)}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\vec{a}.\left(\vec{a}+\vec{b} \right)+\vec{b}.\left(\vec{a}+\vec{b} \right)}
\\ {~\color{magenta}    5    }    &{}    &{}    & {~=~}    &{\left|\vec{a} \right|^2+\vec{a}.\vec{b}+\vec{b}.\vec{a}+\left|\vec{b} \right|^2}
\\ {~\color{magenta}    6    }    &{\Rightarrow}    &{\left|\left(\vec{a}+\vec{b} \right) \right|^2}    & {~=~}    &{\left|\vec{a} \right|^2+2\vec{a}.\vec{b}+\left|\vec{b} \right|^2}
\\ \end{array}}$

3. Consider the middle term in the R.H.S of the above result.
• This term is $\small{2\vec{a}.\vec{b}}$. We know that, the dot product is a real number.
• Any real number will be less than or equal to it's absolute value. So we can write:
$\small{2\vec{a}.\vec{b}~\le~2\left|\vec{a}.\vec{b} \right|}$

4. Now compare $\small{\left[\left|\vec{a} \right|^2+2\vec{a}.\vec{b}+\left|\vec{b} \right|^2 \right]}$ and $\small{\left[\left|\vec{a} \right|^2+2\left|\vec{a}.\vec{b} \right|+\left|\vec{b} \right|^2 \right]}$
• Obviously, the former is less than or equal to the latter.

5. So we can proceed as follows:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{a} \right|^2+2\vec{a}.\vec{b}+\left|\vec{b} \right|^2}    & {~\le~}    &{\left|\vec{a} \right|^2+2\left|\vec{a}.\vec{b} \right|+\left|\vec{b} \right|^2}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left|\left(\vec{a}+\vec{b} \right) \right|^2}    & {~\le~}    &{\left|\vec{a} \right|^2+2\left|\vec{a}.\vec{b} \right|+\left|\vec{b} \right|^2}
\\ \end{array}}$

◼ Remarks:
• 2 (magenta color): Here we replace the L.H.S based on the result from (2)

6. Consider the middle term in the R.H.S of the above result.
• This term is $\small{2\left|\vec{a}.\vec{b} \right|}$.
• Based on Cauchy-Schwartz inequality, this term is less than or equal to $\small{2\left|\vec{a} \right|\left|\vec{b} \right|}$

7. Now compare $\small{\left[\left|\vec{a} \right|^2+2\left|\vec{a}.\vec{b} \right|+\left|\vec{b} \right|^2 \right]}$ and $\small{\left[\left|\vec{a} \right|^2+2\left|\vec{a} \right|\left|\vec{b} \right|+\left|\vec{b} \right|^2 \right]}$
• Obviously, the former is less than or equal to the latter.

8. So we can proceed as follows:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\left(\vec{a}+\vec{b} \right) \right|^2}    & {~\le~}    &{\left|\vec{a} \right|^2+2\left|\vec{a} \right|\left|\vec{b} \right|+\left|\vec{b} \right|^2}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left|\left(\vec{a}+\vec{b} \right) \right|^2}    & {~\le~}    &{\left(\left|\vec{a} \right|+\left|\vec{b} \right| \right)^2}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\left|\left(\vec{a}+\vec{b} \right) \right|}    & {~\le~}    &{\left|\vec{a} \right|+\left|\vec{b} \right| }
\\ \end{array}}$

◼ Remarks:
• 2 (magenta color): Here we apply the identity:
(A+B)² = A² + 2AB + B².
This identity is applicable because, every term in "magenta 1" is a real number.

9. So we can write:
   ♦ Magnitude of the "sum of two vectors"
   ♦ will be less than or equal to
   ♦ the sum of the individual magnitudes

• This is known as the triangle inequality

10. This inequality can be diagrammatically shown as in fig.26.28 below:

Fig.26.28

$\small{\vec{AC}}$ is the sum of $\small{\vec{AB}~\text{and}~\vec{BC}}$. Obviously, length of AC will be smaller than the sum of the lengths of AB and BC. We saw this in our earlier classes. Details here.

11. Consider the case when $\small{\left|\vec{a}+\vec{b} \right|=\left|\vec{a} \right|+\left|\vec{b} \right|}$
• In such a situation, $\small{\left|\vec{AC} \right|=\left|\vec{AB} \right|+\left|\vec{BC} \right|}$
• If the above equality is satisfied, the points A, B and C will be collinear.

12. Note that, we used nonzero vectors to prove the triangle inequality. What if either $\small{\vec{a}~\text{or}~\vec{b}}$ is a zero vector?

The answer can be written in 2 steps:
(i) We have the inequality: $\small{\left|\vec{a}+\vec{b} \right| ~\le~\left|\vec{a} \right|+\left|\vec{b} \right| }$
(ii) Suppose that, $\small{\vec{b}=\vec{0}}$. Then we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\left|\left(\vec{a}+\vec{b} \right) \right|}    & {~\le~}    &{\left|\vec{a} \right|+\left|\vec{b} \right| }
\\ {~\color{magenta}    2    }    &{\Rightarrow}&{\left|\left(\vec{a}+\vec{0} \right) \right|}    & {~\le~}    &{\left|\vec{a} \right|+\left|\vec{0} \right| }
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\left|\vec{a} \right|}    & {~\le~}    &{\left|\vec{a} \right|+0 }
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{\left|\vec{a} \right|}    & {~\le~}    &{\left|\vec{a} \right| }
\\ \end{array}}$

This is true.

---

Now we will see a solved example

Solved example 26.50
Show that the points
$\small{A\left(-2\hat{i}+3\hat{j}+5\hat{k} \right)}$
$\small{B\left(\hat{i}+2\hat{j}+3\hat{k} \right)}$
$\small{C\left(7\hat{i}-\hat{k} \right)}$
are collinear
Solution:
1. We are given the position vectors $\small{\vec{OA},~\vec{OB}~\text{and}~\vec{OC}}$
• The three points form the vertices of a triangle ABC

2. Length of the side AB =
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\left|\vec{AB} \right|}    & {=}    &{\left|\vec{OB} ~-~\vec{OA}\right|}
\\ {~\color{magenta}    2    }    &{}&{}    & {=}    &{\left|3\hat{i}-\hat{j}-2\hat{k} \right|}
\\ {~\color{magenta}    3    }    &{}&{}    & {=}    &{\sqrt{14}}
\\ \end{array}}$

3. Length of the side BC =
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\left|\vec{BC} \right|}    & {=}    &{\left|\vec{OC} ~-~\vec{OB}\right|}
\\ {~\color{magenta}    2    }    &{}&{}    & {=}    &{\left|6\hat{i}-2\hat{j}-4\hat{k} \right|}
\\ {~\color{magenta}    3    }    &{}&{}    & {=}    &{\sqrt{56}~=~2\sqrt{14}}
\\ \end{array}}$

4. Length of the side AC =
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\left|\vec{AC} \right|}    & {=}    &{\left|\vec{OC} ~-~\vec{OA}\right|}
\\ {~\color{magenta}    2    }    &{}&{}    & {=}    &{\left|9\hat{i}-3\hat{j}-6\hat{k} \right|}
\\ {~\color{magenta}    3    }    &{}&{}    & {=}    &{\sqrt{126}~=~3\sqrt{14}}
\\ \end{array}}$

5. Now we analyze the above lengths. We see that:
$\small{\sqrt{14}~+~2\sqrt{14}~=~3\sqrt{14}}$
• That means, $\small{\left|\vec{AB} \right|+\left|\vec{BC} \right|=\left|\vec{AC} \right|}$
• So by triangle inequality, the points A, B and C are collinear

The above solved example helps us to obtain an interesting result. It can be explained in 4 steps:

1. First we write the sum:
$\small{\vec{AB}+\vec{BC}+\vec{CA}}$
= $\small{\left(3\hat{i}-\hat{j}-2\hat{k} \right)+\left(6\hat{i}-2\hat{j}-4\hat{k} \right)+(-1)\left(9\hat{i}-3\hat{j}-6\hat{k} \right)}$
= $\small{0\hat{i}+0\hat{j}+0\hat{k}}$

2. So we can write:
$\small{\vec{AB}+\vec{BC}+\vec{CA}~=~\vec{0}}$

3. Earlier, we saw that:
When sides of a triangle are taken in order, the resultant will be a null vector. See section 26.2

4. In our present case, resultant is a null vector. But the points do not form a triangle.

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In the next section, we will see projection of a vector on a line.

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26.7 - Scalar Product of Two Vectors

In the previous section, we saw vector joining two points. We also saw section formula. In this section, we will see scalar product of two vectors.

We know that, product of two numbers is a number. Similarly, product of two matrices is a matrix. But in the case of vectors, the product depends on the type of multiplication.
   ♦ Scalar multiplication of two vectors gives a scalar
   ♦ Vector multiplication of two vectors gives a vector
In this section, we will see scalar multiplication. It can be explained in 12 steps:

1. The scalar product of two nonzero vectors $\small{\vec{a}~\text{and}~\vec{b}}$ is defined as the product of three items:
(i) The magnitude of $\small{\vec{a}}$, which is: $\small{\left|\vec{a}\right|}$
(ii) The magnitude of $\small{\vec{b}}$, which is: $\small{\left|\vec{b}\right|}$
(iii) $\small{\cos \theta}$
   ♦ Where $\small{\theta}$ is the angle between the two vectors.
• The angle $\small{\theta}$ must be selected carefully. For any two vectors, there will be two angles between them.
   ♦ One which is less than or equal to $\small{\pi}$ radians.
   ♦ The other,which is greater than or equal to $\small{\pi}$ radians.
• This is shown in fig.26.27 below. We must select only the one which is less than or equal to $\small{\pi}$ radians.

Fig.26.27

2. The scalar product of two vectors $\small{\vec{a}~\text{and}~\vec{b}}$ is denoted as $\small{\vec{a}.\vec{b}}$.
• So we can write: $\small{\vec{a}.\vec{b}~=~\left|\vec{a}\right|\,\left|\vec{b}\right|\,\cos \theta}$
3. If either $\small{\vec{a}=\vec{0}}$ or $\small{\vec{b}=\vec{0}}$, then $\small{\theta}$ is not defined. In such a situation, we define $\small{\vec{a}.\vec{b}~=~\vec{0}}$
4. In (1), we saw the three items which are multiplied to obtain the scalar product. Each of those three items, is a scalar. None of them has direction. So the scalar product is a scalar. In other words, scalar product is a real number. Recall that, cosine of any angle, is a real number. It does not have any unit.
5. Suppose that, $\small{\vec{a}~\text{and}~\vec{b}}$ are perpendicular to each other. Then $\small{\theta = \frac{\pi}{2}}$. Consequently, $\small{\cos \theta = \cos\left(\frac{\pi}{2} \right)=0}$. In such a situation, the scalar product will be zero. In fact, the scalar product is zero if and only if the two vectors are perpendicular to each other. We can write:
$\small{\vec{a}.\vec{b}~=~0\iff\vec{a}\perp\vec{b}}$
6. Suppose that, $\small{\theta = 0}$. Then $\small{\cos \theta = \cos\left(0 \right)=1}$. In such a situation, we get:
$\small{\vec{a}.\vec{b}~=~\left|\vec{a}\right|\,\left|\vec{b}\right|(1)~=~\left|\vec{a}\right|\,\left|\vec{b}\right|}$
• Based on this, we can write: $\small{\vec{a}.\vec{a}~=~\left|\vec{a}\right|^2}$.
This is because, the angle between $\small{\vec{a}}$ and itself is zero.
7. Suppose that, $\small{\theta = \pi}$. Then $\small{\cos \theta = \cos\left(\pi \right)=-1}$. In such a situation, we get:
$\small{\vec{a}.\vec{b}~=~\left|\vec{a}\right|\,\left|\vec{b}\right|(-1)~=~-\left|\vec{a}\right|\,\left|\vec{b}\right|}$
• Based on this, we can write: $\small{\vec{a}.\left(-\vec{a} \right)~=~(-1)\left|\vec{a}\right|^2}$.
This is because, the angle between $\small{\vec{a}~\text{and}~-\vec{a}}$ is $\small{\pi}$.
8. Based on (5), we get three interesting results:
   ♦ $\small{\hat{i}.\hat{j}~=~0}$
   ♦ $\small{\hat{j}.\hat{k}~=~0}$
   ♦ $\small{\hat{k}.\hat{i}~=~0}$
9. Based on (6), we get three interesting results:
   ♦ $\small{\hat{i}.\hat{i}~=~\left|\hat{i}\right|^2~=~1^2~=~1}$
   ♦ $\small{\hat{j}.\hat{j}~=~\left|\hat{j}\right|^2~=~1^2~=~1}$
   ♦ $\small{\hat{k}.\hat{k}~=~\left|\hat{k}\right|^2~=~1^2~=~1}$
10. Scalar product can be used to find the angle between two vectors. The method is shown below:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}.\vec{b}}    & {~=~}    &{\left|\vec{a}\right|\,\left|\vec{b}\right|\,\cos \theta}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\cos \theta}    & {~=~}    &{\frac{\vec{a}.\vec{b}}{\left|\vec{a}\right|\,\left|\vec{b}\right|}}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\theta}    & {~=~}    &{\cos^{-1}\left(\frac{\vec{a}.\vec{b}}{\left|\vec{a}\right|\,\left|\vec{b}\right|} \right)}
\\ \end{array}}$
11. Based on (1), we can write:
scalar product involves the multiplication of three real numbers.
• Since they are real numbers, they can be multiplied in any order. So we can write:
$\small{\vec{a}.\vec{b}~=~\vec{b}.\vec{a}}$
• In other words, scalar product is commutative
12. Scalar product of two vectors is also known as dot product of two vectors.

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Two important properties of scalar product

Property I: Distributivity of scalar product over addition
This can be explained as below:
Let $\small{\vec{a},~\vec{b},~\vec{c}}$ be any three vectors. Then we can write:
$\small{\vec{a}\left(\vec{b}+\vec{c} \right)~=~\vec{a}.\vec{b}+\vec{a}.\vec{c}}$

Property II: Distributivity of scalar product over multiplication
This can be explained as below:
Let $\small{\vec{a}~\text{and}~\vec{b}}$ be any two vectors and $\small{\lambda}$ be any scalar. Then we can write:
$\small{\lambda \left(\vec{a}.\vec{b} \right) ~=~\left(\lambda \vec{a}\right).\vec{b} ~=~\vec{a}.\left(\lambda\vec{b} \right)}$

---

Scalar product when vectors are given in component form

Let the two vectors be:
$\small{\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}}$
$\small{\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}}$
Then we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}.\vec{b}}    & {~=~}    &{\left(a_1\hat{i}+a_2\hat{j}+a_3\hat{k} \right).\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{a_1\hat{i}\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \right)}
\\ {~\color{magenta}    {}   }    &{}    &{}    & {}    &{+~a_2\hat{j}\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \right)}
\\ {~\color{magenta}    {}   }    &{}    &{}    & {}    &{+~a_3\hat{k}\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \right)}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{a_1 b_1\left(\hat{i}.\hat{i}\right)+a_1 b_2\left(\hat{i}.\hat{j}\right)+a_1 b_3\left(\hat{i}.\hat{k}\right)}
\\ {~\color{magenta}    {}   }    &{}    &{}    & {}    &{+~a_2 b_1\left(\hat{j}.\hat{i}\right)+a_2 b_2\left(\hat{j}.\hat{j}\right)+a_2 b_3\left(\hat{j}.\hat{k}\right)}
\\ {~\color{magenta}    {}   }    &{}    &{}    & {}    &{+~a_3 b_1\left(\hat{k}.\hat{i}\right)+a_3 b_2\left(\hat{k}.\hat{j}\right)+a_3 b_3\left(\hat{k}.\hat{k}\right)}
\\ {~\color{magenta}    4   }    &{}    &{}    & {~=~}    &{a_1 b_1 + a_2 b_2 + a_3 b_3}
\\ \end{array}}$

---

Now we will see some solved examples.

Solved example 26.35
Find the angle between two vectors $\small{\vec{a}~\text{and}~\vec{b}}$ with magnitudes 1 and 2 respectively and when $\small{\vec{a}.\vec{b}}$ = 1.
Solution:
• We have:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\theta}    & {~=~}    &{\cos^{-1}\left(\frac{\vec{a}.\vec{b}}{\left|\vec{a}\right|\,\left|\vec{b}\right|} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\cos^{-1}\left(\frac{1}{(1)(2)} \right)}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\cos^{-1}\left(\frac{1}{2} \right)}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\frac{\pi}{3}}
\\ \end{array}}$

• Note:
$\small{\theta=\cos^{-1}\left(\frac{1}{2} \right)}$ will give infinite number of solutions. But we want that angle which lies in the interval $\small{\left(0,\pi \right)}$ 

Solved example 26.36
Find the angle between two vectors $\small{\vec{a}~\text{and}~\vec{b}}$ with magnitudes $\small{\sqrt{3}}$ and 2, respectively having $\small{\vec{a}.\vec{b}=\sqrt{6}}$.
Solution:
• We have:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\theta}    & {~=~}    &{\cos^{-1}\left(\frac{\vec{a}.\vec{b}}{\left|\vec{a}\right|\,\left|\vec{b}\right|} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\cos^{-1}\left(\frac{\sqrt{6}}{(\sqrt{3})(2)} \right)}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\cos^{-1}\left(\frac{(\sqrt{3})(\sqrt{2})}{(\sqrt{3})(2)} \right)~=~\cos^{-1}\left(\frac{1}{\sqrt{2}} \right)}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\frac{\pi}{4}}
\\ \end{array}}$
• Note:
$\small{\theta=\cos^{-1}\left(\frac{1}{\sqrt{2}} \right)}$ will give infinite number of solutions. But we want that angle which lies in the interval $\small{\left(0,\pi \right)}$ 

Solved example 26.37
Find the angle $\small{\theta}$ between the vectors $\small{\vec{a}=\hat{i}+\hat{j}-\hat{k}}$ and $\small{\vec{b}=\hat{i}-\hat{j}+\hat{k}}$.
Solution:
1. We have: $\small{\theta=\cos^{-1}\left(\frac{\vec{a}.\vec{b}}{\left|\vec{a}\right|\,\left|\vec{b}\right|} \right)}$
2. So first we need to find $\small{\vec{a}.\vec{b}}$
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}.\vec{b}}    & {~=~}    &{a_1 b_1 + a_2 b_2 + a_3 b_3}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{(1)(1)+(1)(-1)+(-1)(1)}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{1-1-1}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{-1}
\\ \end{array}}$
3. Next we need to find the magnitudes:
$\small{\left|\vec{a}\right|=\sqrt{1^2 + 1^2 + (-1)^2}=\sqrt{3}}$
$\small{\left|\vec{b}\right|=\sqrt{1^2 + (-1)^2 + 1^2}=\sqrt{3}}$
4. Substituting the above values in (1), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\theta}    & {~=~}    &{\cos^{-1}\left(\frac{\vec{a}.\vec{b}}{\left|\vec{a}\right|\,\left|\vec{b}\right|} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\cos^{-1}\left(\frac{-1}{(\sqrt{3})(\sqrt{3})} \right)}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\cos^{-1}\left(\frac{-1}{3} \right)}
\\ \end{array}}$

Solved example 26.38
Find the angle between the vectors $\small{\vec{a}=\hat{i}-2\hat{j}+3\hat{k}}$ and $\small{\vec{b}=3\hat{i}-2\hat{j}+\hat{k}}$.
Solution:
1. We have: $\small{\theta=\cos^{-1}\left(\frac{\vec{a}.\vec{b}}{\left|\vec{a}\right|\,\left|\vec{b}\right|} \right)}$

2. So first we need to find $\small{\vec{a}.\vec{b}}$

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}.\vec{b}}    & {~=~}    &{a_1 b_1 + a_2 b_2 + a_3 b_3}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{(1)(3)+(-2)(-2)+(3)(1)}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{3+4+3}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{10}
\\ \end{array}}$

3. Next we need to find the magnitudes:
$\small{\left|\vec{a}\right|=\sqrt{1^2 + (-2)^2 + 3^2}=\sqrt{14}}$

$\small{\left|\vec{b}\right|=\sqrt{3^2 + (-2)^2 + 1^2}=\sqrt{14}}$

4. Substituting the above values in (1), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\theta}    & {~=~}    &{\cos^{-1}\left(\frac{\vec{a}.\vec{b}}{\left|\vec{a}\right|\,\left|\vec{b}\right|} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\cos^{-1}\left(\frac{10}{(\sqrt{14})(\sqrt{14})} \right)}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\cos^{-1}\left(\frac{10}{14} \right)}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\cos^{-1}\left(\frac{5}{7} \right)}
\\ \end{array}}$

Solved example 26.39
If $\small{\vec{a}=5\hat{i}-\hat{j}-3\hat{k}}$ and $\small{\vec{b}=\hat{i}+3\hat{j}-5\hat{k}}$, then show that the vectors $\small{\left(\vec{a}+\vec{b} \right)~\text{and}~\left(\vec{a}-\vec{b} \right)}$ are perpendicular.
Solution:
1. First we write the sum and difference:
$\small{\vec{c}~=~\vec{a}+\vec{b}~=~6\hat{i}+2\hat{j}-8\hat{k}}$
$\small{\vec{d}~=~\vec{a}-\vec{b}~=~4\hat{i}-4\hat{j}+2\hat{k}}$

2. Next we find the angle $\small{\theta}$ between $\small{\vec{c}~\text{and}~\vec{d}}$

We have: $\small{\theta=\cos^{-1}\left(\frac{\vec{c}.\vec{d}}{\left|\vec{c}\right|\,\left|\vec{d}\right|} \right)}$

3. So we need to find $\small{\vec{c}.\vec{d}}$

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{c}.\vec{d}}    & {~=~}    &{c_1 d_1 + c_2 d_2 + c_3 d_3}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{(6)(4)+(2)(-4)+(-8)(2)}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{24-8-16}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{0}
\\ \end{array}}$

4. There is no need to find the denominator.
Since $\small{\vec{c}.\vec{d}=0}$, we can directly write:
$\small{\theta=\cos^{-1}(0)}$

5. Solving this equation, we get: $\small{\theta~=~\frac{\pi}{2}}$
• Therefore, $\small{\vec{c}~\text{and}~\vec{d}}$ are perpendicular.
• That is., $\small{\left(\vec{a}+\vec{b} \right)~\text{and}~\left(\vec{a}-\vec{b} \right)}$ are perpendicular.

Solved example 26.40
If $\small{\vec{a}}$ is a unit vector and $\small{\left(\vec{x}-\vec{a} \right).\left(\vec{x}+\vec{a} \right) = 8}$, then find $\small{\left|\vec{x}\right|}$.
Solution:
1. First we write the scalar product:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{x}-\vec{a} \right).\left(\vec{x}+\vec{a} \right)}    & {~=~}    &{\left(\vec{x}-\vec{a} \right).\vec{x}~+~\left(\vec{x}-\vec{a} \right).\vec{a}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\vec{x}.\vec{x}-\vec{a}.\vec{x}+\vec{x}.\vec{a}-\vec{a}.\vec{a}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left|\vec{x} \right|^2-\left|\vec{a} \right|\left|\vec{x} \right|\cos \theta + \left|\vec{x} \right|\left|\vec{a} \right|\cos \theta - \left|\vec{a} \right|^2}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\left|\vec{x} \right|^2 - \left|\vec{a} \right|^2}
\\ {~\color{magenta}    5    }    &{}    &{}    & {~=~}    &{\left|\vec{x} \right|^2 - 1}
\\ \end{array}}$

◼ Remarks:
• 3 (magenta color): Here we assume that, the angle between $\small{\vec{x}~\text{and}~\vec{a}}$ is $\small{\theta}$
• 5 (magenta color): Here we use the given information that, $\small{\vec{a}}$ is a unit vector

2. Given that, this product is equal to 8. So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{x}-\vec{a} \right).\left(\vec{x}+\vec{a} \right)}    & {~=~}    &{8}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left|\vec{x} \right|^2 - 1}    & {~=~}    &{8}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\left|\vec{x} \right|^2}    & {~=~}    &{9}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{\left|\vec{x} \right|}    & {~=~}    &{3}
\\ \end{array}}$

◼ Remarks:
• 4 (magenta color): Here we discard the −ve root because, length cannot be −ve.

Solved example 26.41
Find $\small{\left|\left(\vec{a}-\vec{b} \right) \right|}$ if two vectors $\small{\vec{a}~\text{and}~\vec{b}}$ are such that $\small{\left|\vec{a}\right|}$ = 2, $\small{\left|\vec{b}\right|}$ = 3 and $\small{\vec{a}.\vec{b}=4}$.
Solution:
1. First let us write $\small{\left(\vec{a}-\vec{b} \right).\left(\vec{a}-\vec{b} \right)}$:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{a}-\vec{b} \right).\left(\vec{a}-\vec{b} \right)}    & {~=~}    &{\vec{a}.\left(\vec{a}-\vec{b} \right)~-~\vec{b}.\left(\vec{a}-\vec{b} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\vec{a}.\vec{a} - \vec{a}.\vec{b}-\vec{b}.\vec{a}+\vec{b}.\vec{b}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\vec{a}.\vec{a} - 4-4+\vec{b}.\vec{b}}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\left|\vec{a} \right|^2 - 8+\left|\vec{b} \right|^2}
\\ {~\color{magenta}    5    }    &{}    &{}    & {~=~}    &{2^2 - 8+3^2}
\\ {~\color{magenta}    6    }    &{}    &{}    & {~=~}    &{5}
\\ \end{array}}$

◼ Remarks:
• 3 (magenta color): Here we use the given information that,
$\small{\vec{a}.\vec{b}}$ = 4
• 5 (magenta color): Here we use the given information that,
$\small{\left|\vec{a}\right|}$ = 2, $\small{\left|\vec{b}\right|}$ = 3

2. So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{a}-\vec{b} \right).\left(\vec{a}-\vec{b} \right)}    & {~=~}    &{5}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left|\left(\vec{a}-\vec{b} \right) \right|^2}    & {~=~}    &{5}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\left|\left(\vec{a}-\vec{b} \right) \right|}    & {~=~}    &{\sqrt{5}}
\\ \end{array}}$

◼ Remarks:
• 3 (magenta color): Here we discard the −ve root because, length cannot be −ve.

Solved example 26.42
Find $\small{\left|\vec{a}\right|}$ and $\small{\left|\vec{b}\right|}$, if $\small{\left(\vec{a}+\vec{b} \right).\left(\vec{a}-\vec{b} \right)=8}$ and $\small{\left|\vec{a}\right|=8\left|\vec{b} \right|}$.
Solution:
1. First we write the dot product:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{a}+\vec{b} \right).\left(\vec{a}-\vec{b} \right)}    & {~=~}    &{\vec{a}\left(\vec{a}-\vec{b} \right)+\vec{b}\left(\vec{a}-\vec{b} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left|\vec{a} \right|^2-\vec{a}.\vec{b}+\vec{b}.\vec{a}-\left|\vec{b} \right|^2}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left|\vec{a} \right|^2-\left|\vec{b} \right|^2}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{64\left|\vec{b} \right|^2-\left|\vec{b} \right|^2}
\\ {~\color{magenta}    5    }    &{}    &{}    & {~=~}    &{63\left|\vec{b} \right|^2}
\\ \end{array}}$

◼ Remarks:
• 4 (magenta color): Here we use the given information that, $\small{\left|\vec{a}\right|=8\left|\vec{b} \right|}$.

2. We are also given that: $\small{\left(\vec{a}+\vec{b} \right).\left(\vec{a}-\vec{b} \right)=8}$
• Substituting this in the above result, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{a}+\vec{b} \right).\left(\vec{a}-\vec{b} \right)}    & {~=~}    &{63\left|\vec{b} \right|^2}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{63\left|\vec{b} \right|^2}    & {~=~}    &{8}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\left|\vec{b} \right|^2}    & {~=~}    &{\frac{8}{63}~=~\frac{(2)2^2}{(7)3^2}}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{\left|\vec{b} \right|}    & {~=~}    &{\frac{2 \sqrt{2}}{3 \sqrt{7}}}
\\ {~\color{magenta}    5    }    &{\Rightarrow}    &{\left|\vec{a} \right|}    & {~=~}    &{8\,\left|\vec{b} \right|~=~\frac{16 \sqrt{2}}{3 \sqrt{7}}}
\\ \end{array}}$

◼ Remarks:
• 4 (magenta color): Here we discard the −ve root because, length cannot be −ve.

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In the next section, we will see a few more solved examples. We will also see the Cauchy-Schwartz inequality.

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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:
$\small{\left| \vec{P_1 P_2}\right| = \sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left(z_2 - z_1 \right)^2}}$

---

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:
$\small{\left|\vec{BC} \right|^2 = \left|\vec{AB} \right|^2 + \left|\vec{CA} \right|^2}$

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.

---

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

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

---

In the next section, we will see scalar product.

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