26.10 - Cross Product of Two Vectors

In the previous section, we completed a discussion on dot product of two vectors. We saw projection of a vector also. In this section, we will see cross product of two vectors.

Vector (or cross) product of two vectors

We have already seen the basic details about vector product in chapter 7 of our physics classes. Let us recall:
1. We have seen right handed screws and left handed screws in section 7.10.
2. We have seen the direction of the cross product in section 7.11.
3. We have seen the magnitude of the cross product in section 7.12.
4. So we can write:
$\mathbf\small{\vec{a}\times \vec{b}~=~\left|\vec{a} \right|\left|\vec{b} \right|\sin\theta \,\hat{n}}$

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Let us write the important properties of cross product:
1. $\mathbf\small{\vec{a}\times\vec{b}}$ is a vector

2. If $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ are two nonzero vectors, then $\mathbf\small{\vec{a}\times\vec{b}~=~\vec{0}}$ if and only if $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ are parallel (or collinear) to each other. We can write:
$\mathbf\small{\vec{a}\times\vec{b}~=~\vec{0}~\Leftrightarrow~\vec{a}{\parallel} \vec{b}}$
Let us see two specific cases:
Case 1:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times \vec{a}}    & {~=~}    &{\left|\vec{a} \right|\left|\vec{a} \right|\sin(0) \,\hat{n}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left|\vec{a} \right|\left|\vec{a} \right|(0) \,\hat{n}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\vec{0}}
\\ \end{array}}$

Case 2:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times \left(-\vec{a} \right)}    & {~=~}    &{\left|\vec{a} \right|\left|-\vec{a} \right|\sin(\pi) \,\hat{n}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left|\vec{a} \right|\left|-\vec{a} \right|(0) \,\hat{n}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\vec{0}}
\\ \end{array}}$

3. Let $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ be two nonzero vectors. Let the angle $\mathbf\small{\theta}$ between them be $\mathbf\small{\frac{\pi}{2}}$. Then we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times \vec{b}}    & {~=~}    &{\left|\vec{a} \right|\left|\vec{b} \right|\sin(\frac{\pi}{2}) \,\hat{n}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left|\vec{a} \right|\left|\vec{b} \right|(1) \,\hat{n}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left|\vec{a} \right|\left|\vec{b} \right| \,\hat{n}}
\\ \end{array}}$

4. Based on (2) and (3), we get some interesting results:
• Based on (2), we get:
$\mathbf\small{\hat{i}\times\hat{i}~=~\hat{j}\times\hat{j}~=~\hat{k}\times\hat{k}~=~\vec{0}}$
• Based on (3), we get:
    ♦ $\mathbf\small{\hat{i}\times\hat{j}~=~\hat{k}}$
    ♦ $\mathbf\small{\hat{j}\times\hat{k}~=~\hat{i}}$
    ♦ $\mathbf\small{\hat{k}\times\hat{i}~=~\hat{j}}$
Also see fig.7.63 and fig.7.64(a) of section 7.12 of physics notes

5. Let $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ be two nonzero vectors. Let the angle $\mathbf\small{\theta}$ between them be $\mathbf\small{\theta}$. Then we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{a}\times \vec{b} \right|}    & {~=~}    &{\left|\vec{a} \right|\left|\vec{b} \right|\sin\theta}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\sin \theta}    & {~=~}    &{\frac{\left|\vec{a}\times \vec{b} \right|}{\left|\vec{a} \right|\left|\vec{b} \right|}}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{ \theta}    & {~=~}    &{\sin^{-1}\left(\frac{\left|\vec{a}\times \vec{b} \right|}{\left|\vec{a} \right|\left|\vec{b} \right|} \right)}
\\ \end{array}}$

6. It is always true that, vector product is not commutative. This is because, $\mathbf\small{\vec{a}\times \vec{b}~=~-\left(\vec{b}\times \vec{a} \right)}$
• Assume that, both $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ lie on the plane of the paper.
• Also assume that, $\mathbf\small{\vec{a}\times\vec{b}}$ is directed towards the upper side of the plane of the paper.
• Then $\mathbf\small{\vec{b}\times\vec{a}}$ will be directed towards the bottom side of the plane of the paper.
• This is because:
    ♦ For $\mathbf\small{\vec{a}\times\vec{b}}$, we rotate the right handed screw from $\mathbf\small{\vec{a}~\text{to}~\vec{b}}$
    ♦ For $\mathbf\small{\vec{b}\times\vec{a}}$, we rotate the right handed screw from $\mathbf\small{\vec{b}~\text{to}~\vec{a}}$

7. Based on (4) and (6), we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{i}\times \hat{j} }    & {~=~}    &{\hat{k}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\hat{j}\times \hat{i}}    & {~=~}    &{-\left(\hat{i}\times \hat{j} \right)}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\hat{j}\times \hat{i}}    & {~=~}    &{-\hat{k}}
\\ \end{array}}$

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{j}\times \hat{k} }    & {~=~}    &{\hat{i}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\hat{k}\times \hat{j}}    & {~=~}    &{-\left(\hat{j}\times \hat{k} \right)}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\hat{k}\times \hat{j}}    & {~=~}    &{-\hat{i}}
\\ \end{array}}$

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{k}\times \hat{i} }    & {~=~}    &{\hat{j}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\hat{i}\times \hat{k}}    & {~=~}    &{-\left(\hat{k}\times \hat{i} \right)}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\hat{i}\times \hat{k}}    & {~=~}    &{-\hat{j}}
\\ \end{array}}$

8. The fig.26.30 below, shows triangle ABC.

Fig.26.30

• $\mathbf\small{\vec{AC}= \vec{a}~\text{and}~\vec{AB}=\vec{b}}$
• So $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ represent the adjacent sides of the triangle ABC
• We know that, area of the triangle ABC = $\mathbf\small{\frac{1}{2}\left(\left|\vec{AB} \right| \right)\left(CD \right)}$
• But $\mathbf\small{CD = \left|\vec{AC} \right|\sin\theta=\left|\vec{a} \right|\sin\theta}$
• So area of the triangle ABC
=  $\mathbf\small{\frac{1}{2}\left(\left|\vec{AB} \right| \right)\left(\left|\vec{a} \right|\sin\theta \right)}$
=  $\mathbf\small{\frac{1}{2}\left(\left|\vec{b} \right| \right)\left(\left|\vec{a} \right|\sin\theta \right)}$
=  $\mathbf\small{\frac{1}{2}\left(\left|\vec{a}\times\vec{b} \right| \right)}$
• We can write:
Area of the triangle is equal to half of the magnitude of the cross product

9. The fig.26.31 below, shows parallelogram ABCD.

Fig.26.31

• $\mathbf\small{\vec{AD}= \vec{a}~\text{and}~\vec{AB}=\vec{b}}$
• So $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ represent the adjacent sides of the parallelogram ABCD.
• We know that, area of the parallelogram ABCD = $\mathbf\small{\left(\left|\vec{AB} \right| \right)\left(DE \right)}$
• But $\mathbf\small{DE = \left|\vec{AD} \right|\sin\theta=\left|\vec{a} \right|\sin\theta}$
• So area of the parallelogram ABCD
=  $\mathbf\small{\left(\left|\vec{AB} \right| \right)\left( \left|\vec{a} \right| \sin\theta \right)}$
=  $\mathbf\small{\frac{1}{2}\left(\left|\vec{b} \right| \right)\left( \left|\vec{a} \right|\sin\theta \right)}$
=  $\mathbf\small{\left(\left|\vec{a}\times\vec{b} \right| \right)}$
• We can write:
Area of the parallelogram is equal to the magnitude of the cross product

10. Vector product of two vectors is also known as cross product of two vectors.

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Two important properties of cross product

Property I: Distributivity of cross 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:
$\small{\vec{a}\times\left(\vec{b}+\vec{c} \right)~=~\vec{a}\times\vec{b}+\vec{a}\times\vec{c}}$

Property II: Distributivity of cross 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:
$\small{\lambda \left(\vec{a}\times\vec{b} \right) ~=~\left(\lambda \vec{a}\right)\times\vec{b} ~=~\vec{a}\times\left(\lambda\vec{b} \right)}$

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Cross 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:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times\vec{b}}    & {~=~}    &{\left(a_1\hat{i}+a_2\hat{j}+a_3\hat{k} \right)\times\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{a_1\hat{i}\times\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \right)}
\\ {~\color{magenta}    {}   }    &{}    &{}    & {}    &{+~a_2\hat{j}\times\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \right)}
\\ {~\color{magenta}    {}   }    &{}    &{}    & {}    &{+~a_3\hat{k}\times\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \right)}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{a_1 b_1\left(\hat{i}\times\hat{i}\right)+a_1 b_2\left(\hat{i}\times\hat{j}\right)+a_1 b_3\left(\hat{i}\times\hat{k}\right)}
\\ {~\color{magenta}    {}   }    &{}    &{}    & {}    &{+~a_2 b_1\left(\hat{j}\times\hat{i}\right)+a_2 b_2\left(\hat{j}\times\hat{j}\right)+a_2 b_3\left(\hat{j}\times\hat{k}\right)}
\\ {~\color{magenta}    {}   }    &{}    &{}    & {}    &{+~a_3 b_1\left(\hat{k}\times\hat{i}\right)+a_3 b_2\left(\hat{k}\times\hat{j}\right)+a_3 b_3\left(\hat{k}\times\hat{k}\right)}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{a_1 b_1\left(\vec{0}\right)+a_1 b_2\left(\hat{k}\right)+a_1 b_3\left(-\hat{j}\right)}
\\ {~\color{magenta}    {}   }    &{}    &{}    & {}    &{+~a_2 b_1\left(-\hat{k}\right)+a_2 b_2\left(\vec{0}\right)+a_2 b_3\left(\hat{i}\right)}
\\ {~\color{magenta}    {}   }    &{}    &{}    & {}    &{+~a_3 b_1\left(\hat{j}\right)+a_3 b_2\left(-\hat{i}\right)+a_3 b_3\left(\vec{0}\right)}
\\ {~\color{magenta}    5   }    &{}    &{}    & {~=~}    &{\left(a_2 b_3 - a_3 b_2 \right)\hat{i} + \left(a_3 b_1 - a_1 b_3 \right)\hat{j} + \left(a_1 b_2 - a_2 b_1 \right)\hat{k}}
\\ {~\color{magenta}    6   }    &{}    &{}    & {~=~}    &{\left(a_2 b_3 - a_3 b_2 \right)\hat{i} - \left(a_1 b_3 - a_3 b_1 \right)\hat{j} + \left(a_1 b_2 - a_2 b_1 \right)\hat{k}}
\\ \end{array}}$

• The above result is equivalent to the determinant as shown below:

$\vec{a}\times\vec{b}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ a_{1}      &{    a_{2}     }    &{   a_{3}      }    
\\ b_{1}      &{    b_{2}     }    &{ b_{3}        }    
\\ \end{array}\right|$

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Now we will see some solved examples.

Solved example 26.55
Find $\mathbf\small{\left|\vec{a}\times\vec{b} \right|}$, if $\mathbf\small{\vec{a}=2\hat{i}+\hat{j}+\hat{k}~\text{and}~\vec{b}=3\hat{i}+5\hat{j}-2\hat{k}}$
Solution:
1. We have:
$\vec{a}\times\vec{b}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ 2      &{    1     }    &{   3      }    
\\ 3      &{    5     }    &{ -2        }    
\\ \end{array}\right|$

2. So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times\vec{b}}    & {~=~}    &{\left(-2 - 15 \right)\hat{i} - \left(-4 - 9 \right)\hat{j} + \left(10 - 3 \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{-17\hat{i}+13\hat{j}+7\hat{k}}
\\ \end{array}}$

3. Therefore,
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{a}\times\vec{b} \right|}    & {~=~}    &{\sqrt{(-17)^2 + (13)^2 + (7)^2}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\sqrt{507}}
\\ \end{array}}$

Solved example 26.56
Find $\mathbf\small{\left|\vec{a}\times\vec{b} \right|}$, if $\mathbf\small{\vec{a}=\hat{i}-7\hat{j}+7\hat{k}~\text{and}~\vec{b}=3\hat{i}-2\hat{j}+2\hat{k}}$
Solution:
1. We have:
$\vec{a}\times\vec{b}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ 1      &{    -7     }    &{   7      }    
\\ 3      &{    -2     }    &{ 2        }    
\\ \end{array}\right|$

2. So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times\vec{b}}    & {~=~}    &{\left(-2 - 15 \right)\hat{i} - \left(-4 - 9 \right)\hat{j} + \left(10 - 3 \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{-17\hat{i}+13\hat{j}+7\hat{k}}
\\ \end{array}}$

3. Therefore,
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{a}\times\vec{b} \right|}    & {~=~}    &{\sqrt{(19)^2 + (19)^2}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{19\sqrt{2}}
\\ \end{array}}$

Solved example 26.57
Find a unit vector perpendicular to each of the vectors $\mathbf\small{\left(\vec{a}+\vec{b} \right)~\text{and}~\left(\vec{a}-\vec{b} \right)}$, where $\mathbf\small{\vec{a}=\hat{i}+\hat{j}+\hat{k}~\text{and}~\vec{b}=\hat{i}+2\hat{j}+3\hat{k}}$
Solution:
1. We have:
• $\mathbf\small{\vec{c}=\left(\vec{a}+\vec{b} \right)=2\hat{i}+3\hat{j}+4\hat{k}}$
• $\mathbf\small{\vec{d}=\left(\vec{a}-\vec{b} \right)=-\hat{j}-2\hat{k}}$

2. We are asked to find a unit vector perpendicular to both $\mathbf\small{\vec{c}~\text{and}~\vec{d}}$
• We know that:
$\mathbf\small{\vec{c}\times\vec{d}}$ will be perpendicular to both $\mathbf\small{\vec{c}~\text{and}~\vec{d}}$
• So we will first write this cross product. We have:
$\vec{c}\times\vec{d}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ 2      &{    3     }    &{   4      }    
\\ 0      &{    -1     }    &{ -2        }    
\\ \end{array}\right|$

• So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{c}\times\vec{d}}    & {~=~}    &{\left(-6 + 4 \right)\hat{i} - \left(-4 - 0 \right)\hat{j} + \left(-2 - 0 \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{-2\hat{i}+4\hat{j}-2\hat{k}}
\\ \end{array}}$

3. Let $\mathbf\small{\vec{f}~=~\vec{c}\times\vec{d}}$
• Then $\mathbf\small{\hat{f}}$ is a required vector.
We have: $\mathbf\small{\hat{f}=\frac{\vec{f}}{\left|\vec{f} \right|}}$
= $\mathbf\small{\frac{-2\hat{i}+4\hat{j}-2\hat{k}}{\sqrt{(-2)^2 + (4)^2+(-2)^2}}~=~\frac{-2\hat{i}+4\hat{j}-2\hat{k}}{\sqrt{24}}}$
= $\mathbf\small{\frac{-2\hat{i}+4\hat{j}-2\hat{k}}{2\sqrt{6}}~=~\frac{-\hat{i}+2\hat{j}-\hat{k}}{\sqrt{6}}}$

4. Note:
We have a plane in which $\mathbf\small{\vec{c}~\text{and}~\vec{d}}$ lie. The unit vector $\mathbf\small{\hat{f}}$ is perpendicular to that plane. Consequently, $\mathbf\small{-\hat{f}}$ will also be perpendicular to that plane. We would have obtained $\mathbf\small{-\hat{f}}$, if we had started with $\mathbf\small{\vec{d}\times\vec{c}}$ instead of $\mathbf\small{\vec{c}\times\vec{d}}$ in step (2)

Solved example 26.58
Find a unit vector perpendicular to each of the vectors $\mathbf\small{\left(\vec{a}+\vec{b} \right)~\text{and}~\left(\vec{a}-\vec{b} \right)}$, where $\mathbf\small{\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}~\text{and}~\vec{b}=\hat{i}+2\hat{j}-2\hat{k}}$
Solution:
1. We have:
• $\mathbf\small{\vec{c}=\left(\vec{a}+\vec{b} \right)=4\hat{i}+4\hat{j}}$
• $\mathbf\small{\vec{d}=\left(\vec{a}-\vec{b} \right)=-2\hat{i}+4\hat{k}}$

2. We are asked to find a unit vector perpendicular to both $\mathbf\small{\vec{c}~\text{and}~\vec{d}}$
• We know that:
$\mathbf\small{\vec{c}\times\vec{d}}$ will be perpendicular to both $\mathbf\small{\vec{c}~\text{and}~\vec{d}}$
• So we will first write this cross product. We have:
$\vec{c}\times\vec{d}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ 4      &{    4     }    &{   0      }    
\\ -2      &{    0     }    &{ 4        }    
\\ \end{array}\right|$
• So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{c}\times\vec{d}}    & {~=~}    &{\left(16 - 0 \right)\hat{i} - \left(16 - 0 \right)\hat{j} + \left(0+8 \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{16\hat{i}-16\hat{j}+8\hat{k}}
\\ \end{array}}$

3. Let $\mathbf\small{\vec{f}~=~\vec{c}\times\vec{d}}$
• Then $\mathbf\small{\hat{f}}$ is a required unit vector.
We have: $\mathbf\small{\hat{f}=\frac{\vec{f}}{\left|\vec{f} \right|}}$
= $\mathbf\small{\frac{16\hat{i}-16\hat{j}+8\hat{k}}{\sqrt{(16)^2 + (-16)^2+(8)^2}}~=~\frac{16\hat{i}-16\hat{j}+8\hat{k}}{24}}$
= $\mathbf\small{\frac{2\hat{i}-2\hat{j}+\hat{k}}{3}}$

4. Note:
We have a plane in which $\mathbf\small{\vec{c}~\text{and}~\vec{d}}$ lie. The unit vector $\mathbf\small{\hat{f}}$ is perpendicular to that plane. Consequently, $\mathbf\small{-\hat{f}}$ will also be perpendicular to that plane. We would have obtained $\mathbf\small{-\hat{f}}$, if we had started with $\mathbf\small{\vec{d}\times\vec{c}}$ instead of $\mathbf\small{\vec{c}\times\vec{d}}$ in step (2)

Solved example 26.59
Find the area of a triangle having the points A(1,1,1), B(1,2,3) and C(2,3,1) as it's vertices
Solution:
1. Fig.26.32 below shows the rough sketch:

Fig.26.32

2. Based on the rough sketch, we can write:
$\mathbf\small{\vec{a}= \vec{AB}=\hat{j}+2\hat{k}}$
$\mathbf\small{\vec{b}= \vec{AC}=\hat{i}+2\hat{j}}$

3. We have:
$\vec{a}\times\vec{b}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ 0      &{    1     }    &{   2      }    
\\ 1      &{    2     }    &{ 0        }    
\\ \end{array}\right|$

4. So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times\vec{b}}    & {~=~}    &{\left(0 - 4 \right)\hat{i} - \left(0 - 2 \right)\hat{j} + \left(0 - 1 \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{-4\hat{i}+2\hat{j}-\hat{k}}
\\ \end{array}}$

5. Then,
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{a}\times\vec{b} \right|}    & {~=~}    &{\sqrt{(-4)^2 + (2)^2 + (-1)^2}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\sqrt{21}}
\\ \end{array}}$

6. Therefore,
Area of triangle ABC
= $\mathbf\small{\frac{1}{2}\left(\left|\vec{a}\times\vec{b} \right| \right)~=~\frac{\sqrt{21}}{2}}$

Solved example 26.60
Find the area of a triangle with vertices A(1,1,2), B(2,3,5) and C(1,5,5).
Solution:
1. Fig.26.33 below shows the rough sketch:

Fig.26.33

2. Based on the rough sketch, we can write:
$\mathbf\small{\vec{a}= \vec{AB}=\hat{i}+2\hat{j}+3\hat{k}}$
$\mathbf\small{\vec{b}= \vec{AC}=4\hat{j}+3\hat{k}}$

3. We have:
$\vec{a}\times\vec{b}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ 1      &{    2     }    &{   3      }    
\\ 0      &{    4     }    &{ 3        }    
\\ \end{array}\right|$

4. So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times\vec{b}}    & {~=~}    &{\left(6 - 12 \right)\hat{i} - \left(3 - 0 \right)\hat{j} + \left(4 - 0 \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{-6\hat{i}-3\hat{j}+4\hat{k}}
\\ \end{array}}$

5. Then,
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{a}\times\vec{b} \right|}    & {~=~}    &{\sqrt{(-6)^2 + (-3)^2 + (4)^2}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\sqrt{61}}
\\ \end{array}}$

6. Therefore,
Area of triangle ABC
= $\mathbf\small{\frac{1}{2}\left(\left|\vec{a}\times\vec{b} \right| \right)~=~\frac{\sqrt{61}}{2}}$

Solved example 26.61
Find the area of the parallelogram whose adjacent sides are given by the vectors $\small{\vec{a}=3\hat{i}+\hat{j}+4\hat{k}~\text{and}~\vec{b}=\hat{i}-\hat{j}+\hat{k}}$.
Solution:
1. Fig.26.34 below shows the rough sketch:

Fig.26.34

2. We have:
$\vec{a}\times\vec{b}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ 3      &{    1     }    &{   4      }    
\\ 1      &{    -1     }    &{ 1        }    
\\ \end{array}\right|$

3. So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times\vec{b}}    & {~=~}    &{\left(1 + 4 \right)\hat{i} - \left(3 - 4 \right)\hat{j} + \left(-3 - 1 \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{5\hat{i}+\hat{j}-4\hat{k}}
\\ \end{array}}$

4. Then,
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{a}\times\vec{b} \right|}    & {~=~}    &{\sqrt{(5)^2 + (1)^2 + (-4)^2}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\sqrt{42}}
\\ \end{array}}$

5. Therefore,
Area of parallelogram ABCD
= $\mathbf\small{\left|\vec{a}\times\vec{b} \right| ~=~\sqrt{42}}$

Solved example 26.62
Find the area of the parallelogram whose adjacent sides are determined by the vectors $\small{\vec{a}=\hat{i}-\hat{j}+3\hat{k}~\text{and}~\vec{b}=2\hat{i}-7\hat{j}+\hat{k}}$.
Solution:
1. Fig.26.35 below shows the rough sketch:

Fig.26.35

2. We have:
$\vec{a}\times\vec{b}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ 1      &{    -1     }    &{   3      }    
\\ 2      &{    -7     }    &{ 1        }    
\\ \end{array}\right|$

3. So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times\vec{b}}    & {~=~}    &{\left(-1 + 21 \right)\hat{i} - \left(1 - 6 \right)\hat{j} + \left(-7 + 2 \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{20\hat{i}+5\hat{j}-5\hat{k}}
\\ \end{array}}$

4. Then,
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{a}\times\vec{b} \right|}    & {~=~}    &{\sqrt{(20)^2 + (5)^2 + (-5)^2}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\sqrt{450}}
\\ \end{array}}$

5. Therefore,
Area of parallelogram ABCD
= $\mathbf\small{\left|\vec{a}\times\vec{b} \right| ~=~\sqrt{450}~=~15\sqrt{2}}$

Solved example 26.63
If a unit vector $\small{\vec{a}}$ makes angles $\small{\frac{\pi}{3}~\text{with}~\hat{i},~~\frac{\pi}{4}~\text{with}~\hat{j}}$ and an acute angle $\small{\theta}$ with $\small{\hat{k}}$, then find $\small{\theta}$ and hence, the components of $\small{\vec{a}}$.
Solution:
1. $\small{\vec{a}}$ makes an angle $\small{\frac{\pi}{3}~\text{with}~\hat{i}}$.
• $\small{\hat{i}}$ lies along the x-axis. That means, $\small{\vec{a}}$ makes an angle $\small{\frac{\pi}{3}}$ with the x-axis.
• So the x-component of $\small{\vec{a}}$
= $\small{\left|\vec{a} \right|\,\cos\left(\frac{\pi}{3} \right)\hat{i}}$ 
= $\small{(1)\left(\frac{1}{2} \right)\hat{i}}$ 
= $\small{\left(\frac{1}{2} \right)\hat{i}}$  

2. $\small{\vec{a}}$ makes an angle $\small{\frac{\pi}{4}~\text{with}~\hat{j}}$.
• $\small{\hat{j}}$ lies along the y-axis. That means, $\small{\vec{a}}$ makes an angle $\small{\frac{\pi}{4}}$ with the y-axis.
• So the y-component of $\small{\vec{a}}$
= $\small{\left|\vec{a} \right|\,\cos\left(\frac{\pi}{4} \right)\hat{j}}$ 
= $\small{(1)\left(\frac{1}{\sqrt{2}} \right)\hat{j}}$ 
= $\small{\left(\frac{1}{\sqrt{2}} \right)\hat{j}}$  

3. $\small{\vec{a}}$ makes an acute angle $\small{\theta~\text{with}~\hat{k}}$.
• $\small{\hat{k}}$ lies along the z-axis. That means, $\small{\vec{a}}$ makes an acute angle $\small{\theta}$ with the z-axis.
• So the z-component of $\small{\vec{a}}$
= $\small{\left|\vec{a} \right|\,\cos\left(\theta \right)\hat{k}}$ 
= $\small{(1)\cos\left(\theta \right)\hat{k}}$ 
= $\small{\cos\left(\theta \right)\hat{k}}$

4. We wrote the three components. So we can write:
$\small{\vec{a}=\left(\frac{1}{2} \right)\hat{i}+\left(\frac{1}{\sqrt{2}} \right)\hat{j}+\cos\left(\theta \right)\hat{k}}$

5. Given that, $\small{\vec{a}}$ is a unit vector. So we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{a} \right|}    & {~=~}    &{1}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\sqrt{\left(\frac{1}{2} \right)^2+\left(\frac{1}{\sqrt{2}} \right)^2+\left[\cos\left(\theta \right) \right]^2}}    & {~=~}    &{1}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\left(\frac{1}{2} \right)^2+\left(\frac{1}{\sqrt{2}} \right)^2+\left[\cos\left(\theta \right) \right]^2}    & {~=~}    &{1}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{\cos^2\left(\theta \right)}    & {~=~}    &{1-\frac{1}{4}-\frac{1}{2}~=~\frac{1}{4}}
\\ {~\color{magenta}    5    }    &{\Rightarrow}    &{\cos\left(\theta \right)}    & {~=~}    &{\pm\frac{1}{2}}
\\ \end{array}}$

6. So we have two equations:
(i) $\small{\cos \theta~=~\frac{1}{2}}$ 
(ii) $\small{\cos \theta~=~-\frac{1}{2}}$

• Solving the first equation, we get: $\small{\theta~=~\frac{\pi}{3}}$ 
• Solving the second equation, we get: $\small{\theta~=~\frac{2\pi}{3}}$  

• Given that, $\small{\theta}$ is an acute angle. So we can write:
$\small{\theta~=~\frac{\pi}{3}}$

7. So based on step (4), we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}}    & {~=~}    &{\left(\frac{1}{2} \right)\hat{i}+\left(\frac{1}{\sqrt{2}} \right)\hat{j}+\cos\left(\frac{\pi}{3} \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\vec{a}}    & {~=~}    &{\left(\frac{1}{2} \right)\hat{i}+\left(\frac{1}{\sqrt{2}} \right)\hat{j}+\left(\frac{1}{2} \right)\hat{k}}
\\ \end{array}}$

8. Therefore, the components of $\small{\vec{a}}$ are:
$\small{\frac{1}{2},~\frac{1}{\sqrt{2}}~\text{and}~\frac{1}{2}}$

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The link below gives a few more solved examples:

Exercise 10.4

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

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26.9 - Projection of a Vector

In the previous section, we completed a discussion on dot product of two vectors. In this section, we will see projection of a vector.

Projection of a vector on a line

This can be explained in steps:
1. In fig.26.29(i) below, $\small{\vec{AB}}$ makes an angle $\small{\theta}$ with a directed line l. The angle $\small{\theta}$ is measured in the anticlockwise direction.

Fig.26.29

2. Drop a perpendicular (green dashed line) from B on to l. Let C be the foot of the perpendicular. Then length AC will be equal to $\small{\left|\vec{AB} \right|\cos\theta}$.
3. We can think of a vector $\small{\vec{AC}}$.
   ♦ Initial point of this vector is A
   ♦ Terminal point of this vector is C.
• Magnitude of this vector will be $\small{\left|\vec{AB} \right|\cos\theta}$.
◼ $\small{\vec{AC}}$ is called:
Projection vector of $\small{\vec{AB}}$ on the directed line l.
   ♦ It is denoted as $\small{\vec{p}}$
   ♦ It is a vector
◼ $\small{\left|\vec{p} \right|}$ is called:
Projection of $\small{\vec{AB}}$ on the directed line l.
   ♦ It is a distance
   ♦ It is not a vector

◼ So we can write:
   ♦ $\small{\vec{p}}$ is the Projection vector of $\small{\vec{AB}}$ on the directed line l.
   ♦ $\small{\left|\vec{p} \right|}$ is the Projection of $\small{\vec{AB}}$ on the directed line l.

4. Since $\small{\vec{p}}$ is aligned with l, the direction of $\small{\vec{p}}$ has only two possibilities:
   ♦ same as the direction of l
   ♦ opposite to the direction of l

5. In fig.26.29(i) above, $\small{0 < \theta <\frac{\pi}{2}}$. In such a situation, $\small{\vec{p}}$ will have the same direction as l

This can be proved as shown below:
• Let $\small{\hat{p}}$ be the unit vector which has the same direction as l. Multiplying this unit vector by $\small{\left|\vec{p}\right|}$ will give $\small{\vec{p}}$.
• So we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{p} }    & {~=~}    &{\left[\left|\vec{p} \right| \right]\hat{p}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left[\left|\vec{AB} \right|\cos \theta \right]\hat{p}}
\\ \end{array}}$
   ♦ In the interval $\small{\left(0,\frac{\pi}{2} \right) }$, $\small{\cos \theta}$ is +ve. So $\small{\left[\left|\vec{AB} \right|\cos \theta \right]}$ is +ve.
   ♦ $\small{\hat{p}}$ has the same direction as l
• Therefore, $\small{\vec{p}}$ has the same direction as l

6. In fig.26.29(ii) above, $\small{\frac{\pi}{2} < \theta < \pi}$. In such a situation, $\small{\vec{p}}$ will have the opposite direction of l.

This can be proved as shown below:
• As before, we have:
$\small{\vec{p}~=~\left[\left|\vec{AB} \right|\cos \theta \right]\hat{p}}$
   ♦ In the interval $\small{\left(\frac{\pi}{2},\pi \right) }$, $\small{\cos \theta}$ is -`ve. So $\small{\left[\left|\vec{AB} \right|\cos \theta \right]}$ is -`ve.
   ♦ $\small{\hat{p}}$ has the same direction as l
• Therefore, $\small{\vec{p}}$ has the opposite direction of  l
• Recall that, we saw the trigonometric ratios of obtuse angles in section 3.5   

7. In fig.26.29(iii) below, $\small{\pi < \theta < \frac{3\pi}{2}}$. In such a situation, $\small{\vec{p}}$ will have the opposite direction of l

Fig.26.29

This can be proved as shown below:
• As before, we have:
$\small{\vec{p}~=~\left[\left|\vec{AB} \right|\cos \theta \right]\hat{p}}$
   ♦ In the interval $\small{\left(\pi,\frac{3\pi}{2}\right) }$, $\small{\cos \theta}$ is -`ve. So $\small{\left[\left|\vec{AB} \right|\cos \theta \right]}$ is -`ve.
   ♦ $\small{\hat{p}}$ has the same direction as l
• Therefore, $\small{\vec{p}}$ has the opposite direction of  l

8. In fig.26.29(iv) above, $\small{\frac{3\pi}{2} < \theta < 2 \pi}$. In such a situation, $\small{\vec{p}}$ will have the same direction as l

This can be proved as shown below:
• As before, we have:
$\small{\vec{p}~=~\left[\left|\vec{AB} \right|\cos \theta \right]\hat{p}}$
   ♦ In the interval $\small{\left(\frac{3\pi}{2},2\pi \right) }$, $\small{\cos \theta}$ is +ve. So $\small{\left[\left|\vec{AB} \right|\cos \theta \right]}$ is +ve.
   ♦ $\small{\hat{p}}$ has the same direction as l
• Therefore, $\small{\vec{p}}$ has the same direction as l

9. Consider the scalar product $\small{\vec{AB}.\hat{p}}$, where $\small{\hat{p}}$ is the unit vector which has the same direction as l'.
We can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{AB}.\hat{p}}    & {~=~}    &{\left|\vec{AB} \right|\,\left|\hat{p} \right|\cos \theta}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left|\vec{AB} \right|(1)\cos \theta}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left|\vec{AB} \right|\cos \theta}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\text{Projection of}~\vec{AB}~\text{on}~l}
\\ \end{array}}$
• Thus we get a second method to find the projection.
   ♦ First method is to find $\small{\left|\vec{AB} \right|\cos\theta}$
   ♦ Second method is to find the dot product $\small{\vec{AB}.\hat{p}}$
• Remember that, $\small{\vec{AB}.\hat{p}}$ is the dot product of two vectors. So it is a real number. It is the "projection". Not the "projection vector".

10. Suppose that, we want the projection of a vector $\small{\vec{a}~\text{on another vector}~\vec{b}}$. We can make use of the result in (9) above. The steps are as follows:
(i) Assume a directed line l in the same direction as $\small{\vec{b}}$
(ii) Next, we want a unit vector in the same direction as l. Such a unit vector is readily available. It is $\small{\hat{b}}$
(iii) So based on (9), the required projection is: $\small{\vec{a}.\hat{b}}$
(iv) But $\small{\hat{b}~=~\frac{\vec{b}}{\left|\vec{b} \right|}}$
(v) So the required projection = $\small{\vec{a}.\left(\frac{\vec{b}}{\left|\vec{b} \right|} \right)~=~\frac{1}{\left|\vec{b} \right|}\left(\vec{a}.\vec{b} \right)}$
• Remember that, $\small{\frac{1}{\left|\vec{b} \right|}\left(\vec{a}.\vec{b} \right)}$ is the scalar product of two vectors. So it is a real number. It is the "projection". Not the "projection vector".

11. In fig.26.29 above, we considered $\small{\theta}$ to be in the intervals:
$\small{\left(0,\frac{\pi}{2} \right), \left(\frac{\pi}{2},\pi \right), \left(\pi,\frac{3\pi}{2} \right), \left(\frac{3\pi}{2},2\pi \right)}$
We did not consider the cases when $\small{\theta}$ is equal to $\small{0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi}$. We will see those cases now.
(i) $\small{\theta=0}$
Consider fig.26.29(i). We want $\small{\vec{p}~\text{and}~\left|\vec{p} \right|}$ when $\small{\theta=0}$
Based on (5), we have:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{p} }    & {~=~}    &{\left[\left|\vec{AB} \right|\cos \theta \right]\hat{p}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left[\left|\vec{AB} \right|\cos (0) \right]\hat{p}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left[\left|\vec{AB} \right|(1) \right]\hat{p}}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\left(\left|\vec{AB} \right| \right)\hat{p}}
\\ \end{array}}$
• $\small{\hat{p}}$ is a unit vector. It gives us the direction only. In our present case, it has the same direction as l. So we can write:
When $\small{\theta=0}$, the projection vector has the same magnitude as $\small{\vec{AB}}$. And it has the same direction as l.

(ii) $\small{\theta=\frac{\pi}{2}}$
Consider fig.26.29(i or ii). We want $\small{\vec{p}~\text{and}~\left|\vec{p} \right|}$ when $\small{\theta=\frac{\pi}{2}}$
Based on (5), we have:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{p} }    & {~=~}    &{\left[\left|\vec{AB} \right|\cos \theta \right]\hat{p}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left[\left|\vec{AB} \right|\cos \left(\frac{\pi}{2}\right) \right]\hat{p}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left[\left|\vec{AB} \right|(0) \right]\hat{p}}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\left(0 \right)\hat{p}}
\\ \end{array}}$

• $\small{\hat{p}}$ is a unit vector. It gives us the direction only. In our present case, it has the same direction as l. So we can write:
When $\small{\theta=\frac{\pi}{2}}$, the projection vector is a zero vector.   
(iii) $\small{\theta=\pi}$
Consider fig.26.29(ii or iii ). We want $\small{\vec{p}~\text{and}~\left|\vec{p} \right|}$ when $\small{\theta=\pi}$
Based on (5), we have:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{p} }    & {~=~}    &{\left[\left|\vec{AB} \right|\cos \theta \right]\hat{p}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left[\left|\vec{AB} \right|\cos (\pi) \right]\hat{p}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left[\left|\vec{AB} \right|(-1) \right]\hat{p}}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{(-1)\left(\left|\vec{AB} \right| \right)\hat{p}}
\\ \end{array}}$

• $\small{\hat{p}}$ is a unit vector. It gives us the direction only. In our present case, it has the same direction as l. So we can write:
When $\small{\theta=\pi}$, the projection vector has the same magnitude as $\small{\vec{AB}}$. And it has the opposite direction of l.
(iv) $\small{\theta=\frac{3\pi}{2}}$
Consider fig.26.29(iii or iv). We want $\small{\vec{p}~\text{and}~\left|\vec{p} \right|}$ when $\small{\theta=\frac{3\pi}{2}}$
Based on (5), we have:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{p} }    & {~=~}    &{\left[\left|\vec{AB} \right|\cos \theta \right]\hat{p}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left[\left|\vec{AB} \right|\cos \left(\frac{3\pi}{2}\right) \right]\hat{p}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left[\left|\vec{AB} \right|(0) \right]\hat{p}}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\left(0 \right)\hat{p}}
\\ \end{array}}$

• $\small{\hat{p}}$ is a unit vector. It gives us the direction only. In our present case, it has the same direction as l. So we can write:
When $\small{\theta=\frac{3\pi}{2}}$, the projection vector is a zero vector.  
(v) $\small{\theta=2\pi}$
Consider fig.26.29(i). We want $\small{\vec{p}~\text{and}~\left|\vec{p} \right|}$ when $\small{\theta=2\pi}$
Based on (5), we have:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{p} }    & {~=~}    &{\left[\left|\vec{AB} \right|\cos \theta \right]\hat{p}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left[\left|\vec{AB} \right|\cos (2\pi) \right]\hat{p}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left[\left|\vec{AB} \right|(1) \right]\hat{p}}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\left(\left|\vec{AB} \right| \right)\hat{p}}
\\ \end{array}}$

• $\small{\hat{p}}$ is a unit vector. It gives us the direction only. In our present case, it has the same direction as l. So we can write:
When $\small{\theta=2\pi}$, the projection vector has the same magnitude as $\small{\vec{AB}}$. And it has the same direction as l.

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Now we will see the projection when a vector is given in component form. It can be written in 4 steps:
1. Consider a vector $\small{\vec{a}~=~a_1\hat{i}+a_2\hat{j}+a_3\hat{k}}$.
We know that, this vector has three components:
   ♦ $\small{a_1\hat{i}}$, which lies along the x-axis
   ♦ $\small{a_2\hat{j}}$, which lies along the y-axis
   ♦ $\small{a_3\hat{k}}$, which lies along the z-axis
2. Consider the component $\small{a_1\hat{i}}$, which lies along the x-axis.
We know that, magnitude of this vector is $\small{a_1}$.
Since this vector lies on the x-axis, we can write:
   ♦ Projection vector of $\small{\vec{a}}$ on the x-axis is $\small{a_1\hat{i}}$
   ♦ Projection of $\small{\vec{a}}$ on the x-axis is $\small{a_1}$
3. Similarly, we can write:
   ♦ Projection vector of $\small{\vec{a}}$ on the y-axis is $\small{a_2\hat{j}}$
   ♦ Projection of $\small{\vec{a}}$ on the y-axis is $\small{a_2}$
4. Similarly, we can write:
   ♦ Projection vector of $\small{\vec{a}}$ on the z-axis is $\small{a_3\hat{k}}$
   ♦ Projection of $\small{\vec{a}}$ on the z-axis is $\small{a_3}$

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Now we will see the relation between projection and direction cosines. It can be written in 7 steps:
1. Consider a vector $\small{\vec{a}~=~a_1\hat{i}+a_2\hat{j}+a_3\hat{k}}$. We know the direction cosines:
   ♦ $\small{l~=~\frac{a_1}{\left|\vec{a} \right|}}$
   ♦ $\small{m~=~\frac{a_2}{\left|\vec{a} \right|}}$
   ♦ $\small{n~=~\frac{a_3}{\left|\vec{a} \right|}}$
See solved example 26.19 in section 26.5
2. Consider the first direction cosine: $\small{l~=~\frac{a_1}{\left|\vec{a} \right|}}$
• Here $\small{a_1}$ is the projection of $\small{\vec{a}}$ on the x-axis.
3. We have seen a second method to find projection. See step (9) above.
• By that method, the "projection of $\small{\vec{a}}$ on the x-axis" is:
$\small{\vec{a}.\hat{i}}$
4. So the result in (2) becomes: $\small{l~=~\frac{\vec{a}.\hat{i}}{\left|\vec{a} \right|}}$
5. Similarly we can obtain the other two direction cosines also. Let us write all the three together:
• $\small{l~=~\frac{a_1}{\left|\vec{a} \right|}~=~\frac{\vec{a}.\hat{i}}{\left|\vec{a} \right|}}$
• $\small{m~=~\frac{a_2}{\left|\vec{a} \right|}~=~\frac{\vec{a}.\hat{j}}{\left|\vec{a} \right|}}$
• $\small{r~=~\frac{a_3}{\left|\vec{a} \right|}~=~\frac{\vec{a}.\hat{k}}{\left|\vec{a} \right|}}$
6. Suppose that, $\small{\vec{a}}$ is a unit vector.
• Then $\small{\left|\vec{a}\right|=1}$
• In such a situation, step (1) above becomes:
   ♦ $\small{l~=~\frac{a_1}{\left|\vec{a} \right|}~=~\frac{a_1}{1}~=~a_1}$
   ♦ $\small{m~=~\frac{a_2}{\left|\vec{a} \right|}~=~\frac{a_1}{1}~=~a_2}$
   ♦ $\small{n~=~\frac{a_3}{\left|\vec{a} \right|}~=~\frac{a_1}{1}~=~a_3}$
7. So if $\small{\vec{a}}$ is a unit vector, then it's component form can be written as:
$\small{\vec{a}~=~a_1\hat{i}+a_2\hat{j}+a_3\hat{k}~=~l\hat{i}+m\hat{j}+n\hat{k}}$

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Now we will see some solved examples

Solved example 26.51
Show that each of the given three vectors is a unit vector:
$\small{\frac{1}{7}\left(2\hat{i}+3\hat{j}+6\hat{k} \right)}$
$\small{\frac{1}{7}\left(3\hat{i}-6\hat{j}+2\hat{k} \right)}$
$\small{\frac{1}{7}\left(6\hat{i}-3\hat{k} \right)}$
Also show that they are mutually perpendicular to each other.
Solution:
Part (i): Showing that the vectors are unit vectors
• We have seen that:
For a unit vector, the scalar components are same as the direction cosines.
1. Let $\small{\vec{a}=\frac{1}{7}\left(2\hat{i}+3\hat{j}+6\hat{k} \right)}$
• The scalar components are:
   ♦ $\small{a_1 = \frac{2}{7}}$
   ♦ $\small{a_2 = \frac{3}{7}}$
   ♦ $\small{a_3 = \frac{6}{7}}$
• $\small{\left|\vec{a}\right|=\sqrt{\frac{2^2 + 3^2 + 6^2}{7^2}}=\sqrt{\frac{49}{49}}=1}$
• Therefore, the direction cosines are:
   ♦ $\small{l = \frac{a_1}{\left|\vec{a}\right|}=\frac{2/7}{1}=\frac{2}{7}}$
   ♦ $\small{m = \frac{a_2}{\left|\vec{a}\right|}=\frac{3/7}{1}=\frac{3}{7}}$
   ♦ $\small{n = \frac{a_3}{\left|\vec{a}\right|}=\frac{6/7}{1}=\frac{6}{7}}$
• The scalar components are same as the direction cosines. So $\small{\vec{a}}$ is a unit vector.

• The above elaborate steps are written to show the relation between "direction cosines" and "scalar components of a unit vector". In actual practice, to show that $\small{\vec{a}}$ is a unit vector, all we need to do is to show that $\small{\left|\vec{a}\right|=1}$  

2. Let $\small{\vec{b}=\frac{1}{7}\left(3\hat{i}-6\hat{j}+2\hat{k} \right)}$
• The scalar components are:
   ♦ $\small{b_1 = \frac{3}{7}}$
   ♦ $\small{b_2 = \frac{-6}{7}}$
   ♦ $\small{b_3 = \frac{2}{7}}$
• $\small{\left|\vec{b}\right|=\sqrt{\frac{3^2 + (-6)^2 + 2^2}{7^2}}=\sqrt{\frac{49}{49}}=1}$
• Therefore, the direction cosines are:
   ♦ $\small{l = \frac{b_1}{\left|\vec{b}\right|}=\frac{3/7}{1}=\frac{3}{7}}$
   ♦ $\small{m = \frac{b_2}{\left|\vec{b}\right|}=\frac{-6/7}{1}=\frac{-6}{7}}$
   ♦ $\small{n = \frac{b_3}{\left|\vec{b}\right|}=\frac{2/7}{1}=\frac{2}{7}}$
• The scalar components are same as the direction cosines. So $\small{\vec{b}}$ is a unit vector.

3. Let $\small{\vec{c}=\frac{1}{7}\left(6\hat{i}+2\hat{j}-3\hat{k} \right)}$
• The scalar components are:
   ♦ $\small{c_1 = \frac{6}{7}}$
   ♦ $\small{c_2 = \frac{2}{7}}$
   ♦ $\small{c_3 = \frac{-3}{7}}$
• $\small{\left|\vec{c}\right|=\sqrt{\frac{6^2 + 2^2 + (-3)^2}{7^2}}=\sqrt{\frac{49}{49}}=1}$
• Therefore, the direction cosines are:
   ♦ $\small{l = \frac{c_1}{\left|\vec{c}\right|}=\frac{6/7}{1}=\frac{6}{7}}$
   ♦ $\small{m = \frac{c_2}{\left|\vec{c}\right|}=\frac{2/7}{1}=\frac{2}{7}}$
   ♦ $\small{n = \frac{c_3}{\left|\vec{c}\right|}=\frac{-3/7}{1}=\frac{-3}{7}}$
• The scalar components are same as the direction cosines. So $\small{\vec{c}}$ is a unit vector.

Part (ii): Showing that the given three vectors are mutually perpendicular to each other
• We have seen that:
If two vectors are mutually perpendicular to each other, their dot product will be zero.
• For three vectors, three combinations are possible. So we have to calculate three dot products.

1. First we calculate $\small{\vec{a}.\vec{b}}$
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\vec{a}.\vec{b}}    & {=}    &{\left(\frac{2}{7} \right)\left(\frac{3}{7} \right)+\left(\frac{3}{7} \right)\left(\frac{-6}{7} \right)+\left(\frac{6}{7} \right)\left(\frac{2}{7} \right)}
\\ {~\color{magenta}    2    }    &{}&{}    & {=}    &{\frac{6-18+12}{49}=\frac{0}{49}}
\\ {~\color{magenta}    3    }    &{}&{}    & {=}    &{0}
\\ \end{array}}$
• We see that: $\small{\vec{a}.\vec{b}=0}$
So $\small{\vec{a}~\text{and}~\vec{b}}$ are mutually perpendicular to each other.

2. Next we calculate $\small{\vec{b}.\vec{c}}$
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\vec{b}.\vec{c}}    & {=}    &{\left(\frac{3}{7} \right)\left(\frac{6}{7} \right)+\left(\frac{-6}{7} \right)\left(\frac{2}{7} \right)+\left(\frac{2}{7} \right)\left(\frac{-3}{7} \right)}
\\ {~\color{magenta}    2    }    &{}&{}    & {=}    &{\frac{18-12-6}{49}=\frac{0}{49}}
\\ {~\color{magenta}    3    }    &{}&{}    & {=}    &{0}
\\ \end{array}}$
• We see that: $\small{\vec{b}.\vec{c}=0}$
So $\small{\vec{b}~\text{and}~\vec{c}}$ are mutually perpendicular to each other.

3. Finally we calculate $\small{\vec{c}.\vec{a}}$
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\vec{c}.\vec{a}}    & {=}    &{\left(\frac{6}{7} \right)\left(\frac{2}{7} \right)+\left(\frac{2}{7} \right)\left(\frac{3}{7} \right)+\left(\frac{-3}{7} \right)\left(\frac{6}{7} \right)}
\\ {~\color{magenta}    2    }    &{}&{}    & {=}    &{\frac{12+6-18}{49}=\frac{0}{49}}
\\ {~\color{magenta}    3    }    &{}&{}    & {=}    &{0}
\\ \end{array}}$
• We see that: $\small{\vec{c}.\vec{a}=0}$
So $\small{\vec{c}~\text{and}~\vec{a}}$ are mutually perpendicular to each other.

Solved example 26.52
Find the projection of the vector $\small{\vec{a}=2\hat{i}+3\hat{j}+2\hat{k}}$ on the vector $\small{\vec{b}=\hat{i}+2\hat{j}+\hat{k}}$
Solution:
1. We have:
Projection of $\small{\vec{a}~\text{on}~\vec{b}=\frac{1}{\left|\vec{b} \right|}\left(\vec{a}.\vec{b} \right)}$
2. Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\frac{1}{\left|\vec{b} \right|}\left(\vec{a}.\vec{b} \right)}    & {=}    &{\left(\frac{1}{\sqrt{1^2 + 2^2+1^2}} \right)\left[\left(2 \right)\left(1 \right)+\left(3 \right)\left(2 \right)+\left(2 \right)\left(1 \right) \right]}
\\ {~\color{magenta}    2    }    &{}&{}    & {=}    &{\frac{2+6+2}{\sqrt{6}}=\frac{10}{\sqrt{6}}}
\\ {~\color{magenta}    3    }    &{}&{}    & {=}    &{\frac{5\sqrt{6}}{3}}
\\ \end{array}}$

Solved example 26.53
Find the projection of the vector $\small{\vec{a}=\hat{i}-\hat{j}}$ on the vector $\small{\vec{b}=\hat{i}+\hat{j}}$
Solution:
1. We have:
Projection of $\small{\vec{a}~\text{on}~\vec{b}=\frac{1}{\left|\vec{b} \right|}\left(\vec{a}.\vec{b} \right)}$
2. Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\frac{1}{\left|\vec{b} \right|}\left(\vec{a}.\vec{b} \right)}    & {=}    &{\left(\frac{1}{\sqrt{1^2 +1^2}} \right)\left[\left(1 \right)\left(1 \right)+\left(-1 \right)\left(1 \right) \right]}
\\ {~\color{magenta}    2    }    &{}&{}    & {=}    &{\frac{0}{\sqrt{2}}}
\\ {~\color{magenta}    3    }    &{}&{}    & {=}    &{0}
\\ \end{array}}$

Solved example 26.54
Find the projection of the vector $\small{\vec{a}=\hat{i}+3\hat{j}+7\hat{k}}$ on the vector $\small{\vec{b}=7\hat{i}-\hat{j}+8\hat{k}}$
Solution:
1. We have:
Projection of $\small{\vec{a}~\text{on}~\vec{b}=\frac{1}{\left|\vec{b} \right|}\left(\vec{a}.\vec{b} \right)}$
2. Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\frac{1}{\left|\vec{b} \right|}\left(\vec{a}.\vec{b} \right)}    & {=}    &{\left(\frac{1}{\sqrt{7^2 + (-1)^2+8^2}} \right)\left[\left(1 \right)\left(7 \right)+\left(3 \right)\left(-1 \right)+\left(7 \right)\left(8 \right) \right]}
\\ {~\color{magenta}    2    }    &{}&{}    & {=}    &{\frac{7-3+56}{\sqrt{114}}}
\\ {~\color{magenta}    3    }    &{}&{}    & {=}    &{\frac{60}{\sqrt{114}}}
\\ \end{array}}$

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The link below gives a few more solved examples:

Exercise 10.3

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In the next section, we will see cross product.

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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 $\small{\left(\vec{a}+\lambda \vec{b} \right)}$ is perpendicular to $\small{\vec{c}}$, then find the value of $\small{\lambda}$.
Solution:
1.First we write the sum:
$\small{\left(\vec{a}+\lambda \vec{b} \right)}$
= $\small{\left(2-\lambda \right)\hat{i}+\left(2+2\lambda \right)\hat{j}+\left(3+\lambda \right)\hat{k}}$ 

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:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{3 + 2\vec{a}.\vec{b} + 2\vec{b}.\vec{c} + 2\vec{c}.\vec{a}}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{3 + 2\left(\vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} \right)}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{ 2\left(\vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} \right)}    & {~=~}    &{-3}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{ \vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} }    & {~=~}    &{\frac{-3}{2}}
\\ \end{array}}$

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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.

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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 4 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.

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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 $\small{\vec{OA},~\vec{OB}~\text{and}~\vec{OC}}$
• Let us assume that, 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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