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.
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.
In the next section, we will see a few more solved examples. We will also see the Cauchy-Schwartz inequality.
Previous
Contents
Next
Copyright©2026 Higher secondary mathematics.blogspot.com
No comments:
Post a Comment