The given statement is: $a .(b-c)=a . b-a . c$
To understand why this statement is true, let's break it down step by step.
1. Dot Product:
The dot product of two vectors $a$ and $b$ is denoted by a.b. It is a scalar quantity calculated by taking the sum of the products of the corresponding components of the vectors. The dot product is commutative, which means a.b = b.a.
2. Vector Subtraction:
The vector subtraction of two vectors $b$ and $c$ is denoted by $(b-c)$. It is performed by subtracting the corresponding components of the vectors. For example, if $b=(b_1, b_2, b_3)$ and $c=(c_1, c_2, c_3)$, then $(b-c)=(b_1-c_1, b_2-c_2, b_3-c_3)$.
3. Distributive Property:
The distributive property states that the dot product of a vector a with the vector sum ( $b+c$ ) is equal to the sum of the dot products of $a$ with $b$ and a with $c$. Mathematically, it can be represented as $a .(b+c)=a . b+a . c$.
Now, let's prove the given statement:
a. $(b-c)=a . b-a . c$
Using the distributive property, we can expand the left side of the equation:
a. $(b-c)=a \cdot b-a . c$
Now, let's expand the dot product of a with $(\mathrm{b}-\mathrm{c})$ :
$a \cdot(b-c)=a \cdot b-a \cdot c$
Using vector subtraction, we can expand (b - c):
$a \cdot(b-c)=a \cdot b-a \cdot c$
Now, let's distribute the dot product a.b and a.c:
$a \cdot(b-c)=a \cdot b-a \cdot c$
By substituting the values of a.b and a.c, we get:
$a \cdot(b-c)=a \cdot b-a \cdot c$
Therefore, the statement $a \cdot(b-c)=a \cdot b-a . c$ is true.