Question

If A B C are three vectors show that A*(B+C)=A*B+B*C

Answer 1

Answer on Question #68130 – Math – Linear Algebra

Question

If $A, B, C$ are three vectors, show that $A \cdot (B + C) = A \cdot B + B \cdot C$.

Solution

The given property is not valid. Let's take for example

$$A = \left[ \begin{array}{c} 2 \\ 0 \end{array} \right], \qquad B = \left[ \begin{array}{c} 0 \\ 1 \end{array} \right], \qquad C = \left[ \begin{array}{c} 1 \\ 1 \end{array} \right].$$

Then we obtain

$$B + C = \left[ \begin{array}{c} 1 \\ 2 \end{array} \right], \qquad A \cdot (B + C) = 2 \cdot 1 + 0 \cdot 2 = 2$$

and

$$A \cdot B = 2 \cdot 0 + 0 \cdot 1 = 0, \quad B \cdot C = 0 \cdot 1 + 1 \cdot 1 = 1, \quad A \cdot B + B \cdot C = 1.$$

And we see, that $A \cdot (B + C) \neq A \cdot B + B \cdot C$.

But for three arbitrary vectors $A, B, C$ the distributive property of dot product over addition is valid, namely

$$A \cdot (B + C) = A \cdot B + A \cdot C.$$

We now prove it. Let

$$A = \left[ \begin{array}{c} a _ {1} \\ a _ {2} \\ \vdots \\ a _ {n} \end{array} \right], \qquad B = \left[ \begin{array}{c} b _ {1} \\ b _ {2} \\ \vdots \\ b _ {n} \end{array} \right], \qquad C = \left[ \begin{array}{c} c _ {1} \\ c _ {2} \\ \vdots \\ c _ {n} \end{array} \right].$$

Then

$$B + C = \left[ \begin{array}{c} b _ {1} \\ b _ {2} \\ \vdots \\ b _ {n} \end{array} \right] + \left[ \begin{array}{c} c _ {1} \\ c _ {2} \\ \vdots \\ c _ {n} \end{array} \right] = \left[ \begin{array}{c} b _ {1} + c _ {1} \\ b _ {2} + c _ {2} \\ \vdots \\ b _ {n} + c _ {n} \end{array} \right],$$ $$\begin{array}{l} A \cdot (B + C) = \left[ \begin{array}{c} a _ {1} \\ a _ {2} \\ \vdots \\ a _ {n} \end{array} \right] \cdot \left[ \begin{array}{c} b _ {1} + c _ {1} \\ b _ {2} + c _ {2} \\ \vdots \\ b _ {n} + c _ {n} \end{array} \right] = a _ {1} \cdot (b _ {1} + c _ {1}) + a _ {2} \cdot (b _ {2} + c _ {2}) + \dots + a _ {n} \cdot (b _ {n} + c _ {n}) \\ = a _ {1} \cdot b _ {1} + a _ {1} \cdot c _ {1} + a _ {2} \cdot b _ {2} + a _ {2} \cdot c _ {2} + \dots + a _ {n} \cdot b _ {n} + a _ {n} \cdot c _ {n} \\ = \left(a _ {1} \cdot b _ {1} + a _ {2} \cdot b _ {2} + \dots + a _ {n} \cdot b _ {n}\right) + \left(a _ {1} \cdot c _ {1} + a _ {2} \cdot c _ {2} + \dots + a _ {n} \cdot c _ {n}\right) = A \cdot B + A \cdot C. \end{array}$$

The distribution property for dot product of vectors is proved.

The cross product is distributive over addition as well, namely for three arbitrary vectors $A, B, C$

$$A \times (B + C) = A \times B + A \times C.$$

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