Can 2 times the dot product of 2 complex numbers equal to this?

**x3bnm** · August 27th 2012, 07:04 AM

There's a section in Gibson's "Elementary Geometry of Differentiable Curves:An Undergraduate Introduction" book on page 6:

"Example 1.4: Recall that the component of a vector $a$ in the direction of a unit vector $b$ is the vector $(a \cdot b)b$ . (Example 1.1)
It is useful to express this in complex notation. Note that for any vectors $a,b$ we have $2(a \cdot b) = a \overline{b} + b \overline{a}$ : in particular when
b is a unit vector (i.e. $b \overline{b} = \left | b \right |^2 = 1$ ) we have $2(a \cdot b)b = a + \overline{a} b^2$ ."

What I don't understand is why and how is: $2(a \cdot b) = a \overline{b} + b \overline{a}$ ?

The author said(given above) and I quote "for any vectors" this is true.

Can anyone kindly shed light on this and explain a bit why this is so?

**Plato** · August 27th 2012, 07:19 AM

Originally Posted by x3bnm

"Example 1.4: Recall that the component of a vector $a$ in the direction of a unit vector $b$ is the vector $(a \cdot b)b$ . (Example 1.1)
It is useful to express this in complex notation. Note that for any vectors $a,b$ we have $2(a \cdot b) = a \overline{b} + b \overline{a}$ : in particular when
b is a unit vector (i.e. $b \overline{b} = \left | b \right |^2 = 1$ ) we have $2(a \cdot b)b = a + \overline{a} b^2$ ."
What I don't understand is why and how is: $2(a \cdot b) = a \overline{b} + b \overline{a}$ ?

Suppose that $a=x+iy~\&~b=u+iv$ .
You can show that $a\overline{b}+\overline{a}b=2(xu+yv)$ .
But as vectors $a\cdot b=xu+yv.$

**Vlasev** · August 27th 2012, 07:23 AM

We have

$a\overline{b}+\overline{a}b = a\overline{b}+\overline{a\overline{b}} = 2Re(a\overline{b})$

The real part is just

$a_1b_1-a_2(-b_2) = a_1b_1+a_2b_2 = a\cdot b$

**x3bnm** · August 27th 2012, 07:34 AM

Thanks a lot Plato and Vlasev. That makes sense now. Again Thanks.

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