Proving distributive nature dot product

Question

In my physics textbook on the chapter vectors it asks me to prove that $$\mathbf {a \cdot (b+c)} = \mathbf {a \cdot b+ a \cdot c} .$$ So I thought to prove it as follows:

Since $$\mathbf {x \cdot y} =xy \cos \theta$$ ($\theta$ being the angle between the two vectors)

Therefore $$\mathbf {a \cdot (b+c)} = a(| \mathbf {b+c}|) \cos \theta$$

Now I can't go any further. Can somebody tell me if its possible to prove that through the above stated method? If so how?

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EDIT I know how to do it the way user did it but I'm just wondering whether the way I'm trying to prove it is possible to do(if so how?), even if it's hardest way to do it!

Answer 1

You're overthinking it: $$a\cdot(b+c)=\sum_ia_i(b+c)_i=\sum_ia_i(b_i+c_i)=\sum_ia_ib_i+\sum_ia_ic_i=a\cdot b+a\cdot c.$$

Answer 2

Let

• $\vec a=a_x \hat i+a_y \hat j$
• $\vec b=b_x \hat i+b_y \hat j$
• $\vec c=c_x \hat i+c_y \hat j$

then since $\hat i\cdot \hat i=1$, $\hat j\cdot \hat j=1$, $\hat i\cdot \hat j=0$

$$\vec a\cdot(\vec b+\vec c)=(a_x \hat i+a_y \hat j)[(b_x+c_x)\hat i+(b_y+c_y)\hat j]=(a_xb_x+a_xc_x)+(a_yb_y+a_yc_y)=$$

$$=(a_xb_x+a_yb_y )+(a_xc_x+a_yc_y )=\vec a \cdot \vec b+\vec a \cdot \vec c$$

The result can be extended in a similar way for vectors in $\mathbb R^3$.