Linear Algebra – How to show dot product is distributive for coplanar vectors using A · B = ABcos(?)

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I am asked to use the equation $\mathbf{A} \cdot \mathbf{B} = AB\cos(\theta)$ to show that the dot product is distributive when the three vectors are coplanar. It seems that the typical way in which this is proven is with reference to a diagram, as was done in this answer; however, I want to try and prove this without relying on a diagram.

This equation is still valid for more than 2 vectors, since we know that the sum of two vectors (say, $\mathbf{B}$ and $\mathbf{C}$) is itself a vector:

$$\mathbf{A} \cdot (\mathbf{B} + \mathbf{C}) = A(B + C)\cos(\theta)$$

So let $\mathbf{A}, \mathbf{B}, \mathbf{C}$ be coplanar vectors. Since we have that 2 vectors span a plane, this implies that one of the vectors must be a linear combination of the other two. Let that vector be $\mathbf{C} = \alpha \mathbf{A} + \beta \mathbf{B} = \alpha(a_1, a_2, a_3) + \beta(b_1, b_2, b_3)$, where $\alpha, \beta \in \mathbb{R}$.

So we have that

$$\begin{align} \mathbf{A} \cdot ( \mathbf{B} + \mathbf{C}) &= A(B + C) \cos(\theta) \\ &= \sqrt{{a_1}^2 + {a_2}^2 + {a_3}^2}\sqrt{(b_1 + c_1)^2 + (b_2 + c_2)^2 + (b_3 + c_3)^2} \cos(\theta) \end{align}$$

We can continue by factoring out the squares under the second square root, but I do not see how this would progress the proof?

What am I missing? Did I make an error, or is my reasoning so far correct? I would greatly appreciate it if people would please take the time to clarify this.

Best Answer

Let $\beta = \angle (\vec A,\vec B)$, $\gamma = \angle (\vec A, \vec C)$.

Take an orthonormal basis for the plane with $\hat e_1 = \vec A / A$ and write $\vec B = B(\cos\beta \hat e_1 + \sin \beta\hat e_2), \vec C = C(\cos\gamma \hat e_1 + \sin\gamma\hat e_2)$.

Then

$\vec B + \vec C = (B\cos\beta + C\cos\gamma)\hat e_1 + (B\sin\beta + C\sin\gamma) \hat e_2$

and

$\cos \angle (\vec A, \vec B + \vec C) = (B\cos\beta + C\cos\gamma) / \|\vec B + \vec C\|$.

Hence $\vec A\cdot (\vec B + \vec C) = A\|\vec B + \vec C\|\cos\angle(\vec A, \vec B + \vec C) = AB\cos\beta + AC\cos\gamma = \vec A\cdot\vec B + \vec A\cdot \vec C$

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