Showing that $\vec{a}\times\vec{c}=\vec{b}\times\vec{c} \implies \vec{c}\cdot\vec{a}-\vec{c}\cdot\vec{b}=\pm\|\vec{c}\|\cdot\|\vec{a}-\vec{b}\|$

Question

For vectors $\vec{a},\vec{b},\vec{c}\in\mathbb{R}^{3}$, how do I show that:

If $\vec{a}\times\vec{c}=\vec{b}\times\vec{c} \implies \vec{c}\cdot\vec{a}-\vec{c}\cdot\vec{b}=\pm\|\vec{c}\|\cdot\|\vec{a}-\vec{b}\|$

I began by using the geometric definitions of the cross and dot products:

$$\vec{v}\times\vec{w}=\|\vec{v}\|\|\vec{w}\|\sin{\theta}\:\hat{n} \\ \vec{v}\cdot\vec{w}=\|\vec{v}\|\|\vec{w}\|\cos{\theta}$$

Therefore giving us:

If $\vec{a}\times\vec{c}=\vec{b}\times\vec{c}$, then $$\|\vec{a}\|\|\vec{c}\|\sin{\theta_{1}}\:\hat{n}_{1}=\|\vec{b}\|\|\vec{c}\|\sin{\theta_{2}}\:\hat{n}_{2} \implies \hat{n}_{1}=\pm\hat{n}_{2}$$

Therefore we can write:

$$\|\vec{a}\|\sin{\theta_{1}}=\pm\|\vec{b}\|\sin{\theta_{2}}$$

Using the geometric definition for the dot product we therefore have:

$$\vec{c}\cdot\vec{a}-\vec{c}\cdot\vec{b}=\|\vec{a}\|\|\vec{c}\|\cos{\theta_{1}}-\|\vec{b}\|\|\vec{c}\|\cos{\theta_{2}}=\|\vec{c}\|(\|\vec{a}\|\cos{\theta_{1}}-\|\vec{b}\|\cos{\theta_{2}})$$

However, I'm unsure of how to proceed further

Answer 1

Here's how I would do it.

First, $(a-b)\times c = 0$, so $$\|(a-b)\times c\|^2 = 0.$$

Now I will use the identity $$\|x \times y\|^2 = \|x\|^2\|y\|^2-(x\cdot y)^2.$$ (You can verify this identity by expanding both sides using your angle expressions for dot and cross product.)

In this case, $x = a-b$ and $y=c$, so $$0 = \|a-b\|^2\|c\|^2 - [(a-b)\cdot c]^2$$ or $$[(a-b)\cdot c]^2 = \|c\|^2\|a-b\|^2.$$

Taking the square root of both sides gives me your expression.

Answer 2

Hint: $c \times(a-b)=|c||a-b|sin(\alpha)=0 $ so $|c||a-b|=0$ or $sin(\alpha)=0 $

$c.a-c.b = c.(a-b)$