value of $\vec{AC}\cdot \vec{BD}$

Question

If $A,B,C,D$ be $4$ points in a space and satisfy $|\vec{AB}|=3,|\vec{BC}|=7,|\vec{CD}|=11,|\vec{DA}|=9,$ Then value of $\vec{AC}\cdot \vec{BD}=$

what i try

Let position vector of $A(0)\;,\;B(\vec{b})\;,\; C(\vec{c}),D(\vec{d})$

Then $\vec{AC}\cdot \vec{BD}=|\vec{AC}||\vec{BD}|\cos \theta$

where $\theta$ ia an angle between $\vec{AC}$ and $\vec{BD}$

How do i solve it help me please

Answer 1

Use

$$\mathbf {AC} = \mathbf {AB} - \mathbf {CB} = \mathbf {AD} - \mathbf {CD}$$ $$\mathbf {BD} = \mathbf {AD} - \mathbf {AB} = \mathbf {CD} - \mathbf {CB}$$

to evaluate $\mathbf {AC}\cdot \mathbf {BD} $

$$2\mathbf {AC}\cdot \mathbf {BD}= \mathbf {AC}\cdot (\mathbf {AD} - \mathbf {AB}) + \mathbf {AC}\cdot (\mathbf {CD} - \mathbf {CB})$$ $$= \mathbf {AC}\cdot (\mathbf {AD} + \mathbf {CD}) - \mathbf {AC}\cdot (\mathbf {AB} + \mathbf {CB})$$ $$= (\mathbf {AD} - \mathbf {CD})\cdot (\mathbf {AD} + \mathbf {CD}) - (\mathbf {AB} - \mathbf {CB} )\cdot (\mathbf {AB} + \mathbf {CB})$$ $$= |\mathbf {AD}|^2- |\mathbf {CD}|^2 -|\mathbf {AB}|^2 + |\mathbf {CB}|^2$$ $$= 9^2- 11^2 -3^2 + 7^2=0$$

Answer 2

There was actually a sign mistake in the last step of my calculation, now it's fixed.

After giving it a thought, I noticed that $$3^2+11^2=\|\vec{AB}\|^2+\|\vec{CD}\|^2=\|\vec{BC}\|^2+\|\vec{AD}\|^2=7^2+9^2$$ thus $$\|\vec{AB}\|^2-\|\vec{AD}\|^2=\|\vec{BC}\|^2-\|\vec{CD}\|^2$$ this implies that $$(\vec{AB}+\vec{AD})\cdot(\vec{AB}-\vec{AD})=(\vec{BC}+\vec{CD})(\vec{BC}-\vec{CD})$$ and now by rewriting the expression we end up with \begin{align*} (\vec{AB}+\vec{AD})\cdot (\vec{DB})&=(\vec{BC}-\vec{CD})\vec{BD}\\ (-\vec{AB}-\vec{AD})\cdot (\vec{BD})&=(\vec{BC}-\vec{CD})\vec{BD}\\ 0&=(\vec{BC}-\vec{CD}\color{red}{+}\vec{AB}\color{red}{+}\vec{AD})\cdot \vec{BD}\\ 0&=(\vec{AB}+\vec{BC}+\vec{AD}+\vec{DC})\cdot \vec{BD}\\ 0&=2\vec{AC}\cdot \vec{BD} \end{align*} which gives the result.

Thi can analogously be obtained from \begin{align*} \|\vec{AB}\|^2-\|\vec{BC}\|^2&=\|\vec{AD}\|^2-\|\vec{CD}\|^2\\ \end{align*}