mriteyl

2022-09-07

If A,B,C,D be 4 points in a space and satisfy $|\stackrel{\to }{AB}|=3,|\stackrel{\to }{BC}|=7,|\stackrel{\to }{CD}|=11,|\stackrel{\to }{DA}|=9,$ Then value of $\stackrel{\to }{AC}\cdot \stackrel{\to }{BD}=$
what i try
Let position vector of $A\left(0\right)\phantom{\rule{thickmathspace}{0ex}},\phantom{\rule{thickmathspace}{0ex}}B\left(\stackrel{\to }{b}\right)\phantom{\rule{thickmathspace}{0ex}},\phantom{\rule{thickmathspace}{0ex}}C\left(\stackrel{\to }{c}\right),D\left(\stackrel{\to }{d}\right)$
Then $\stackrel{\to }{AC}\cdot \stackrel{\to }{BD}=|\stackrel{\to }{AC}||\stackrel{\to }{BD}|\mathrm{cos}\theta$
where $\theta$ ia an angle between $\stackrel{\to }{AC}$ and $\stackrel{\to }{BD}$
How do i solve it help me please

Abigayle Lynn

Use
$\mathbf{A}\mathbf{C}=\mathbf{A}\mathbf{B}-\mathbf{C}\mathbf{B}=\mathbf{A}\mathbf{D}-\mathbf{C}\mathbf{D}$
$\mathbf{B}\mathbf{D}=\mathbf{A}\mathbf{D}-\mathbf{A}\mathbf{B}=\mathbf{C}\mathbf{D}-\mathbf{C}\mathbf{B}$
to evaluate $\mathbf{A}\mathbf{C}\cdot \mathbf{B}\mathbf{D}$
$2\mathbf{A}\mathbf{C}\cdot \mathbf{B}\mathbf{D}=\mathbf{A}\mathbf{C}\cdot \left(\mathbf{A}\mathbf{D}-\mathbf{A}\mathbf{B}\right)+\mathbf{A}\mathbf{C}\cdot \left(\mathbf{C}\mathbf{D}-\mathbf{C}\mathbf{B}\right)$
$=\mathbf{A}\mathbf{C}\cdot \left(\mathbf{A}\mathbf{D}+\mathbf{C}\mathbf{D}\right)-\mathbf{A}\mathbf{C}\cdot \left(\mathbf{A}\mathbf{B}+\mathbf{C}\mathbf{B}\right)$
$=\left(\mathbf{A}\mathbf{D}-\mathbf{C}\mathbf{D}\right)\cdot \left(\mathbf{A}\mathbf{D}+\mathbf{C}\mathbf{D}\right)-\left(\mathbf{A}\mathbf{B}-\mathbf{C}\mathbf{B}\right)\cdot \left(\mathbf{A}\mathbf{B}+\mathbf{C}\mathbf{B}\right)$
$=|\mathbf{A}\mathbf{D}{|}^{2}-|\mathbf{C}\mathbf{D}{|}^{2}-|\mathbf{A}\mathbf{B}{|}^{2}+|\mathbf{C}\mathbf{B}{|}^{2}$
$={9}^{2}-{11}^{2}-{3}^{2}+{7}^{2}=0$

Litzy Downs

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
${3}^{2}+{11}^{2}=‖\stackrel{\to }{AB}{‖}^{2}+‖\stackrel{\to }{CD}{‖}^{2}=‖\stackrel{\to }{BC}{‖}^{2}+‖\stackrel{\to }{AD}{‖}^{2}={7}^{2}+{9}^{2}$
thus
$‖\stackrel{\to }{AB}{‖}^{2}-‖\stackrel{\to }{AD}{‖}^{2}=‖\stackrel{\to }{BC}{‖}^{2}-‖\stackrel{\to }{CD}{‖}^{2}$
this implies that
$\left(\stackrel{\to }{AB}+\stackrel{\to }{AD}\right)\cdot \left(\stackrel{\to }{AB}-\stackrel{\to }{AD}\right)=\left(\stackrel{\to }{BC}+\stackrel{\to }{CD}\right)\left(\stackrel{\to }{BC}-\stackrel{\to }{CD}\right)$
and now by rewriting the expression we end up with
$\begin{array}{rl}\left(\stackrel{\to }{AB}+\stackrel{\to }{AD}\right)\cdot \left(\stackrel{\to }{DB}\right)& =\left(\stackrel{\to }{BC}-\stackrel{\to }{CD}\right)\stackrel{\to }{BD}\\ \left(-\stackrel{\to }{AB}-\stackrel{\to }{AD}\right)\cdot \left(\stackrel{\to }{BD}\right)& =\left(\stackrel{\to }{BC}-\stackrel{\to }{CD}\right)\stackrel{\to }{BD}\\ 0& =\left(\stackrel{\to }{BC}-\stackrel{\to }{CD}+\stackrel{\to }{AB}+\stackrel{\to }{AD}\right)\cdot \stackrel{\to }{BD}\\ 0& =\left(\stackrel{\to }{AB}+\stackrel{\to }{BC}+\stackrel{\to }{AD}+\stackrel{\to }{DC}\right)\cdot \stackrel{\to }{BD}\\ 0& =2\stackrel{\to }{AC}\cdot \stackrel{\to }{BD}\end{array}$
which gives the result.
Thi can analogously be obtained from
$\begin{array}{rl}‖\stackrel{\to }{AB}{‖}^{2}-‖\stackrel{\to }{BC}{‖}^{2}& =‖\stackrel{\to }{AD}{‖}^{2}-‖\stackrel{\to }{CD}{‖}^{2}\end{array}$

Do you have a similar question?