Question

If the three successive vertices of a parallelogram have the position vectors as, $A(-3,-2,0);B(3,-3,1)$ and $C(5,0,2).$ Then find :
the angle between $\overrightarrow{AC}$ and $\overrightarrow{BD}$

Answer 1

$A(-3,-2,0)B(3,-3,1)C(2,0,2)$

same $ABCD$ is a parallelogram

$D_x=A_x+C_x+B_x$

$\Rightarrow D=\left(\right(-3+5-3),(-2+10.\left(9\right)),D(2,1\left)\right)$

$D=(-1,1,1)$

angle between $\overrightarrow{AC}$ and $\overrightarrow{BC}$:

$\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=(5\hat{i}+2\hat{k})-(-3\hat{i}-2\hat{j})$

$\overrightarrow{AC}=8\hat{i}+2\hat{j}+2\hat{k}$

$\overrightarrow{BD}=\overrightarrow{OD}-\overrightarrow{OB}=(-\hat{i}+\hat{j}+\hat{k})-(3\hat{i}-3\hat{j}+\hat{k})$

$=(4\hat{i}+4\hat{j})$

$|\overrightarrow{AC}+\overrightarrow{BD}|=(8\hat{i}+2\hat{j}+2\hat{k})+(-4\hat{i}+4\hat{j})$

$=(4\hat{i}-\hat{j}+2\hat{k})$

$(\overrightarrow{AC}+\overrightarrow{BD}{)}^2=|\overrightarrow{AC}{|}^2+|\overrightarrow{BD}{|}^2+2|AC||\overrightarrow{BD}|\cos (\overrightarrow{AC},\overrightarrow{BD})$

$|AC|=\sqrt{64+4+4}=\sqrt{72}$

$|BD|=\sqrt{16+16}=-\sqrt{32}$

$|\overrightarrow{AC}+\overrightarrow{BD}|=\sqrt{16+36+7}=\sqrt{56}$

$(\sqrt{56}{)}^2=(\sqrt{72}{)}^2+(\sqrt{32}{)}^2+2\sqrt{\left(32\right)\left(72\right)}\cos \theta$

$56=72+32+2\sqrt{\left(64\right)\left(36\right)}\cos \theta$

$-48=2\times 8\times 6\cos \theta$

$\cos \theta =-1/2$

$\theta ={120}^o$