Given parallelepiped $ABCDA_1B_1C_1D_1$ find distance between $AC_1$ and $BD$.
$A = (-1,2,-3)$, $B = (1,4,1)$, $D= (3,2,-7)$, $A_1= (-7,0,0)$
So what I am doing is I tried to find coordinate of $C$ then to find coordinate of $C_1$.So in order to find coordinate of $C_1$ we need to equal length of $CC_1$ to $AA_1$ but I can't do anything after this. Maybe I am on the wrong way and there is much easier solution?Any help
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In a parallelepiped, there is a special relation:
$$ \vec{C}_{1}=\vec{B}+\vec{D}+\vec{A}_{1}-2\vec{A} $$
After you find $\vec{C}_{1}$, next we need to find the minimum distance.
To do this, find a vector that is orthogonal to both $\vec{C}_{1}-\vec{A}$ and $\vec{B}-\vec{D}$ (hint: cross product). Normalize this vector. Finally, find the dot product of this normalized vector to $\vec{B}-\vec{A}$.
Since this is a parallelipiped, you can use relationship between equal sides: use $C-B=D-A$ to get $C$. Then use $C_1-C=A_1-A$. So $$C-(1,4,1)=(3,2,-7)-(-1,2,-3)$$ $$C=(5,4,-3)$$ I hope you can continue from here.