MHB What Is the Distance Between Lines HO and PB in a Cuboid?

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SUMMARY

The discussion centers on calculating the distance between the lines HO and PB in a cuboid ABCD.EFGH, where AB = 4 cm, BC = 3 cm, and CG = 5 cm. The initial calculation using the parallelogram formula yielded $$\frac{1}{5}\sqrt5$$ cm, which was later confirmed to be correct after clarifying that OBPH is the correct parallelogram. The potential answers provided were $$5\sqrt3$$ cm, $$5\sqrt2$$ cm, $$\sqrt5$$ cm, $$\frac{5}{2}\sqrt2$$ cm, and $$\frac{5}{3}\sqrt3$$ cm.

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Monoxdifly
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In an ABCD.EFGH cuboid with AB = 4 cm, BC = 3 cm, and CG = 5 cm there is a parallelogram OBFPH with O is located at the center of ABCD and P is located at the center of EFGH. The distance between the lines HO and PB is ...
A. $$5\sqrt3$$ cm
B. $$5\sqrt2$$ cm
C. $$\sqrt5$$ cm
D. $$\frac{5}{2}\sqrt2$$ cm
E. $$\frac{5}{3}\sqrt3$$ cm

By making use of the parallelogram formula, I got $$\frac{1}{5}\sqrt5$$. Do you guys get the same answer as me or any of the options?
 
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Monoxdifly said:
In an ABCD.EFGH cuboid with AB = 4 cm, BC = 3 cm, and CG = 5 cm there is a parallelogram OBFPH with O is located at the center of ABCD and P is located at the center of EFGH. The distance between the lines HO and PB is ...
A. $$5\sqrt3$$ cm
B. $$5\sqrt2$$ cm
C. $$\sqrt5$$ cm
D. $$\frac{5}{2}\sqrt2$$ cm
E. $$\frac{5}{3}\sqrt3$$ cm

By making use of the parallelogram formula, I got $$\frac{1}{5}\sqrt5$$. Do you guys get the same answer as me or any of the options?
OBFPH is not a parallelogram. Did you mean OBPH? If so, then I agree with your answer $\frac15\sqrt5$. But maybe you misread the question?
 
Sorry, I meant OBPH.
 

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