MHB Finding the Distance from Point P to AC on an ABCD.EFGH Cube

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The discussion centers on finding the distance from point P to line AC in an ABCD.EFGH cube with a side length of 8 cm. Initially, there was confusion regarding the placement of point P, which was clarified to be within AH. The coordinates for the vertices of the cube were established, leading to the calculation of point P as (8,6,6). The distance from P to AC was determined using calculus, resulting in a minimum distance of 3√6 cm. Alternative methods, such as algebraic manipulation and trigonometry, were also mentioned as valid approaches to solving the problem.
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Given an ABCD.EFGH cube whose its side length is 8 cm. The point P is within AB so that AP = 3PH. The distance between P to AC is ...
A. $$2\sqrt3$$ cm
B. $$3\sqrt3$$ cm
C. $$2\sqrt6$$ cm
D. $$3\sqrt6$$ cm
E. $$4\sqrt6$$ cm

So AH is $$8\sqrt2$$ cm and PH = $$6\sqrt2$$. What do I do now? Is the triangle HPC a right triangle?
 
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Monoxdifly said:
Given an ABCD.EFGH cube whose its side length is 8 cm. The point P is within AB so that AP = 3PH. The distance between P to AC is ...
A. $$2\sqrt3$$ cm
B. $$3\sqrt3$$ cm
C. $$2\sqrt6$$ cm
D. $$3\sqrt6$$ cm
E. $$4\sqrt6$$ cm

So AH is $$8\sqrt2$$ cm and PH = $$6\sqrt2$$. What do I do now? Is the triangle HPC a right triangle?
This sounds wrong. If P lies on AB then it must be closer to A than to H. So the condition AP = 3PH is impossible.
 
Sorry, I meant P is within AH. Sigh, why did I make a lot of typos last night?
 
My method is to use coordinates, taking the vertices of the cube as
$A = (8,0,0),$
$B = (0,0,0),$
$C = (0,8,0),$
$D = (8,8,0),$
$E = (8,0,8),$
$F = (0,0,8),$
$G = (0,8,8),$
$H = (8,8,8).$
(Of course, that is not the only way to assign the vertices. I was working from a sketch in which it was convenient to take $B$ as the origin.)

The point $P$ is then $(8,6,6)$. A point on $AC$ is given by $(t,8-t,0)$. The distance $d$ from that point to $P$ satisfies $d^2 = (t-8)^2 + (2-t)^2 + 6^2$. If you minimise that by calculus, you find that $t=5$, and $d = 3\sqrt6$.

Edit. If you don't want to use calculus, you can do it algebraically by completing the square: $$d^2 = (t-8)^2 + (2-t)^2 + 6^2 = 2t^2 - 20t + 104 = 2(t-5)^2 + 54.$$ That clearly has minimum value 54, when $t=5$. So $d = \sqrt{54} = 3\sqrt6$.
 
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I ended up solving the problem with trigonometry instead of algebra and calculus. Thanks, anyway. It's always nice to see different approaches.
 
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