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

• MHB
• Monoxdifly
In summary, given an ABCD.EFGH cube whose side length is 8 cm, the point P is within AB so that AP = 3PH. The distance between P to AC is 3\sqrt6 cm.
Monoxdifly
MHB
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. $$\displaystyle 2\sqrt3$$ cm
B. $$\displaystyle 3\sqrt3$$ cm
C. $$\displaystyle 2\sqrt6$$ cm
D. $$\displaystyle 3\sqrt6$$ cm
E. $$\displaystyle 4\sqrt6$$ cm

So AH is $$\displaystyle 8\sqrt2$$ cm and PH = $$\displaystyle 6\sqrt2$$. What do I do now? Is the triangle HPC a right triangle?

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. $$\displaystyle 2\sqrt3$$ cm
B. $$\displaystyle 3\sqrt3$$ cm
C. $$\displaystyle 2\sqrt6$$ cm
D. $$\displaystyle 3\sqrt6$$ cm
E. $$\displaystyle 4\sqrt6$$ cm

So AH is $$\displaystyle 8\sqrt2$$ cm and PH = $$\displaystyle 6\sqrt2$$. 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 $d = \sqrt{54} = 3\sqrt6$.

Last edited:
I ended up solving the problem with trigonometry instead of algebra and calculus. Thanks, anyway. It's always nice to see different approaches.

## 1. What is the formula for finding the distance from Point P to AC on a cube?

The formula for finding the distance from Point P to AC on a cube is the Pythagorean theorem, which states that the square of the hypotenuse (longest side) of a right triangle is equal to the sum of the squares of the other two sides.

## 2. How do you determine the coordinates of Point P on the cube?

The coordinates of Point P on the cube can be determined by using the coordinates of the other three vertices of the triangle. Point P will be located at the midpoint of the line segment connecting the two points on opposite faces of the cube.

## 3. Can the distance from Point P to AC be negative?

No, the distance from Point P to AC cannot be negative. Distance is a measure of length, which is always a positive value.

## 4. How does the orientation of the cube affect the distance from Point P to AC?

The orientation of the cube does not affect the distance from Point P to AC. As long as the coordinates of the vertices and Point P remain the same, the distance will be the same regardless of the cube's orientation.

## 5. Is there a way to find the distance from Point P to AC without using the Pythagorean theorem?

No, the Pythagorean theorem is the only formula that can be used to find the distance from Point P to AC on a cube. Other methods, such as using trigonometric functions, are not applicable in this scenario.

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