Cube Distance Problem: Finding A1 To C1P Distance

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Let's dive into this geometry problem involving a cube, a point on one of its edges, and the distance from a vertex to a line. This problem might seem tricky at first, but we'll break it down step by step to make it super clear. We're going to find the distance from vertex A1 to the line C1P in a cube ABCDA1B1C1D1. Point P is located on the edge AD such that the ratio of AP to AD is 1:4, and the length of the cube's edge is given as 'a'.

Setting Up the Problem

Alright, guys, first things first, let's visualize the cube. Imagine a cube ABCDA1B1C1D1. Now, picture point P sitting on the edge AD. We know that AP is one-fourth the length of AD. Since the cube's edge length is 'a', this means AP = a/4 and PD = 3a/4. Our mission is to find the shortest distance from vertex A1 to the line C1P. This distance is simply the perpendicular distance from the point A1 to the line C1P. To solve this problem effectively, we will use a combination of spatial reasoning and the distance formula in three dimensions.

Visualizing the Cube and Key Points

To get a handle on this, let's think about the geometry. We can set up a coordinate system to make things easier. Place the cube in a 3D coordinate system with vertex A at the origin (0, 0, 0). Then:

  • A (0, 0, 0)
  • B (a, 0, 0)
  • D (0, a, 0)
  • A1 (0, 0, a)
  • C1 (a, a, a)

Since AP:AD = 1:4, the coordinates of point P can be calculated. P lies on AD, so its y-coordinate will be (1/4) * a, and its x and z coordinates will be 0. Thus:

  • P (0, a/4, 0)

Finding the Vectors

Now, let's find the vectors we need. The vector C1P and the vector A1P will be crucial for our calculations. Remember, a vector from point A to point B is found by subtracting the coordinates of A from the coordinates of B.

  • C1P: This vector goes from C1 to P. C1 has coordinates (a, a, a), and P has coordinates (0, a/4, 0). So, C1P = P - C1 = (0 - a, a/4 - a, 0 - a) = (-a, -3a/4, -a).
  • A1P: This vector goes from A1 to P. A1 has coordinates (0, 0, a), and P has coordinates (0, a/4, 0). So, A1P = P - A1 = (0 - 0, a/4 - 0, 0 - a) = (0, a/4, -a).

Calculating the Distance

To find the distance from A1 to the line C1P, we'll use the formula that involves the cross product of two vectors. The distance (d) from a point to a line is given by:

d = |A1P x C1P| / |C1P|

Where:

  • A1P x C1P is the cross product of vectors A1P and C1P.
  • |A1P x C1P| is the magnitude of the cross product.
  • |C1P| is the magnitude of vector C1P.

Step 1: Calculate the Cross Product (A1P x C1P)

The cross product of two vectors (A1P = (x1, y1, z1) and C1P = (x2, y2, z2)) is given by:

A1P x C1P = (y1z2 - z1y2, z1x2 - x1z2, x1y2 - y1x2)

In our case, A1P = (0, a/4, -a) and C1P = (-a, -3a/4, -a). Plugging in the values:

A1P x C1P = ((a/4)(-a) - (-a)(-3a/4), (-a)(-a) - (0)(-a), (0)(-3a/4) - (a/4)(-a))

Simplifying:

A1P x C1P = (-a^2/4 - 3a^2/4, a^2 - 0, 0 + a^2/4)

A1P x C1P = (-a^2, a^2, a^2/4)

Step 2: Calculate the Magnitude of the Cross Product |A1P x C1P|

The magnitude of a vector (x, y, z) is given by √(x^2 + y^2 + z^2). So,

|A1P x C1P| = √((-a2)2 + (a2)2 + (a2/4)2)

|A1P x C1P| = √(a^4 + a^4 + a^4/16)

|A1P x C1P| = √(2a^4 + a^4/16)

|A1P x C1P| = √(32a^4/16 + a^4/16)

|A1P x C1P| = √(33a^4/16)

|A1P x C1P| = (a^2/4)√33

Step 3: Calculate the Magnitude of Vector C1P |C1P|

We have C1P = (-a, -3a/4, -a). The magnitude is:

|C1P| = √((-a)^2 + (-3a/4)^2 + (-a)^2)

|C1P| = √(a^2 + 9a^2/16 + a^2)

|C1P| = √(2a^2 + 9a^2/16)

|C1P| = √(32a^2/16 + 9a^2/16)

|C1P| = √(41a^2/16)

|C1P| = (a/4)√41

Step 4: Calculate the Distance (d)

Now we have all the pieces to calculate the distance:

d = |A1P x C1P| / |C1P|

d = ((a^2/4)√33) / ((a/4)√41)

d = (a^2√33) / (a√41)

d = a√33 / √41

To rationalize the denominator, multiply the numerator and denominator by √41:

d = (a√33 * √41) / (√41 * √41)

d = (a√1353) / 41

d = a√(33 * 41) / 41

d = a√1353 / 41

d = a * (√1353 / 41)

Step 5: Simplify the Expression

Let's try to simplify √1353. Prime factorization of 1353 gives us 3 * 11 * 41. So,

√1353 = √(3 * 11 * 41)

Since there are no perfect squares, we can't simplify it further. Thus, the distance is:

d = a * (√1353 / 41)

However, looking back at the given options, we need to express our answer in the form a * (some number). Let's revisit our calculation of the distance:

d = a√33 / √41

d = a (√33 / √41) * (√41 / √41)

d = a (√(33 * 41) / 41)

d = a √1353 / 41

Now, let's see if any of the answer choices match our result:

  1. a * 31/43
  2. a * 33/47
  3. a * 33/41
  4. a * 35/46
  5. a * 29/43

Our calculated distance is a * (√1353 / 41). Let's approximate √1353. Since 36^2 = 1296 and 37^2 = 1369, √1353 is close to 36.78.

So, our distance is approximately a * (36.78 / 41) β‰ˆ a * 0.897.

Now, let's calculate the values for the answer choices:

  1. 31/43 β‰ˆ 0.721
  2. 33/47 β‰ˆ 0.702
  3. 33/41 β‰ˆ 0.805
  4. 35/46 β‰ˆ 0.761
  5. 29/43 β‰ˆ 0.674

None of these exactly match 0.897. Let's double-check our calculations. A small mistake can throw off the entire answer. Going back through each step, we ensure the cross product, magnitudes, and final division are all correct. Upon careful review, no arithmetic errors are immediately apparent.

Reassessing the Calculations

The distance d = a (√1353 / 41) appears correct based on our steps. However, the numerical approximation doesn't directly match any of the given options. It's possible there's a need for further simplification or a clever manipulation we haven't yet considered.

Another approach is to square our result and compare it to the squared values of the options. Our result squared is:

(a√1353 / 41)^2 = a^2 * (1353 / 1681)

Now let's square the options and see if any are close:

  1. (31/43)^2 β‰ˆ 0.52
  2. (33/47)^2 β‰ˆ 0.49
  3. (33/41)^2 = 1089 / 1681 β‰ˆ 0.648
  4. (35/46)^2 β‰ˆ 0.579
  5. (29/43)^2 β‰ˆ 0.454

Our squared result is 1353/1681 β‰ˆ 0.805. Option 3 squared, 1089/1681 β‰ˆ 0.648, is not close. This suggests that our calculated answer a * (√1353 / 41) is indeed correct, and there might be a typo in the provided options.

Final Answer

After careful calculations and double-checking each step, the distance from vertex A1 to line C1P is:

d = a * (√1353 / 41)

Based on our approximation, this is closest to option 3, a * 33/41, though not an exact match. It is plausible that option 3 is the intended answer with a slight numerical discrepancy due to rounding or a minor error in the original problem statement or answer choices.

Guys, geometry problems can be real head-scratchers, but breaking them down step by step and carefully reviewing your work is the key to success!