If the distance between the planes `8x+12y-14z=2 and 4x+6y-7z=2` can be expressed in the form `(1)/(sqrt(N))`, where N is natural, then the value of `(N(N+1))/(2)` is

Options

A. `4950`
B. `5050`
C. `5150`
D. `5151`

Answer

To find the distance between the planes given by the equations \(8x + 12y - 14z = 2\) and \(4x + 6y - 7z = 2\), we can follow these steps:

### Step 1: Identify the equations of the planes
The equations of the planes are:
1. \(8x + 12y - 14z = 2\) (Equation 1)
2. \(4x + 6y - 7z = 2\) (Equation 2)

### Step 2: Rewrite the equations in a comparable form
We can rewrite Equation 1 to have the same coefficients as Equation 2. Dividing the entire first equation by 2 gives:
\[ 4x + 6y - 7z = 1 \]
Now we can denote this as Equation 3:
3. \(4x + 6y - 7z = 1\)

### Step 3: Confirm that the planes are parallel
Both planes can be expressed in the form \(Ax + By + Cz = D\) where the normal vectors are proportional:
- Normal vector of Plane 1: \((8, 12, -14)\)
- Normal vector of Plane 2: \((4, 6, -7)\)

Since \((8, 12, -14) = 2(4, 6, -7)\), the planes are indeed parallel.

### Step 4: Calculate the distance between the parallel planes
The distance \(D\) between two parallel planes given by \(Ax + By + Cz = D_1\) and \(Ax + By + Cz = D_2\) is given by the formula:
\[ D = \frac{|D_2 - D_1|}{\sqrt{A^2 + B^2 + C^2}} \]
For our planes:
- \(D_1 = 1\) (from Equation 3)
- \(D_2 = 2\) (from Equation 1)

Substituting into the formula gives:
\[ D = \frac{|2 - 1|}{\sqrt{4^2 + 6^2 + 7^2}} = \frac{1}{\sqrt{16 + 36 + 49}} = \frac{1}{\sqrt{101}} \]

### Step 5: Express the distance in the required form
We have found that:
\[ D = \frac{1}{\sqrt{101}} \]
This means \(N = 101\).

### Step 6: Calculate \(\frac{N(N+1)}{2}\)
Now we need to calculate:
\[ \frac{N(N+1)}{2} = \frac{101 \times (101 + 1)}{2} = \frac{101 \times 102}{2} = \frac{10302}{2} = 5151 \]

### Final Answer
Thus, the value of \(\frac{N(N+1)}{2}\) is \(5151\).

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