If `vec(A)=hati-3hatj+4hatk,vec(B)=6hati+4hatj-8hatk,vec( C )=5hati+2hatj-5hatk` and a vector `vec( R )` satisfies `vec( R )xx vec(B)=vec( C )xx vec(B),vec( R ).vec(A)=0`, then `(|vec(B)|)/(|vec( R )-vec( C )|)` is equal to ______________

Answer

To solve the problem, we will follow the steps outlined in the video transcript.

### Step 1: Define the vectors
We have the following vectors:
- \(\vec{A} = \hat{i} - 3\hat{j} + 4\hat{k}\)
- \(\vec{B} = 6\hat{i} + 4\hat{j} - 8\hat{k}\)
- \(\vec{C} = 5\hat{i} + 2\hat{j} - 5\hat{k}\)

### Step 2: Use the given conditions
We know that:
1. \(\vec{R} \times \vec{B} = \vec{C} \times \vec{B}\)
2. \(\vec{R} \cdot \vec{A} = 0\)

### Step 3: Cross product condition
From the first condition, we can rewrite it as:
\[ \vec{R} \times \vec{B} - \vec{C} \times \vec{B} = \vec{0} \]
Crossing both sides with \(\vec{A}\):
\[ \vec{A} \times (\vec{R} \times \vec{B}) = \vec{A} \times (\vec{C} \times \vec{B}) \]

### Step 4: Apply the vector triple product identity
Using the vector triple product identity \(\vec{X} \times (\vec{Y} \times \vec{Z}) = (\vec{X} \cdot \vec{Z})\vec{Y} - (\vec{X} \cdot \vec{Y})\vec{Z}\), we get:
\[ (\vec{A} \cdot \vec{B}) \vec{R} - (\vec{A} \cdot \vec{R}) \vec{B} = (\vec{A} \cdot \vec{C}) \vec{B} - (\vec{A} \cdot \vec{B}) \vec{C} \]

### Step 5: Substitute \(\vec{R} \cdot \vec{A} = 0\)
Since \(\vec{R} \cdot \vec{A} = 0\), we can simplify:
\[ (\vec{A} \cdot \vec{B}) \vec{R} = (\vec{A} \cdot \vec{C}) \vec{B} + (\vec{A} \cdot \vec{B}) \vec{C} \]

### Step 6: Solve for \(\vec{R}\)
Rearranging gives:
\[ \vec{R} = \frac{(\vec{A} \cdot \vec{C}) \vec{B}}{(\vec{A} \cdot \vec{B})} \]

### Step 7: Calculate \(\vec{A} \cdot \vec{B}\) and \(\vec{A} \cdot \vec{C}\)
Calculating:
- \(\vec{A} \cdot \vec{B} = (1)(6) + (-3)(4) + (4)(-8) = 6 - 12 - 32 = -38\)
- \(\vec{A} \cdot \vec{C} = (1)(5) + (-3)(2) + (4)(-5) = 5 - 6 - 20 = -21\)

### Step 8: Substitute values into \(\vec{R}\)
Substituting these values into the equation for \(\vec{R}\):
\[ \vec{R} = \frac{-21 \vec{B}}{-38} = \frac{21}{38} \vec{B} \]

### Step 9: Find \(|\vec{R} - \vec{C}|\)
Now we need to find \(|\vec{R} - \vec{C}|\):
\[ \vec{R} - \vec{C} = \frac{21}{38} \vec{B} - \vec{C} \]

### Step 10: Calculate the modulus
To find the modulus, we will need to calculate \(|\vec{B}|\) and \(|\vec{R} - \vec{C}|\).

1. Calculate \(|\vec{B}|\):
\[ |\vec{B}| = \sqrt{6^2 + 4^2 + (-8)^2} = \sqrt{36 + 16 + 64} = \sqrt{116} = 2\sqrt{29} \]

2. Calculate \(|\vec{R} - \vec{C}|\):
We can express \(|\vec{R} - \vec{C}|\) in terms of \(|\vec{B}|\):
\[ |\vec{R} - \vec{C}| = \left|\frac{21}{38} \vec{B} - \vec{C}\right| \]

### Step 11: Find the ratio
Finally, we need to find:
\[ \frac{|\vec{B}|}{|\vec{R} - \vec{C}|} \]

### Final Calculation
After substituting and simplifying, we find:
\[ \frac{|\vec{B}|}{|\vec{R} - \vec{C}|} = \frac{38}{21} \]

Thus, the final answer is:
\[ \frac{|\vec{B}|}{|\vec{R} - \vec{C}|} = \frac{38}{21} \]