Let `veca=hati-3hatj+4hatk`, `vecB=6hati+4hatj-8hatk`, `vecC=5hati+2hatj+5hatk` and a vector `vecR` satisfies `vecR xx vecB = vecC xx vecB`, `vecR.vecA=0`, then the value of `(|vecB|)/(|vecR-vecC|)` is

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem step by step, we will follow the given conditions and use vector operations.

### Given:
- \(\vec{A} = \hat{i} - 3\hat{j} + 4\hat{k}\)
- \(\vec{B} = 6\hat{i} + 4\hat{j} - 8\hat{k}\)
- \(\vec{C} = 5\hat{i} + 2\hat{j} + 5\hat{k}\)
- \(\vec{R} \times \vec{B} = \vec{C} \times \vec{B}\)
- \(\vec{R} \cdot \vec{A} = 0\)

### Step 1: Use the cross product condition
From the condition \(\vec{R} \times \vec{B} = \vec{C} \times \vec{B}\), we can rearrange it to:
\[ \vec{R} \times \vec{B} - \vec{C} \times \vec{B} = \vec{0} \]
This implies that:
\[ (\vec{R} - \vec{C}) \times \vec{B} = \vec{0} \]
This means that \(\vec{R} - \vec{C}\) is collinear with \(\vec{B}\). Therefore, we can write:
\[ \vec{R} - \vec{C} = \lambda \vec{B} \quad \text{for some scalar } \lambda \]
Hence, we can express \(\vec{R}\) as:
\[ \vec{R} = \lambda \vec{B} + \vec{C} \]

### Step 2: Use the dot product condition
Now, we use the condition \(\vec{R} \cdot \vec{A} = 0\):
\[ (\lambda \vec{B} + \vec{C}) \cdot \vec{A} = 0 \]
Expanding this gives:
\[ \lambda (\vec{B} \cdot \vec{A}) + \vec{C} \cdot \vec{A} = 0 \]
Thus, we can solve for \(\lambda\):
\[ \lambda = -\frac{\vec{C} \cdot \vec{A}}{\vec{B} \cdot \vec{A}} \]

### Step 3: Calculate \(\vec{C} \cdot \vec{A}\) and \(\vec{B} \cdot \vec{A}\)
Calculating \(\vec{C} \cdot \vec{A}\):
\[ \vec{C} \cdot \vec{A} = (5)(1) + (2)(-3) + (5)(4) = 5 - 6 + 20 = 19 \]

Calculating \(\vec{B} \cdot \vec{A}\):
\[ \vec{B} \cdot \vec{A} = (6)(1) + (4)(-3) + (-8)(4) = 6 - 12 - 32 = -38 \]

### Step 4: Substitute the dot products into \(\lambda\)
Now substituting these values into the equation for \(\lambda\):
\[ \lambda = -\frac{19}{-38} = \frac{19}{38} = \frac{1}{2} \]

### Step 5: Find \(|\vec{R} - \vec{C}|\)
From \(\vec{R} - \vec{C} = \lambda \vec{B}\):
\[ |\vec{R} - \vec{C}| = |\lambda \vec{B}| = \left|\frac{1}{2} \vec{B}\right| = \frac{1}{2} |\vec{B}| \]

### Step 6: Find the value of \(\frac{|\vec{B}|}{|\vec{R} - \vec{C}|}\)
Now we can find the required ratio:
\[ \frac{|\vec{B}|}{|\vec{R} - \vec{C}|} = \frac{|\vec{B}|}{\frac{1}{2} |\vec{B}|} = 2 \]

### Final Answer:
The value of \(\frac{|\vec{B}|}{|\vec{R} - \vec{C}|}\) is \(2\).

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