The distance between the line `r=2hat(i)-2hat(j)+3hat(k)+lambda(hat(i)-hat(j)+4hat(k))` and the plane `rcdot(hat(i)+5hat(j)+hat(k))=5,` is

Options

A. `(10)/(3)`
B. `(3)/(10)`
C. `(10)/(3sqrt(3))`
D. `(10)/(9)`

Answer

To find the distance between the given line and the plane, we will follow a systematic approach.

### Step-by-Step Solution

1. **Identify the Line and Plane Equations:**
- The line is given by:
\[ \mathbf{r} = 2\hat{i} - 2\hat{j} + 3\hat{k} + \lambda(\hat{i} - \hat{j} + 4\hat{k}) \]
- The plane is given by:
\[ \mathbf{r} \cdot (\hat{i} + 5\hat{j} + \hat{k}) = 5 \]

2. **Convert the Line Equation to Symmetric Form:**
- The line can be expressed in symmetric form as:
\[ \frac{x - 2}{1} = \frac{y + 2}{-1} = \frac{z - 3}{4} \]

3. **Identify Direction Vector and Point on the Line:**
- The direction vector of the line, \(\mathbf{A}\), is:
\[ \mathbf{A} = \hat{i} - \hat{j} + 4\hat{k} \]
- A point on the line, \(\mathbf{B}\), is:
\[ \mathbf{B} = 2\hat{i} - 2\hat{j} + 3\hat{k} \]

4. **Identify Normal Vector of the Plane:**
- The normal vector of the plane, \(\mathbf{n}\), is:
\[ \mathbf{n} = \hat{i} + 5\hat{j} + \hat{k} \]

5. **Check if the Line is Parallel to the Plane:**
- Calculate the dot product \(\mathbf{A} \cdot \mathbf{n}\):
\[ \mathbf{A} \cdot \mathbf{n} = (1)(1) + (-1)(5) + (4)(1) = 1 - 5 + 4 = 0 \]
- Since the dot product is zero, the line is parallel to the plane.

6. **Calculate the Distance from the Line to the Plane:**
- The formula for the distance \(d\) from a point to a plane is given by:
\[ d = \frac{|\mathbf{n} \cdot \mathbf{B} - c|}{|\mathbf{n}|} \]
- Here, \(c\) is the constant from the plane equation. From the plane equation \(\mathbf{r} \cdot \mathbf{n} = 5\), we have \(c = 5\).

7. **Calculate \(\mathbf{n} \cdot \mathbf{B}\):**
- Compute \(\mathbf{n} \cdot \mathbf{B}\):
\[ \mathbf{n} \cdot \mathbf{B} = (1)(2) + (5)(-2) + (1)(3) = 2 - 10 + 3 = -5 \]

8. **Substitute into the Distance Formula:**
- Now substitute the values into the distance formula:
\[ d = \frac{|-5 - 5|}{|\mathbf{n}|} = \frac{|-10|}{|\mathbf{n}|} \]

9. **Calculate the Magnitude of \(\mathbf{n}\):**
- The magnitude of \(\mathbf{n}\) is:
\[ |\mathbf{n}| = \sqrt{1^2 + 5^2 + 1^2} = \sqrt{1 + 25 + 1} = \sqrt{27} = 3\sqrt{3} \]

10. **Final Calculation of Distance:**
- Substitute back to find \(d\):
\[ d = \frac{10}{3\sqrt{3}} = \frac{10\sqrt{3}}{9} \]

### Final Answer
The distance between the line and the plane is:
\[ \frac{10}{3\sqrt{3}} \]