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

Answer

To find the distance between the given line and the plane, we can follow these steps:

### Step 1: Identify the line and the 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:
\[ \mathbf{r} \cdot (\hat{i} + 5\hat{j} + \hat{k}) = 5 \]

### Step 2: Determine the direction ratios of the line and the normal vector of the plane
The direction ratios of the line can be extracted from the vector part:
\[ \mathbf{d} = \hat{i} - \hat{j} + 4\hat{k} \]
The normal vector of the plane can be derived from the coefficients of the plane equation:
\[ \mathbf{n} = \hat{i} + 5\hat{j} + \hat{k} \]

### Step 3: Check if the line is parallel to the plane
To check if the line is parallel to the plane, we can see if the direction vector of the line is perpendicular to the normal vector of the plane by taking their dot product:
\[ \mathbf{d} \cdot \mathbf{n} = (1)(1) + (-1)(5) + (4)(1) = 1 - 5 + 4 = 0 \]
Since the dot product is zero, the line is parallel to the plane.

### Step 4: Find a point on the line
To find a point on the line, we can set \( \lambda = 0 \):
\[ \mathbf{A} = (2, -2, 3) \]

### Step 5: Write the equation of the plane in standard form
The equation of the plane can be rearranged as:
\[ x + 5y + z - 5 = 0 \]
Here, \( A = 1, B = 5, C = 1, D = 5 \).

### Step 6: Use the distance formula from a point to a plane
The distance \( d \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz - D = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + Cz_0 - D|}{\sqrt{A^2 + B^2 + C^2}} \]
Substituting \( (x_0, y_0, z_0) = (2, -2, 3) \):
\[ d = \frac{|1(2) + 5(-2) + 1(3) - 5|}{\sqrt{1^2 + 5^2 + 1^2}} \]

### Step 7: Simplify the numerator
Calculating the numerator:
\[ = |2 - 10 + 3 - 5| = |-10| = 10 \]

### Step 8: Simplify the denominator
Calculating the denominator:
\[ \sqrt{1 + 25 + 1} = \sqrt{27} = 3\sqrt{3} \]

### Step 9: Calculate the distance
Putting it all together:
\[ d = \frac{10}{3\sqrt{3}} = \frac{10\sqrt{3}}{9} \]

### Final Answer
The distance between the line and the plane is:
\[ \frac{10\sqrt{3}}{9} \text{ units} \]

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