The plane whose vector equation is `vecr.(hati+2hatj-hatk)=1` and the line whose vector equation is `vecr=(-hati+hatj+hatk)+lamda(2hati+hatj+4hatk)` are parallel. Find the distance between them.

Options

A. `(2)/(sqrt5)` units
B. `(1)/(sqrt6)` unit
C. `(3)/(sqrt6)` units
D. `(4)/(sqrt6)` units

Answer

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

### Step 1: Identify the normal vector of the plane and the point on the line
The vector equation of the plane is given as:
\[ \vec{r} \cdot (\hat{i} + 2\hat{j} - \hat{k}) = 1 \]
From this equation, we can identify the normal vector \(\vec{n}\) of the plane as:
\[ \vec{n} = \hat{i} + 2\hat{j} - \hat{k} \]

The line is given by the vector equation:
\[ \vec{r} = (-\hat{i} + \hat{j} + \hat{k}) + \lambda(2\hat{i} + \hat{j} + 4\hat{k}) \]
From this equation, we can see that the point \(\vec{A}\) on the line is:
\[ \vec{A} = -\hat{i} + \hat{j} + \hat{k} \]

### Step 2: Find the value of \(d\) from the plane equation
The equation of the plane can be rewritten in the form:
\[ \vec{r} \cdot \vec{n} = d \]
Here, \(d = 1\).

### Step 3: Calculate the distance from the point to the plane
The formula for the distance \(D\) from a point \(\vec{A}\) to a plane defined by \(\vec{r} \cdot \vec{n} = d\) is given by:
\[ D = \frac{|\vec{A} \cdot \vec{n} - d|}{|\vec{n}|} \]

### Step 4: Calculate \(\vec{A} \cdot \vec{n}\)
Now, we need to calculate the dot product \(\vec{A} \cdot \vec{n}\):
\[ \vec{A} = -\hat{i} + \hat{j} + \hat{k} \]
\[ \vec{n} = \hat{i} + 2\hat{j} - \hat{k} \]
Calculating the dot product:
\[ \vec{A} \cdot \vec{n} = (-1)(1) + (1)(2) + (1)(-1) = -1 + 2 - 1 = 0 \]

### Step 5: Substitute into the distance formula
Now substituting into the distance formula:
\[ D = \frac{|0 - 1|}{|\vec{n}|} = \frac{1}{|\vec{n}|} \]

### Step 6: Calculate \(|\vec{n}|\)
Next, we need to calculate the magnitude of \(\vec{n}\):
\[ |\vec{n}| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \]

### Step 7: Final calculation of distance
Now substituting back into the distance formula:
\[ D = \frac{1}{\sqrt{6}} \]

Thus, the distance between the given plane and line is:
\[ \boxed{\frac{1}{\sqrt{6}}} \]