The distance of the plane `overset(to) (r ) ((2)/(7) hat(i) + (3)/(7) hat(j) - (6)/(7) hat(k) )= 1` from the origins is.

Options

A. 7 units
B. `(1)/(7)` units
C. `1` unit
D. none of these

Answer

To find the distance of the plane given by the equation

\[ \mathbf{r} \cdot \left(\frac{2}{7} \hat{i} + \frac{3}{7} \hat{j} - \frac{6}{7} \hat{k}\right) = 1 \]

from the origin, we can follow these steps:

### Step 1: Identify the normal vector of the plane
The equation of the plane can be expressed in the form

\[ \mathbf{r} \cdot \mathbf{n} = d \]

where \(\mathbf{n}\) is the normal vector and \(d\) is the distance from the origin. Here, the normal vector \(\mathbf{n}\) is

\[ \mathbf{n} = \frac{2}{7} \hat{i} + \frac{3}{7} \hat{j} - \frac{6}{7} \hat{k} \]

### Step 2: Calculate the magnitude of the normal vector
To check if \(\mathbf{n}\) is a unit vector, we need to calculate its magnitude:

\[ |\mathbf{n}| = \sqrt{\left(\frac{2}{7}\right)^2 + \left(\frac{3}{7}\right)^2 + \left(-\frac{6}{7}\right)^2} \]

Calculating each term:

\[ \left(\frac{2}{7}\right)^2 = \frac{4}{49}, \quad \left(\frac{3}{7}\right)^2 = \frac{9}{49}, \quad \left(-\frac{6}{7}\right)^2 = \frac{36}{49} \]

Adding these together:

\[ |\mathbf{n}| = \sqrt{\frac{4}{49} + \frac{9}{49} + \frac{36}{49}} = \sqrt{\frac{49}{49}} = \sqrt{1} = 1 \]

### Step 3: Use the distance formula
Since \(\mathbf{n}\) is a unit vector, the distance \(d\) from the origin to the plane is equal to the constant on the right side of the equation, which is 1.

Thus, we conclude:

\[ d = 1 \]

### Final Answer
The distance of the plane from the origin is \(1\).

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