The distance of the point (1,-1,1) from `vec(r)*(3 hat(i) - 4 hat(j)- 12 hat(k)) + 13 = 0 ` is

Options

A. `(8)/(13)` units
B. `(8)/(3)` units
C. `(13)/(3)` units
D. `(13)/(8)` units

Answer

To find the distance of the point \( P(1, -1, 1) \) from the plane given by the equation \( \vec{r} \cdot (3 \hat{i} - 4 \hat{j} - 12 \hat{k}) + 13 = 0 \), we can follow these steps:

### Step 1: Identify the normal vector and constant
The equation of the plane can be rewritten in the form \( \vec{r} \cdot \vec{n} = d \), where \( \vec{n} \) is the normal vector and \( d \) is a constant. From the given equation:

\[ \vec{r} \cdot (3 \hat{i} - 4 \hat{j} - 12 \hat{k}) = -13 \]

We can identify:
- Normal vector \( \vec{n} = 3 \hat{i} - 4 \hat{j} - 12 \hat{k} \)
- Constant \( d = -13 \)

### Step 2: Write the position vector of the point
The position vector \( \vec{a} \) of the point \( P(1, -1, 1) \) is:

\[ \vec{a} = 1 \hat{i} - 1 \hat{j} + 1 \hat{k} \]

### Step 3: Calculate the distance from the point to the plane
The distance \( D \) from a point to a plane is given by the formula:

\[ D = \frac{|\vec{a} \cdot \vec{n} - d|}{|\vec{n}|} \]

#### Step 3.1: Calculate \( \vec{a} \cdot \vec{n} \)
Now, we compute the dot product \( \vec{a} \cdot \vec{n} \):

\[ \vec{a} \cdot \vec{n} = (1 \hat{i} - 1 \hat{j} + 1 \hat{k}) \cdot (3 \hat{i} - 4 \hat{j} - 12 \hat{k}) \]

Calculating the dot product:

\[ = 1 \cdot 3 + (-1) \cdot (-4) + 1 \cdot (-12) \]
\[ = 3 + 4 - 12 = -5 \]

#### Step 3.2: Substitute into the distance formula
Now substitute \( \vec{a} \cdot \vec{n} \) and \( d \) into the distance formula:

\[ D = \frac{|-5 - (-13)|}{|\vec{n}|} \]
\[ = \frac{|-5 + 13|}{|\vec{n}|} \]
\[ = \frac{|8|}{|\vec{n}|} \]

### Step 4: Calculate the magnitude of the normal vector \( |\vec{n}| \)
The magnitude of the normal vector \( \vec{n} \) is calculated as follows:

\[ |\vec{n}| = \sqrt{3^2 + (-4)^2 + (-12)^2} \]
\[ = \sqrt{9 + 16 + 144} = \sqrt{169} = 13 \]

### Step 5: Final calculation of distance
Now substitute \( |\vec{n}| \) back into the distance formula:

\[ D = \frac{8}{13} \]

Thus, the distance of the point \( (1, -1, 1) \) from the plane is:

\[ \boxed{\frac{8}{13}} \]