The perpendicular distance from the origin to the plane through the point (2,3,-1) and perpendicular to the vector `3hati-4hatj+7hatk` is

Options

A. `13/(sqrt(74))`
B. `(-13)/(sqrt(74))`
C. `13`
D. none of these

Explanation

To find the perpendicular distance from the origin to the plane that passes through the point (2, 3, -1) and is perpendicular to the vector \( \vec{n} = 3\hat{i} - 4\hat{j} + 7\hat{k} \), we can follow these steps:

### Step 1: Find the equation of the plane
The general equation of a plane can be given by:

\[ ax + by + cz = d \]

where \( (a, b, c) \) are the components of the normal vector to the plane. In this case, the normal vector is \( \vec{n} = (3, -4, 7) \). Thus, the equation of the plane can be written as:

\[ 3x - 4y + 7z = d \]

### Step 2: Substitute the point into the plane equation
We know the plane passes through the point \( (2, 3, -1) \). We can substitute these coordinates into the plane equation to find \( d \):

\[ 3(2) - 4(3) + 7(-1) = d \]

Calculating this gives:

\[ 6 - 12 - 7 = d \]
\[ d = 6 - 12 - 7 = -13 \]

So the equation of the plane is:

\[ 3x - 4y + 7z = -13 \]

### Step 3: Use the distance formula from a point to a plane
The formula for the distance \( D \) from a point \( (x_0, y_0, z_0) \) to the plane \( ax + by + cz = d \) is given by:

\[ D = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}} \]

For the origin \( (0, 0, 0) \), we have \( x_0 = 0, y_0 = 0, z_0 = 0 \). Substituting into the formula gives:

\[ D = \frac{|3(0) - 4(0) + 7(0) + 13|}{\sqrt{3^2 + (-4)^2 + 7^2}} \]

### Step 4: Simplify the distance calculation
Calculating the numerator:

\[ D = \frac{|0 + 0 + 0 + 13|}{\sqrt{9 + 16 + 49}} = \frac{13}{\sqrt{74}} \]

### Step 5: Final result
Thus, the perpendicular distance from the origin to the plane is:

\[ D = \frac{13}{\sqrt{74}} \]