The equation of the plane which is at a distance of 5 units from the origin and whose normal has the d.c.'s `(6)/(7),-(2)/(7),-(3)/(7)` is

Options

A. 6x – 2y – 3z = 35
B. 6x + 2y - 3z = 50
C. 3x – 12y + 4z = 26
D. `2x+y+z=3sqrt11`

Answer

To find the equation of the plane that is at a distance of 5 units from the origin and has a normal with direction cosines \( \frac{6}{7}, -\frac{2}{7}, -\frac{3}{7} \), we can follow these steps:

### Step 1: Identify the normal vector
The direction cosines give us the components of the normal vector \( \vec{n} \). The normal vector \( \vec{n} \) can be expressed as:
\[ \vec{n} = 6\hat{i} - 2\hat{j} - 3\hat{k} \]

### Step 2: Write the equation of the plane
The general equation of a plane can be given in the form:
\[ ax + by + cz = d \]
where \( (a, b, c) \) are the components of the normal vector and \( d \) is the distance from the origin.

### Step 3: Substitute the normal vector components
From our normal vector \( \vec{n} = (6, -2, -3) \), we have:
- \( a = 6 \)
- \( b = -2 \)
- \( c = -3 \)

### Step 4: Use the distance from the origin
The distance \( d \) from the origin to the plane is given as 5 units. The equation of the plane can also be expressed using the formula:
\[ \frac{ax + by + cz}{\sqrt{a^2 + b^2 + c^2}} = d \]
Substituting the values:
\[ \frac{6x - 2y - 3z}{\sqrt{6^2 + (-2)^2 + (-3)^2}} = 5 \]

### Step 5: Calculate the magnitude of the normal vector
Calculating \( \sqrt{6^2 + (-2)^2 + (-3)^2} \):
\[ \sqrt{36 + 4 + 9} = \sqrt{49} = 7 \]

### Step 6: Substitute back into the equation
Now substituting back into the equation:
\[ \frac{6x - 2y - 3z}{7} = 5 \]
Multiplying both sides by 7 gives:
\[ 6x - 2y - 3z = 35 \]

### Final Equation of the Plane
Thus, the equation of the plane is:
\[ 6x - 2y - 3z = 35 \]