Find the vector equation of a plane which is at a distance of 4 units from the origin and which has `(2hati-3hatj+6hatk)` as the normal vector.

Options

A. `vecr.(2hati-3hatj+6hatk)=28`
B. `vecr.(2hati+3hatj+6hatk)=28`
C. `vecr.(2hati-3hatj+6hatk)=0`
D. `vecr.(2hati-3hatj+6hatk)=-28`

Answer

To find the vector equation of a plane that is at a distance of 4 units from the origin and has the normal vector \( \vec{n} = 2\hat{i} - 3\hat{j} + 6\hat{k} \), we can follow these steps:

### Step 1: Identify the normal vector
The normal vector of the plane is given as:
\[ \vec{n} = 2\hat{i} - 3\hat{j} + 6\hat{k} \]

### Step 2: Calculate the magnitude of the normal vector
The magnitude of the normal vector \( \vec{n} \) is calculated as follows:
\[ |\vec{n}| = \sqrt{(2)^2 + (-3)^2 + (6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]

### Step 3: Find the unit normal vector
The unit normal vector \( \hat{n} \) is obtained by dividing the normal vector by its magnitude:
\[ \hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{2\hat{i} - 3\hat{j} + 6\hat{k}}{7} = \frac{2}{7}\hat{i} - \frac{3}{7}\hat{j} + \frac{6}{7}\hat{k} \]

### Step 4: Use the formula for the equation of the plane
The equation of the plane can be expressed using the unit normal vector and the distance from the origin. The formula is:
\[ \hat{n} \cdot \vec{r} = d \]
where \( d \) is the distance from the origin, which is given as 4 units. Thus, we have:
\[ \hat{n} \cdot \vec{r} = 4 \]

### Step 5: Substitute the values into the equation
Substituting the unit normal vector into the equation:
\[ \left(\frac{2}{7}\hat{i} - \frac{3}{7}\hat{j} + \frac{6}{7}\hat{k}\right) \cdot \vec{r} = 4 \]

### Step 6: Multiply through by 7 to eliminate the fraction
To simplify, multiply both sides by 7:
\[ \vec{r} \cdot (2\hat{i} - 3\hat{j} + 6\hat{k}) = 28 \]

### Final Equation of the Plane
Thus, the vector equation of the plane is:
\[ \vec{r} \cdot (2\hat{i} - 3\hat{j} + 6\hat{k}) = 28 \]

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