Find the equation of the plane which is at a distance of `sqrt(29)` units from origin and the perpendicular vector from origin to this plane is `(4hati-2hatj+3hatk)`.

Answer

To find the equation of the plane that is at a distance of \(\sqrt{29}\) units from the origin and has a normal vector \(\vec{n} = 4\hat{i} - 2\hat{j} + 3\hat{k}\), we can follow these steps:

### Step 1: Identify the components
The normal vector \(\vec{n}\) is given as:
\[ \vec{n} = 4\hat{i} - 2\hat{j} + 3\hat{k} \]
The distance \(D\) from the origin to the plane is given as:
\[ D = \sqrt{29} \]

### Step 2: Find the magnitude of the normal vector
The magnitude of the normal vector \(\vec{n}\) is calculated as follows:
\[ |\vec{n}| = \sqrt{(4)^2 + (-2)^2 + (3)^2} \]
Calculating each term:
\[ = \sqrt{16 + 4 + 9} = \sqrt{29} \]

### Step 3: Find the unit normal vector
The unit normal vector \(\hat{n}\) is given by:
\[ \hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{4\hat{i} - 2\hat{j} + 3\hat{k}}{\sqrt{29}} \]

### Step 4: Write the general equation of the plane
The general equation of a plane can be expressed as:
\[ \vec{r} \cdot \hat{n} = D \]
Substituting \(\hat{n}\) and \(D\):
\[ \vec{r} \cdot \left(\frac{4\hat{i} - 2\hat{j} + 3\hat{k}}{\sqrt{29}}\right) = \sqrt{29} \]

### Step 5: Clear the fraction
To eliminate the fraction, multiply both sides by \(\sqrt{29}\):
\[ \vec{r} \cdot (4\hat{i} - 2\hat{j} + 3\hat{k}) = 29 \]

### Step 6: Write the final equation of the plane
The equation of the plane can be expressed as:
\[ 4x - 2y + 3z = 29 \]

Thus, the equation of the plane is:
\[ 4x - 2y + 3z = 29 \]