Find the equation of plane in Cartesian form which is at a distance of 5 unit from origin and its normal vector from origin is parallel to a vector formed by joining points A(1, 2, 3) and B(3, -4, -6).

Answer

To find the equation of the plane in Cartesian form that is at a distance of 5 units from the origin and has a normal vector parallel to the vector formed by joining points A(1, 2, 3) and B(3, -4, -6), we can follow these steps:

### Step 1: Find the vector AB
First, we need to find the vector that joins points A and B. The vector AB can be calculated as follows:

\[ \vec{AB} = \vec{B} - \vec{A} = (3 - 1) \hat{i} + (-4 - 2) \hat{j} + (-6 - 3) \hat{k} \]

Calculating the components:

\[ \vec{AB} = 2 \hat{i} - 6 \hat{j} - 9 \hat{k} \]

### Step 2: Determine the unit normal vector
Since the normal vector of the plane is parallel to vector AB, we can denote the normal vector \(\vec{n}\) as \(\vec{AB}\). To find the unit normal vector \(\hat{n}\), we need to calculate the magnitude of \(\vec{AB}\):

\[ |\vec{AB}| = \sqrt{(2)^2 + (-6)^2 + (-9)^2} = \sqrt{4 + 36 + 81} = \sqrt{121} = 11 \]

Now, we can find the unit normal vector:

\[ \hat{n} = \frac{\vec{AB}}{|\vec{AB}|} = \frac{2 \hat{i} - 6 \hat{j} - 9 \hat{k}}{11} = \frac{2}{11} \hat{i} - \frac{6}{11} \hat{j} - \frac{9}{11} \hat{k} \]

### Step 3: Use the equation of the plane
The equation of a plane in vector form can be expressed as:

\[ \vec{r} \cdot \hat{n} = d \]

where \(d\) is the distance from the origin to the plane. Given that the distance is 5 units, we have \(d = 5\).

### Step 4: Substitute into the equation
Let \(\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}\). The dot product becomes:

\[ \vec{r} \cdot \hat{n} = \left(x \hat{i} + y \hat{j} + z \hat{k}\right) \cdot \left(\frac{2}{11} \hat{i} - \frac{6}{11} \hat{j} - \frac{9}{11} \hat{k}\right) \]

Calculating the dot product:

\[ \frac{2}{11} x - \frac{6}{11} y - \frac{9}{11} z = 5 \]

### Step 5: Clear the fraction
To eliminate the fraction, multiply the entire equation by 11:

\[ 2x - 6y - 9z = 55 \]

### Final Equation
Thus, the equation of the plane in Cartesian form is:

\[ 2x - 6y - 9z = 55 \]

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