Any first degree equation in x, y and z represents a plane i.e., `ax+by+cz+d=0` is the general equation of a plane. If p be the length of perpendicular from the origin to a plane and d.c. of this normal is `lt l, m n gt`, then the equation of the plane in the normal form is `lx+my+nz=p_(1)`.
Vector equation of a plane passing through a point having position vector `vec(a)` and normal to vector `vec(n)` is
`(vec(r)-vec(a))vec(n)=0" or "vec(r).vec(n)=vec(a).vec(n)`
Suppose a vector `vec(n)` of magnitude `2sqrt(3)` such that it makes equal acual angles with the co-ordinate axes. If `vec(n)` is a normal to the plane containing the point `(1,-1,2)`.
Cartesian equation of the plane is

Options

A. `x-y+z-1=0`
B. `x+y+z=2`
C. `3x-7y+z=7`
D. None of these

Answer

To find the Cartesian equation of the plane that contains the point (1, -1, 2) and has a normal vector of magnitude \(2\sqrt{3}\) making equal angles with the coordinate axes, we can follow these steps:

### Step 1: Determine the Normal Vector
Since the normal vector \(\vec{n}\) makes equal angles with the coordinate axes, we can express it in terms of its direction cosines. Let the direction cosines be \(l = m = n\).

Given that the magnitude of the normal vector is \(2\sqrt{3}\), we can write:
\[ |\vec{n}| = \sqrt{l^2 + m^2 + n^2} = 2\sqrt{3} \]
Since \(l = m = n\), we have:
\[ |\vec{n}| = \sqrt{3l^2} = 2\sqrt{3} \]
Squaring both sides gives:
\[ 3l^2 = 12 \implies l^2 = 4 \implies l = 2 \text{ (taking positive value since direction cosines are positive)} \]
Thus, \(l = m = n = 2\).

### Step 2: Write the Normal Vector
The normal vector can be expressed as:
\[ \vec{n} = 2\hat{i} + 2\hat{j} + 2\hat{k} \]

### Step 3: Use the Point-Normal Form of the Plane Equation
The point-normal form of the equation of a plane is given by:
\[ (\vec{r} - \vec{a}) \cdot \vec{n} = 0 \]
Where \(\vec{a}\) is the position vector of the point (1, -1, 2):
\[ \vec{a} = 1\hat{i} - 1\hat{j} + 2\hat{k} \]
And \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\).

### Step 4: Substitute into the Plane Equation
Substituting \(\vec{r}\), \(\vec{a}\), and \(\vec{n}\) into the equation:
\[ [(x\hat{i} + y\hat{j} + z\hat{k}) - (1\hat{i} - 1\hat{j} + 2\hat{k})] \cdot (2\hat{i} + 2\hat{j} + 2\hat{k}) = 0 \]
This simplifies to:
\[ [(x - 1)\hat{i} + (y + 1)\hat{j} + (z - 2)\hat{k}] \cdot (2\hat{i} + 2\hat{j} + 2\hat{k}) = 0 \]

### Step 5: Compute the Dot Product
Calculating the dot product:
\[ 2(x - 1) + 2(y + 1) + 2(z - 2) = 0 \]
Simplifying this gives:
\[ 2x - 2 + 2y + 2 + 2z - 4 = 0 \]
\[ 2x + 2y + 2z - 4 = 0 \]

### Step 6: Divide by 2
Dividing the entire equation by 2:
\[ x + y + z - 2 = 0 \]
Thus, the Cartesian equation of the plane is:
\[ x + y + z = 2 \]