Equation of the line passing through (1,1,1) and perpendicular to the plane 2x+3y-z-5=0 is

Options

A. `(x-1)/2=(y-1)/3 =(z-1)/1`
B. `(x-1)/2=(y-1)/3=(z-1)/-1`
C. `(x-1)/2=(y-1)/-1=(z-1)/1`
D. None of the above

Answer

To find the equation of the line passing through the point (1, 1, 1) and perpendicular to the plane given by the equation \(2x + 3y - z - 5 = 0\), we can follow these steps:

### Step 1: Identify the normal vector of the plane
The normal vector of a plane given by the equation \(Ax + By + Cz + D = 0\) is represented by the coefficients of \(x\), \(y\), and \(z\). For the plane \(2x + 3y - z - 5 = 0\), the coefficients are:
- \(A = 2\)
- \(B = 3\)
- \(C = -1\)

Thus, the normal vector \(\vec{n}\) is:
\[ \vec{n} = (2, 3, -1) \]

### Step 2: Use the point and the direction ratios to write the line equation
The line we need to find passes through the point \((1, 1, 1)\) and has direction ratios equal to the normal vector of the plane. Therefore, the direction ratios of the line are also \((2, 3, -1)\).

### Step 3: Write the parametric equations of the line
Using the point \((x_1, y_1, z_1) = (1, 1, 1)\) and the direction ratios \((a, b, c) = (2, 3, -1)\), we can write the parametric equations of the line as follows:
\[ \frac{x - 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{-1} \]

### Step 4: Rearranging the equation
To express the equation in a more standard form, we can rearrange it:
\[ \frac{x - 1}{2} = \frac{y - 1}{3} = \frac{1 - z}{1} \]

### Final Equation of the Line
Thus, the equation of the line can be expressed as:
\[ \frac{x - 1}{2} = \frac{y - 1}{3} = \frac{1 - z}{1} \]