What is the equation of the plane which passes through the z-axis and is perpendicular to the line `(x-a)/(costheta)=(y+2)/(sintheta)=(z-3)/0?`a. `x+yt a ntheta=0` b. `y+x t a ntheta=0` c. `xcostheta-ysintheta=0`d. `xsintheta-ycostheta=0`

Options

A. `x+ytantheta=0`
B. `y+xtantheta=0`
C. `xcostheta-ysintheta=0`
D. `xsintheta-ycostheta=0`

Explanation

To find the equation of the plane that passes through the z-axis and is perpendicular to the given line, we can follow these steps:

### Step 1: Identify the direction ratios of the line
The equation of the line is given as:
\[ \frac{x - a}{\cos \theta} = \frac{y + 2}{\sin \theta} = \frac{z - 3}{0} \]
From this equation, we can identify the direction ratios of the line. The direction ratios are:
- For \(x\): \(\cos \theta\)
- For \(y\): \(\sin \theta\)
- For \(z\): \(0\)

Thus, the direction ratios of the line are \((\cos \theta, \sin \theta, 0)\).

### Step 2: Determine the normal vector to the plane
Since the plane is perpendicular to the line, the normal vector of the plane will be the same as the direction ratios of the line. Therefore, the normal vector \( \vec{n} \) of the plane is:
\[ \vec{n} = (\cos \theta, \sin \theta, 0) \]

### Step 3: Use the point-normal form of the plane equation
The plane passes through the z-axis, which can be represented by the point \((0, 0, z)\) for any \(z\). We can use the point-normal form of the plane equation:
\[ n_1(x - x_0) + n_2(y - y_0) + n_3(z - z_0) = 0 \]
Here, \((x_0, y_0, z_0) = (0, 0, 0)\) (a point on the z-axis) and \((n_1, n_2, n_3) = (\cos \theta, \sin \theta, 0)\).

Substituting these values into the equation gives:
\[ \cos \theta (x - 0) + \sin \theta (y - 0) + 0(z - 0) = 0 \]
This simplifies to:
\[ \cos \theta x + \sin \theta y = 0 \]

### Step 4: Rearranging the equation
We can rearrange this equation to express it in a more standard form:
\[ \cos \theta x + \sin \theta y = 0 \]
If we divide the entire equation by \(\cos \theta\) (assuming \(\cos \theta \neq 0\)), we get:
\[ x + \tan \theta y = 0 \]

### Conclusion
Thus, the equation of the plane is:
\[ x + \tan \theta y = 0 \]
This corresponds to option (a).