Find the equation of the plane containing the line `(y)/(b)+(z)/(c)=1,x=0`, and parallel to the line `(x)/(a)-(z)/(c)1,y=0`.

Explanation

To find the equation of the plane containing the line \(\frac{y}{b} + \frac{z}{c} = 1, x = 0\) and parallel to the line \(\frac{x}{a} - \frac{z}{c} = 1, y = 0\), we can follow these steps:

### Step 1: Identify the direction ratios of the lines

The first line can be expressed in parametric form. From the equation \(\frac{y}{b} + \frac{z}{c} = 1\) and \(x = 0\), we can set:
- \(y = bt\)
- \(z = c(1 - t)\)
- \(x = 0\)

This gives us the parametric equations for the first line:
\[ (0, bt, c(1 - t)) \]

The direction ratios of the first line are:
- \(0\) (for \(x\))
- \(b\) (for \(y\))
- \(-c\) (for \(z\))

Thus, the direction ratios of the first line are \((0, b, -c)\).

For the second line, from \(\frac{x}{a} - \frac{z}{c} = 1\) and \(y = 0\), we can set:
- \(x = a(s + 1)\)
- \(y = 0\)
- \(z = c(s)\)

This gives us the parametric equations for the second line:
\[ (a(s + 1), 0, cs) \]

The direction ratios of the second line are:
- \(a\) (for \(x\))
- \(0\) (for \(y\))
- \(-c\) (for \(z\))

Thus, the direction ratios of the second line are \((a, 0, -c)\).

### Step 2: Find the normal vector to the plane

The plane that contains the first line and is parallel to the second line will have a normal vector that is perpendicular to both direction ratios. We can find this normal vector by taking the cross product of the two direction ratios.

Let:
\[ \mathbf{d_1} = (0, b, -c) \quad \text{and} \quad \mathbf{d_2} = (a, 0, -c) \]

The cross product \(\mathbf{d_1} \times \mathbf{d_2}\) is calculated as follows:
\[ \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & b & -c \\ a & 0 & -c \end{vmatrix} \]

Calculating the determinant:
\[ \mathbf{n} = \mathbf{i}(b \cdot (-c) - 0 \cdot (-c)) - \mathbf{j}(0 \cdot (-c) - (-c) \cdot a) + \mathbf{k}(0 \cdot 0 - b \cdot a) \]
\[ = -bc \mathbf{i} + ac \mathbf{j} - ab \mathbf{k} \]
Thus, the normal vector \(\mathbf{n} = (-bc, ac, -ab)\).

### Step 3: Write the equation of the plane

The general equation of a plane with normal vector \((A, B, C)\) passing through a point \((x_0, y_0, z_0)\) is given by:
\[ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \]

We can take a point on the first line. When \(t = 0\), we have:
\[ (0, 0, c) \]
Thus, \(x_0 = 0\), \(y_0 = 0\), and \(z_0 = c\).

Substituting into the plane equation:
\[ -bc(x - 0) + ac(y - 0) - ab(z - c) = 0 \]
This simplifies to:
\[ -bcx + acy - abz + abc = 0 \]

### Step 4: Rearranging the equation

Rearranging gives us:
\[ bcx - acy + abz = abc \]

### Final Equation of the Plane

The final equation of the plane is:
\[ bcx - acy + abz = abc \]