The equation of the plane passing through `(2,3,-4) and (1,-1,3)` and parallel to x-axis is
4(B)The equation of plane passing through `(0,1,0)` and perpendicular to `y=0`, then the perpendicular distance from `(0,0,0)` to the plane is zero.

Options

A. `7y-4z-5=0`
B. `4y-7z-5=0`
C. `4y+7z+5=0`
D. `7y+4z-5=0`

Answer

To find the equation of the plane passing through the points \( (2, 3, -4) \) and \( (1, -1, 3) \) and parallel to the x-axis, we can follow these steps:

### Step 1: Identify the direction vector
First, we need to find the direction vector between the two points \( A(2, 3, -4) \) and \( B(1, -1, 3) \).

The direction vector \( \vec{AB} \) can be calculated as:
\[ \vec{AB} = B - A = (1 - 2, -1 - 3, 3 - (-4)) = (-1, -4, 7) \]

### Step 2: Determine the normal vector
Since the plane is parallel to the x-axis, the normal vector \( \vec{n} \) to the plane will have no component in the x-direction. Thus, we can represent the normal vector as:
\[ \vec{n} = (0, n_y, n_z) \]
To find \( n_y \) and \( n_z \), we can use the direction vector \( \vec{AB} \):
\[ \vec{n} = (0, -1, -4) \]

### Step 3: Use the point-normal form of the plane equation
The equation of a plane in point-normal form is given by:
\[ n_x(x - x_0) + n_y(y - y_0) + n_z(z - z_0) = 0 \]
Substituting \( (x_0, y_0, z_0) = (2, 3, -4) \) and \( \vec{n} = (0, -1, -4) \):
\[ 0(x - 2) - 1(y - 3) - 4(z + 4) = 0 \]
This simplifies to:
\[ -y + 3 - 4z - 16 = 0 \]
Rearranging gives:
\[ y + 4z - 19 = 0 \]

### Step 4: Final equation of the plane
Thus, the equation of the plane is:
\[ y + 4z - 19 = 0 \]

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### Step 5: Find the equation of the second plane
For the second part of the question, we need to find the equation of the plane that passes through the point \( (0, 1, 0) \) and is perpendicular to the plane \( y = 0 \) (the x-z plane).

Since the plane is perpendicular to the x-z plane, it will be parallel to the y-axis. The equation of a plane parallel to the y-axis can be expressed as:
\[ x = k \quad \text{or} \quad z = k \]
Since the point through which the plane passes is \( (0, 1, 0) \), we can express the equation of the plane as:
\[ y = 1 \]

### Step 6: Perpendicular distance from the origin to the plane
The perpendicular distance \( d \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
For the plane \( y = 1 \), we can rewrite it as:
\[ 0x + 1y + 0z - 1 = 0 \]
Here, \( A = 0, B = 1, C = 0, D = -1 \). The point is \( (0, 0, 0) \):
\[ d = \frac{|0 \cdot 0 + 1 \cdot 0 + 0 \cdot 0 - 1|}{\sqrt{0^2 + 1^2 + 0^2}} = \frac{|-1|}{1} = 1 \]

Thus, the perpendicular distance from the origin \( (0, 0, 0) \) to the plane \( y = 1 \) is \( 1 \).

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