consider the planes `P_1:2x-y+z=6 and P_2:x+2y-z=4` having normal `vec(N)_1 and vec(N)_2` respectively . The distance of the origin from the plane passing through the point (1,1,1) and whose normal is perpendicular to `N_1 and N_2` is

Options

A. `7/sqrt5` units
B. `sqrt(7/5)` units
C. `sqrt(3/5)` units
D. `14/sqrt35` units

Answer

To find the distance of the origin from the plane that passes through the point (1,1,1) and whose normal is perpendicular to the normals of the given planes \( P_1: 2x - y + z = 6 \) and \( P_2: x + 2y - z = 4 \), we will follow these steps:

### Step 1: Identify the normals of the planes
The normal vector of plane \( P_1 \) is given by:
\[ \vec{N}_1 = \langle 2, -1, 1 \rangle \]
The normal vector of plane \( P_2 \) is given by:
\[ \vec{N}_2 = \langle 1, 2, -1 \rangle \]

### Step 2: Find the normal of the desired plane
The normal of the desired plane is perpendicular to both \( \vec{N}_1 \) and \( \vec{N}_2 \). We can find this normal by calculating the cross product \( \vec{N}_1 \times \vec{N}_2 \).

Calculating the cross product:
\[ \vec{N} = \vec{N}_1 \times \vec{N}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 2 & -1 \end{vmatrix} \]

Expanding this determinant:
\[ \vec{N} = \hat{i} \begin{vmatrix} -1 & 1 \\ 2 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -1 \\ 1 & 2 \end{vmatrix} \]

Calculating the minors:
\[ \vec{N} = \hat{i}((-1)(-1) - (1)(2)) - \hat{j}((2)(-1) - (1)(1)) + \hat{k}((2)(2) - (-1)(1)) \]
\[ = \hat{i}(1 - 2) - \hat{j}(-2 - 1) + \hat{k}(4 + 1) \]
\[ = \hat{i}(-1) + \hat{j}(3) + \hat{k}(5) \]
Thus, the normal vector of the desired plane is:
\[ \vec{N} = \langle -1, 3, 5 \rangle \]

### Step 3: Write the equation of the plane
The equation of a plane can be written in the form:
\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]
where \( (x_0, y_0, z_0) \) is a point on the plane and \( \langle a, b, c \rangle \) is the normal vector.

Using the point \( (1, 1, 1) \) and the normal vector \( \langle -1, 3, 5 \rangle \):
\[ -1(x - 1) + 3(y - 1) + 5(z - 1) = 0 \]
Expanding this:
\[ -x + 1 + 3y - 3 + 5z - 5 = 0 \]
\[ -x + 3y + 5z - 7 = 0 \]
Rearranging gives:
\[ -x + 3y + 5z = 7 \]

### Step 4: Find the distance from the origin to the plane
The distance \( d \) from a point \( (x_0, y_0, z_0) \) to the plane \( ax + by + cz = d \) is given by:
\[ d = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}} \]
For the origin \( (0, 0, 0) \) and the plane \( -x + 3y + 5z = 7 \):
\[ d = \frac{|-1(0) + 3(0) + 5(0) - 7|}{\sqrt{(-1)^2 + 3^2 + 5^2}} \]
\[ = \frac{|0 - 7|}{\sqrt{1 + 9 + 25}} = \frac{7}{\sqrt{35}} \]

### Final Result
Thus, the distance of the origin from the plane is:
\[ \frac{7}{\sqrt{35}} = \frac{7}{5\sqrt{7}} = \frac{\sqrt{7}}{5} \]