Equation of plane passing through (2,0,5) and parallel to the vectors i-j+k and 3i + 2j - k is

Options

A. x-4y + 5z=27
B. x-4y - 5z+23=0
C. 4x-y+5z=33
D. 4x+ y - 5z=33

Answer

To find the equation of a plane that passes through the point (2, 0, 5) and is parallel to the vectors \( \mathbf{a} = \mathbf{i} - \mathbf{j} + \mathbf{k} \) and \( \mathbf{b} = 3\mathbf{i} + 2\mathbf{j} - \mathbf{k} \), we can follow these steps:

### Step 1: Identify the point and the direction vectors
The plane passes through the point \( P(2, 0, 5) \) and is parallel to the vectors:
- \( \mathbf{a} = (1, -1, 1) \)
- \( \mathbf{b} = (3, 2, -1) \)

### Step 2: Find the normal vector to the plane
The normal vector \( \mathbf{n} \) to the plane can be found by taking the cross product of the two direction vectors \( \mathbf{a} \) and \( \mathbf{b} \).

\[ \mathbf{n} = \mathbf{a} \times \mathbf{b} \]

Calculating the cross product:

\[ \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 3 & 2 & -1 \end{vmatrix} \]

Expanding this determinant:

\[ \mathbf{n} = \mathbf{i} \begin{vmatrix} -1 & 1 \\ 2 & -1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 1 \\ 3 & -1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix} \]

Calculating the 2x2 determinants:

\[ \begin{vmatrix} -1 & 1 \\ 2 & -1 \end{vmatrix} = (-1)(-1) - (1)(2) = 1 - 2 = -1 \]

\[ \begin{vmatrix} 1 & 1 \\ 3 & -1 \end{vmatrix} = (1)(-1) - (1)(3) = -1 - 3 = -4 \]

\[ \begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix} = (1)(2) - (-1)(3) = 2 + 3 = 5 \]

Thus, the normal vector is:

\[ \mathbf{n} = -1\mathbf{i} + 4\mathbf{j} + 5\mathbf{k} = (-1, 4, 5) \]

### Step 3: Write the equation of the plane
The equation of a plane in the form \( Ax + By + Cz + D = 0 \) can be written using the normal vector \( (A, B, C) \) and the point \( (x_0, y_0, z_0) \):

\[ -1(x - 2) + 4(y - 0) + 5(z - 5) = 0 \]

Expanding this:

\[ -1(x - 2) + 4y + 5(z - 5) = 0 \]

\[ -x + 2 + 4y + 5z - 25 = 0 \]

Rearranging gives:

\[ -x + 4y + 5z - 23 = 0 \]

Thus, the equation of the plane is:

\[ x - 4y - 5z + 23 = 0 \]

### Final Answer:
The equation of the plane is:

\[ x - 4y - 5z + 23 = 0 \]

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