A line L lies in the plane `2x-y-z=4` such that it is perpendicular to the line `(x-2)/(2)=(y-3)/(1)=(z-4)/(5)`. The line L passes through the point of intersection of the given line and given plane. Which of the following points does not satisfy line L?

Options

A. `(-5, -(1)/(2), -(27)/(2))`
B. `(3, 1, 6)`
C. `(0, (29)/(2), (-37)/(2))`
D. `(-4, (5)/(2), (-29)/(2))`

Answer

To solve the problem step by step, we need to find the line \( L \) that lies in the plane \( 2x - y - z = 4 \) and is perpendicular to the given line defined by the equations \( \frac{x - 2}{2} = \frac{y - 3}{1} = \frac{z - 4}{5} \). The line \( L \) will pass through the point of intersection of the given line and the plane.

### Step 1: Find the point of intersection of the line and the plane.

We can express the line in parametric form. Let \( t \) be the parameter:
\[ x = 2 + 2t, \quad y = 3 + t, \quad z = 4 + 5t \]

Now, we substitute these values into the equation of the plane:
\[ 2(2 + 2t) - (3 + t) - (4 + 5t) = 4 \]
Expanding this:
\[ 4 + 4t - 3 - t - 4 - 5t = 4 \]
Combining like terms:
\[ 4t - 6t + 4 - 3 - 4 = 4 \]
This simplifies to:
\[ -2t - 3 = 4 \]
Solving for \( t \):
\[ -2t = 7 \implies t = -\frac{7}{2} \]

### Step 2: Substitute \( t \) back to find the coordinates of the intersection point.

Now we substitute \( t = -\frac{7}{2} \) back into the parametric equations:
\[ x = 2 + 2\left(-\frac{7}{2}\right) = 2 - 7 = -5 \]
\[ y = 3 + \left(-\frac{7}{2}\right) = 3 - \frac{7}{2} = \frac{6}{2} - \frac{7}{2} = -\frac{1}{2} \]
\[ z = 4 + 5\left(-\frac{7}{2}\right) = 4 - \frac{35}{2} = \frac{8}{2} - \frac{35}{2} = -\frac{27}{2} \]

Thus, the point of intersection is:
\[ (-5, -\frac{1}{2}, -\frac{27}{2}) \]

### Step 3: Determine the direction vectors.

The normal vector \( \mathbf{n} \) of the plane \( 2x - y - z = 4 \) is:
\[ \mathbf{n} = \langle 2, -1, -1 \rangle \]

The direction vector \( \mathbf{b} \) of the given line is:
\[ \mathbf{b} = \langle 2, 1, 5 \rangle \]

### Step 4: Find the direction vector of line \( L \).

Since line \( L \) is perpendicular to the given line, we can find the direction vector of line \( L \) by taking the cross product of \( \mathbf{n} \) and \( \mathbf{b} \):
\[ \mathbf{d} = \mathbf{n} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & -1 \\ 2 & 1 & 5 \end{vmatrix} \]

Calculating the determinant:
\[ \mathbf{d} = \mathbf{i}((-1)(5) - (-1)(1)) - \mathbf{j}(2(5) - (-1)(2)) + \mathbf{k}(2(1) - (-1)(2)) \]
\[ = \mathbf{i}(-5 + 1) - \mathbf{j}(10 + 2) + \mathbf{k}(2 + 2) \]
\[ = \mathbf{i}(-4) - \mathbf{j}(12) + \mathbf{k}(4) \]
\[ = \langle -4, -12, 4 \rangle \]

### Step 5: Write the equation of line \( L \).

Using the point of intersection and the direction vector, the equation of line \( L \) can be expressed as:
\[ \frac{x + 5}{-4} = \frac{y + \frac{1}{2}}{-12} = \frac{z + \frac{27}{2}}{4} \]

### Step 6: Check which point does not satisfy line \( L \).

To determine which point does not satisfy line \( L \), we need to check the given options against the derived equation of line \( L \).