The line joining the points (2, 1, 3) and (4, -2, 5) cuts the plane `2x+y-z=3`.
Where does the line cut the plane ?

Options

A. (0, -4, -1)
B. (0, -4, 1)
C. (1,4,0)
D. (0,4,1)

Explanation

To find the point where the line joining the points (2, 1, 3) and (4, -2, 5) intersects the plane given by the equation \(2x + y - z = 3\), we can follow these steps:

### Step 1: Find the direction ratios of the line
The direction ratios of the line can be found using the coordinates of the two points. The direction ratios can be calculated as follows:

\[ \text{Direction ratios} = (x_2 - x_1, y_2 - y_1, z_2 - z_1) = (4 - 2, -2 - 1, 5 - 3) = (2, -3, 2) \]

### Step 2: Write the parametric equations of the line
Using the point (2, 1, 3) as the starting point and the direction ratios, we can write the parametric equations of the line:

\[ x = 2 + 2t, \quad y = 1 - 3t, \quad z = 3 + 2t \]

where \(t\) is a parameter.

### Step 3: Substitute the parametric equations into the plane equation
The equation of the plane is given by \(2x + y - z = 3\). We substitute the parametric equations into this plane equation:

\[ 2(2 + 2t) + (1 - 3t) - (3 + 2t) = 3 \]

### Step 4: Simplify the equation
Now, simplify the equation:

\[ 4 + 4t + 1 - 3t - 3 - 2t = 3 \]

Combine like terms:

\[ (4t - 3t - 2t) + (4 + 1 - 3) = 3 \]

This simplifies to:

\[ -1 = 3 \]

### Step 5: Solve for \(t\)
This equation simplifies to:

\[ -1 + 3 = 0 \quad \Rightarrow \quad 0 = 0 \]

This means that the line lies entirely in the plane, and thus, it intersects the plane at every point along the line.

### Step 6: Find a specific intersection point
To find a specific intersection point, we can choose a value for \(t\). Let’s take \(t = 0\):

\[ x = 2 + 2(0) = 2, \quad y = 1 - 3(0) = 1, \quad z = 3 + 2(0) = 3 \]

So, one intersection point is:

\[ (2, 1, 3) \]

### Conclusion
The line joining the points (2, 1, 3) and (4, -2, 5) cuts the plane \(2x + y - z = 3\) at every point along the line, and a specific intersection point is \((2, 1, 3)\).

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