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. a) (0,-4,-1)`
B. b) (0,-4,1)`
C. c) (1,4,0)`
D. d) (0,4,1)`

Answer

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 parametric equations of the line
The line joining the points \((2, 1, 3)\) and \((4, -2, 5)\) can be expressed in parametric form. The direction ratios of the line can be calculated as follows:

\[ \text{Direction ratios} = (4 - 2, -2 - 1, 5 - 3) = (2, -3, 2) \]

Using the point \((2, 1, 3)\) as a reference point, we can write the parametric equations of the line:

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

### Step 2: Substitute the parametric equations into the plane equation
Now, we substitute these parametric equations into the equation of the plane \(2x + y - z = 3\):

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

### Step 3: Simplify the equation
Expanding and simplifying the equation gives:

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

Combining like terms:

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

### Step 4: Solve for \(t\)
Rearranging the equation to solve for \(t\):

\[ -t = 3 - 2 \]
\[ -t = 1 \implies t = -1 \]

### Step 5: Find the coordinates of the intersection point
Now that we have \(t = -1\), we can substitute this value back into the parametric equations to find the coordinates of the intersection point:

\[ x = 2 + 2(-1) = 2 - 2 = 0 \]
\[ y = 1 - 3(-1) = 1 + 3 = 4 \]
\[ z = 3 + 2(-1) = 3 - 2 = 1 \]

Thus, the point of intersection is \((0, 4, 1)\).

### Final Answer
The line cuts the plane at the point \((0, 4, 1)\).

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