The line passing through the points `(1,2,-1) and (3,-1,2)` meets the yz plane at which one of the following points?

Options

A. `(0,-7/2,5/2)`
B. `(0,7/2,1/7)`
C. `(0,-7/2,-5/2)`
D. `(0,7/2,-5/2)`

Answer

To find the point where the line passing through the points \( (1, 2, -1) \) and \( (3, -1, 2) \) meets the YZ plane, we can follow these steps:

### Step 1: Determine the direction vector of the line
The direction vector \( \mathbf{d} \) can be found by subtracting the coordinates of the first point from the second point.

\[ \mathbf{d} = (3 - 1, -1 - 2, 2 - (-1)) = (2, -3, 3) \]

### Step 2: Write the parametric equations of the line
Using the first point \( (1, 2, -1) \) and the direction vector \( (2, -3, 3) \), we can express the parametric equations of the line as:

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

### Step 3: Find where the line intersects the YZ plane
The YZ plane is defined by the equation \( x = 0 \). We need to set the equation for \( x \) to zero and solve for \( \lambda \):

\[ 1 + 2\lambda = 0 \]

Solving for \( \lambda \):

\[ 2\lambda = -1 \implies \lambda = -\frac{1}{2} \]

### Step 4: Substitute \( \lambda \) back into the equations for \( y \) and \( z \)
Now we substitute \( \lambda = -\frac{1}{2} \) into the equations for \( y \) and \( z \):

For \( y \):

\[ y = 2 - 3\left(-\frac{1}{2}\right) = 2 + \frac{3}{2} = \frac{4}{2} + \frac{3}{2} = \frac{7}{2} \]

For \( z \):

\[ z = -1 + 3\left(-\frac{1}{2}\right) = -1 - \frac{3}{2} = -\frac{2}{2} - \frac{3}{2} = -\frac{5}{2} \]

### Step 5: Write the coordinates of the intersection point
The coordinates of the point where the line meets the YZ plane are:

\[ (0, \frac{7}{2}, -\frac{5}{2}) \]

### Final Answer
Thus, the line meets the YZ plane at the point \( \left(0, \frac{7}{2}, -\frac{5}{2}\right) \).

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