The line `(x-2)/(3)=(y+1)/(2)=(z-1)/(-1)` intersects the curve `xy=c^2, z=0,` if c is equal to

Options

A. `pm1`
B. `pm(1)/(3)`
C. `pmsqrt(5)`
D. None of these

Answer

To find the value of \( c \) for which the line intersects the curve \( xy = c^2 \) and \( z = 0 \), we will follow these steps:

### Step 1: Write the equation of the line
The line is given by the equation:
\[ \frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{-1} \]

### Step 2: Set \( z = 0 \)
Since we need to find the intersection with the plane where \( z = 0 \), we substitute \( z = 0 \) into the line's equation:
\[ \frac{x - 2}{3} = \frac{y + 1}{2} = \frac{0 - 1}{-1} \]
This simplifies to:
\[ \frac{x - 2}{3} = \frac{y + 1}{2} = 1 \]

### Step 3: Solve for \( x \) and \( y \)
From the equation \( \frac{x - 2}{3} = 1 \):
\[ x - 2 = 3 \implies x = 5 \]

From the equation \( \frac{y + 1}{2} = 1 \):
\[ y + 1 = 2 \implies y = 1 \]

### Step 4: Substitute \( x \) and \( y \) into the curve equation
Now we have \( x = 5 \) and \( y = 1 \). We substitute these values into the curve equation \( xy = c^2 \):
\[ 5 \cdot 1 = c^2 \implies c^2 = 5 \]

### Step 5: Solve for \( c \)
Taking the square root of both sides, we find:
\[ c = \pm \sqrt{5} \]

### Final Answer
Thus, the value of \( c \) is:
\[ c = \pm \sqrt{5} \]

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