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 solve the problem, we need to find the value of \( c \) such that the line given by the equation

\[ \frac{x-2}{3} = \frac{y+1}{2} = \frac{z-1}{-1} \]

intersects with the curve defined by

\[ xy = c^2 \quad \text{and} \quad z = 0. \]

### Step 1: Parameterize the line

Let \( t \) be the parameter such that:

\[ \frac{x-2}{3} = t, \quad \frac{y+1}{2} = t, \quad \frac{z-1}{-1} = t. \]

From these equations, we can express \( x, y, z \) in terms of \( t \):

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

### Step 2: Set \( z = 0 \)

To find the intersection with the plane \( z = 0 \), we set the equation for \( z \) to zero:

\[ -t + 1 = 0 \implies t = 1. \]

### Step 3: Substitute \( t \) into \( x \) and \( y \)

Now, substitute \( t = 1 \) back into the equations for \( x \) and \( y \):

\[ x = 3(1) + 2 = 5, \]
\[ y = 2(1) - 1 = 1. \]

### Step 4: Calculate \( c^2 \)

Now that we have the coordinates \( (x, y) = (5, 1) \), we can substitute these values into the equation of the curve \( xy = c^2 \):

\[ 5 \cdot 1 = c^2 \implies c^2 = 5. \]

### Step 5: Solve for \( c \)

Taking the square root of both sides, we find:

\[ c = \sqrt{5} \quad \text{or} \quad c = -\sqrt{5}. \]

Since \( c \) is typically taken to be non-negative in this context, we conclude:

\[ c = \sqrt{5}. \]

### Final Answer

Thus, the value of \( c \) is:

\[ c = \sqrt{5}. \]

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