Given vectors a = (3, -1, 5) and b = (1, 2, -3). A vector c which is perpendicular to z-axis and satisfying `c*a=9" and "c*b=-4` is

Options

A. (2, -2, 0)
B. (4, -2, 0)
C. (2, -3, 0)
D. (1, 2, 4)

Answer

To solve the problem, we need to find the vector \( \mathbf{c} \) that is perpendicular to the z-axis and satisfies the conditions \( \mathbf{c} \cdot \mathbf{a} = 9 \) and \( \mathbf{c} \cdot \mathbf{b} = -4 \).

Given:
- \( \mathbf{a} = (3, -1, 5) \)
- \( \mathbf{b} = (1, 2, -3) \)

Since \( \mathbf{c} \) is perpendicular to the z-axis, its z-component must be 0. We can express \( \mathbf{c} \) as:
\[ \mathbf{c} = (x, y, 0) \]

### Step 1: Set up the dot product equations
Using the dot product for the first condition \( \mathbf{c} \cdot \mathbf{a} = 9 \):
\[ (x, y, 0) \cdot (3, -1, 5) = 9 \]
This expands to:
\[ 3x - y + 0 = 9 \]
So, we have our first equation:
\[ 3x - y = 9 \quad \text{(Equation 1)} \]

### Step 2: Set up the second dot product equation
Now, using the second condition \( \mathbf{c} \cdot \mathbf{b} = -4 \):
\[ (x, y, 0) \cdot (1, 2, -3) = -4 \]
This expands to:
\[ x + 2y + 0 = -4 \]
So, we have our second equation:
\[ x + 2y = -4 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we have a system of two equations:
1. \( 3x - y = 9 \)
2. \( x + 2y = -4 \)

We can solve this system using substitution or elimination. Let's use substitution.

From Equation 1, we can express \( y \) in terms of \( x \):
\[ y = 3x - 9 \]

Now, substitute \( y \) into Equation 2:
\[ x + 2(3x - 9) = -4 \]
Expanding this gives:
\[ x + 6x - 18 = -4 \]
Combining like terms:
\[ 7x - 18 = -4 \]
Adding 18 to both sides:
\[ 7x = 14 \]
Dividing by 7:
\[ x = 2 \]

### Step 4: Find \( y \)
Now substitute \( x = 2 \) back into the expression for \( y \):
\[ y = 3(2) - 9 = 6 - 9 = -3 \]

### Step 5: Write the final vector \( \mathbf{c} \)
Since \( z = 0 \), we can now write the vector \( \mathbf{c} \):
\[ \mathbf{c} = (2, -3, 0) \]

### Final Answer
The vector \( \mathbf{c} \) is:
\[ \mathbf{c} = (2, -3, 0) \]