Given the vectors `a=3i-j+5k" and "b=i+2j-3k`. A vector c which is perpendicular to the z-axis and satisfies `c*a=9" and "c*b=-4` is

Options

A. `2i-3j`
B. `-2i+3j`
C. `-4i-4j`
D. `i-j+k`

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:
- \( \mathbf{a} = 3\mathbf{i} - \mathbf{j} + 5\mathbf{k} \)
- \( \mathbf{b} = \mathbf{i} + 2\mathbf{j} - 3\mathbf{k} \)

Since \( \mathbf{c} \) is perpendicular to the z-axis, it can be expressed as:
\[ \mathbf{c} = \alpha \mathbf{i} + \beta \mathbf{j} \]
where \( \alpha \) and \( \beta \) are the components of \( \mathbf{c} \) in the x and y directions respectively.

### Step 1: Use the first condition \( \mathbf{c} \cdot \mathbf{a} = 9 \)
Calculating the dot product:
\[ \mathbf{c} \cdot \mathbf{a} = (\alpha \mathbf{i} + \beta \mathbf{j}) \cdot (3\mathbf{i} - \mathbf{j} + 5\mathbf{k}) = 3\alpha - \beta + 0 \]
Setting this equal to 9:
\[ 3\alpha - \beta = 9 \quad \text{(Equation 1)} \]

### Step 2: Use the second condition \( \mathbf{c} \cdot \mathbf{b} = -4 \)
Calculating the dot product:
\[ \mathbf{c} \cdot \mathbf{b} = (\alpha \mathbf{i} + \beta \mathbf{j}) \cdot (\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}) = \alpha + 2\beta + 0 \]
Setting this equal to -4:
\[ \alpha + 2\beta = -4 \quad \text{(Equation 2)} \]

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

We can solve these equations simultaneously. From Equation 1, we can express \( \beta \) in terms of \( \alpha \):
\[ \beta = 3\alpha - 9 \]

Substituting this expression for \( \beta \) into Equation 2:
\[ \alpha + 2(3\alpha - 9) = -4 \]
Expanding this:
\[ \alpha + 6\alpha - 18 = -4 \]
Combining like terms:
\[ 7\alpha - 18 = -4 \]
Adding 18 to both sides:
\[ 7\alpha = 14 \]
Dividing by 7:
\[ \alpha = 2 \]

Now substituting \( \alpha = 2 \) back into the expression for \( \beta \):
\[ \beta = 3(2) - 9 = 6 - 9 = -3 \]

### Step 4: Write the vector \( \mathbf{c} \)
Now that we have \( \alpha \) and \( \beta \):
\[ \mathbf{c} = 2\mathbf{i} - 3\mathbf{j} \]

### Final Answer
Thus, the vector \( \mathbf{c} \) is:
\[ \mathbf{c} = 2\mathbf{i} - 3\mathbf{j} \]