A vector c perpendicular to the vectors `2i+3j-k" and "i-2j+3k` satisfying `c.(2i-j+k)=-6` is

Options

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

Answer

To find the vector \( \mathbf{c} \) that is perpendicular to the vectors \( \mathbf{a} = 2\mathbf{i} + 3\mathbf{j} - \mathbf{k} \) and \( \mathbf{b} = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k} \), and satisfies the condition \( \mathbf{c} \cdot (2\mathbf{i} - \mathbf{j} + \mathbf{k}) = -6 \), we can follow these steps:

### Step 1: Assume the vector \( \mathbf{c} \)
Let \( \mathbf{c} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k} \).

### Step 2: Set up the perpendicularity conditions
Since \( \mathbf{c} \) is perpendicular to \( \mathbf{a} \) and \( \mathbf{b} \), we can set up the following equations based on the dot product being zero:

1. \( \mathbf{c} \cdot \mathbf{a} = 0 \)
\[ (x\mathbf{i} + y\mathbf{j} + z\mathbf{k}) \cdot (2\mathbf{i} + 3\mathbf{j} - \mathbf{k}) = 0 \]
This gives us:
\[ 2x + 3y - z = 0 \quad \text{(Equation 1)} \]

2. \( \mathbf{c} \cdot \mathbf{b} = 0 \)
\[ (x\mathbf{i} + y\mathbf{j} + z\mathbf{k}) \cdot (\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}) = 0 \]
This gives us:
\[ x - 2y + 3z = 0 \quad \text{(Equation 2)} \]

### Step 3: Set up the condition from the third equation
We also have the condition:
\[ \mathbf{c} \cdot (2\mathbf{i} - \mathbf{j} + \mathbf{k}) = -6 \]
This gives us:
\[ (2x - y + z = -6) \quad \text{(Equation 3)} \]

### Step 4: Solve the equations
Now we have three equations:
1. \( 2x + 3y - z = 0 \) (Equation 1)
2. \( x - 2y + 3z = 0 \) (Equation 2)
3. \( 2x - y + z = -6 \) (Equation 3)

#### Step 4.1: Solve Equation 1 for \( z \)
From Equation 1:
\[ z = 2x + 3y \quad \text{(Substituting into other equations)} \]

#### Step 4.2: Substitute \( z \) into Equation 2
Substituting \( z \) into Equation 2:
\[ x - 2y + 3(2x + 3y) = 0 \]
This simplifies to:
\[ x - 2y + 6x + 9y = 0 \implies 7x + 7y = 0 \implies x + y = 0 \implies y = -x \quad \text{(Equation 4)} \]

#### Step 4.3: Substitute \( y \) into Equation 1
Substituting \( y = -x \) into Equation 1:
\[ 2x + 3(-x) - z = 0 \implies 2x - 3x - z = 0 \implies -x - z = 0 \implies z = -x \quad \text{(Equation 5)} \]

### Step 5: Substitute \( y \) and \( z \) into Equation 3
Substituting \( y = -x \) and \( z = -x \) into Equation 3:
\[ 2x - (-x) + (-x) = -6 \implies 2x + x - x = -6 \implies 2x = -6 \implies x = -3 \]

### Step 6: Find \( y \) and \( z \)
Using \( x = -3 \):
\[ y = -(-3) = 3, \quad z = -(-3) = 3 \]

### Step 7: Write the vector \( \mathbf{c} \)
Thus, the vector \( \mathbf{c} \) is:
\[ \mathbf{c} = -3\mathbf{i} + 3\mathbf{j} + 3\mathbf{k} \]

### Final Answer
\[ \mathbf{c} = -3\mathbf{i} + 3\mathbf{j} + 3\mathbf{k} \]

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