Let `a=hati - hatj, b=hati + hatj + hatk` and c be a vector such that `a xx c + b=0` and a.c =4, then `|c|^(2)` is equal to .

Options

A. 8
B. `19/2`
C. 9
D. `17/2`

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the vectors
Given:
- \( \mathbf{a} = \hat{i} - \hat{j} \)
- \( \mathbf{b} = \hat{i} + \hat{j} + \hat{k} \)

Let \( \mathbf{c} = x \hat{i} + y \hat{j} + z \hat{k} \).

### Step 2: Use the dot product condition
We know that \( \mathbf{a} \cdot \mathbf{c} = 4 \).

Calculating the dot product:
\[ \mathbf{a} \cdot \mathbf{c} = (1)(x) + (-1)(y) = x - y \]
Setting this equal to 4 gives us:
\[ x - y = 4 \quad \text{(Equation 1)} \]

### Step 3: Use the cross product condition
We also have the condition \( \mathbf{a} \times \mathbf{c} + \mathbf{b} = 0 \), which implies:
\[ \mathbf{a} \times \mathbf{c} = -\mathbf{b} \]

Calculating \( \mathbf{a} \times \mathbf{c} \) using the determinant method:
\[ \mathbf{a} \times \mathbf{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ x & y & z \end{vmatrix} \]

Expanding this determinant:
\[ = \hat{i} \begin{vmatrix} -1 & 0 \\ y & z \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 0 \\ x & z \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -1 \\ x & y \end{vmatrix} \]
\[ = \hat{i} (-1 \cdot z - 0 \cdot y) - \hat{j} (1 \cdot z - 0 \cdot x) + \hat{k} (1 \cdot y - (-1) \cdot x) \]
\[ = -z \hat{i} - z \hat{j} + (y + x) \hat{k} \]

Setting this equal to \( -\mathbf{b} = -(\hat{i} + \hat{j} + \hat{k}) \):
\[ -z \hat{i} - z \hat{j} + (y + x) \hat{k} = -\hat{i} - \hat{j} - \hat{k} \]

### Step 4: Equate coefficients
From the above equation, we can equate coefficients:
1. For \( \hat{i} \): \( -z = -1 \) → \( z = 1 \)
2. For \( \hat{j} \): \( -z = -1 \) → \( z = 1 \) (consistent)
3. For \( \hat{k} \): \( y + x = -1 \) → \( y + x = -1 \quad \text{(Equation 2)} \)

### Step 5: Solve the equations
Now we have two equations:
1. \( x - y = 4 \) (Equation 1)
2. \( x + y = -1 \) (Equation 2)

Adding these two equations:
\[ (x - y) + (x + y) = 4 - 1 \]
\[ 2x = 3 \quad \Rightarrow \quad x = \frac{3}{2} \]

Substituting \( x \) back into Equation 2:
\[ \frac{3}{2} + y = -1 \quad \Rightarrow \quad y = -1 - \frac{3}{2} = -\frac{5}{2} \]

### Step 6: Write the vector \( \mathbf{c} \)
Now we have:
- \( x = \frac{3}{2} \)
- \( y = -\frac{5}{2} \)
- \( z = 1 \)

Thus, \( \mathbf{c} = \frac{3}{2} \hat{i} - \frac{5}{2} \hat{j} + 1 \hat{k} \).

### Step 7: Calculate \( |\mathbf{c}|^2 \)
Now we calculate the magnitude squared of \( \mathbf{c} \):
\[ |\mathbf{c}|^2 = \left(\frac{3}{2}\right)^2 + \left(-\frac{5}{2}\right)^2 + (1)^2 \]
\[ = \frac{9}{4} + \frac{25}{4} + 1 \]
\[ = \frac{9}{4} + \frac{25}{4} + \frac{4}{4} = \frac{38}{4} = \frac{19}{2} \]

### Final Answer
Thus, \( |\mathbf{c}|^2 = \frac{19}{2} \).

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