Let `veca=hati-hatj,vecb=hati+hatj+hatk` and `vecc` be a vector such that `vecaxxvecc+vecb=vec0` and `veca.vecc=4`, then `|vecc|^(2)` is equal to:

Options

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

Explanation

To solve the problem step by step, we will follow the given conditions and use vector algebra.

### Step 1: Define the vectors
Let:
\[ \vec{a} = \hat{i} - \hat{j} \]
\[ \vec{b} = \hat{i} + \hat{j} + \hat{k} \]
Let \(\vec{c} = x \hat{i} + y \hat{j} + z \hat{k}\).

### Step 2: Use the cross product condition
We are given that:
\[ \vec{a} \times \vec{c} + \vec{b} = \vec{0} \]
This implies:
\[ \vec{a} \times \vec{c} = -\vec{b} \]

### Step 3: Calculate the cross product \(\vec{a} \times \vec{c}\)
Using the determinant form for the cross product:
\[ \vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ x & y & z \end{vmatrix} \]

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

### Step 4: Set the cross product equal to \(-\vec{b}\)
Setting \(\vec{a} \times \vec{c} = -\vec{b}\):
\[ -z \hat{i} - z \hat{j} + (y + x) \hat{k} = -(\hat{i} + \hat{j} + \hat{k}) \]
This gives us the following equations:
1. \(-z = -1 \implies z = 1\)
2. \(-z = -1 \implies z = 1\)
3. \(y + x = -1\)

### Step 5: Use the dot product condition
We are also given:
\[ \vec{a} \cdot \vec{c} = 4 \]
Calculating the dot product:
\[ \vec{a} \cdot \vec{c} = (1)(x) + (-1)(y) + (0)(z) = x - y = 4 \]

### Step 6: Solve the system of equations
Now we have two equations:
1. \(x + y = -1\) (from the cross product)
2. \(x - y = 4\) (from the dot product)

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

Substituting \(x\) back into the first equation:
\[ \frac{3}{2} + y = -1 \implies y = -1 - \frac{3}{2} = -\frac{5}{2} \]

### Step 7: Find \(|\vec{c}|^2\)
Now we have:
\[ x = \frac{3}{2}, \quad y = -\frac{5}{2}, \quad z = 1 \]
The magnitude squared of \(\vec{c}\) is:
\[ |\vec{c}|^2 = x^2 + y^2 + z^2 = \left(\frac{3}{2}\right)^2 + \left(-\frac{5}{2}\right)^2 + 1^2 \]
\[ = \frac{9}{4} + \frac{25}{4} + 1 = \frac{9}{4} + \frac{25}{4} + \frac{4}{4} = \frac{38}{4} = \frac{19}{2} \]

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

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