Given `veca=xhati+yhatj+2hatk,vecb=hati-hatj+hatk , vecc=hati+2hatj, veca botvecb,veca.vecc=4` then find the value of `[(veca, vecb, vecc)]`.

Options

A. `[vecavecbvecc]^(2)=|veca|`
B. `[vecavecbvecc]=|veca|`
C. `[vecavecbvecc]=0`
D. `[vecavecbvecc]=0`

Explanation

To solve the problem step by step, let's break it down:

### Given:
- \(\vec{a} = x \hat{i} + y \hat{j} + 2 \hat{k}\)
- \(\vec{b} = \hat{i} - \hat{j} + \hat{k}\)
- \(\vec{c} = \hat{i} + 2 \hat{j}\)
- \(\vec{a} \perp \vec{b}\) (which means \(\vec{a} \cdot \vec{b} = 0\))
- \(\vec{a} \cdot \vec{c} = 4\)

### Step 1: Find the dot product \(\vec{a} \cdot \vec{b}\)
Since \(\vec{a} \perp \vec{b}\), we have:
\[ \vec{a} \cdot \vec{b} = 0 \]
Calculating the dot product:
\[ \vec{a} \cdot \vec{b} = (x \hat{i} + y \hat{j} + 2 \hat{k}) \cdot (\hat{i} - \hat{j} + \hat{k}) = x(1) + y(-1) + 2(1) \]
This simplifies to:
\[ x - y + 2 = 0 \quad \text{(1)} \]

### Step 2: Find the dot product \(\vec{a} \cdot \vec{c}\)
We also know:
\[ \vec{a} \cdot \vec{c} = 4 \]
Calculating this dot product:
\[ \vec{a} \cdot \vec{c} = (x \hat{i} + y \hat{j} + 2 \hat{k}) \cdot (\hat{i} + 2 \hat{j}) = x(1) + y(2) + 2(0) \]
This simplifies to:
\[ x + 2y = 4 \quad \text{(2)} \]

### Step 3: Solve the system of equations
Now we have two equations:
1. \(x - y + 2 = 0\)
2. \(x + 2y = 4\)

From equation (1):
\[ x - y = -2 \implies x = y - 2 \quad \text{(3)} \]

Substituting (3) into (2):
\[ (y - 2) + 2y = 4 \]
This simplifies to:
\[ 3y - 2 = 4 \]
\[ 3y = 6 \implies y = 2 \]

Substituting \(y = 2\) back into (3):
\[ x = 2 - 2 = 0 \]

### Step 4: Substitute values back into \(\vec{a}\)
Now substituting \(x\) and \(y\) back into \(\vec{a}\):
\[ \vec{a} = 0 \hat{i} + 2 \hat{j} + 2 \hat{k} = 2 \hat{j} + 2 \hat{k} \]

### Step 5: Find the scalar triple product \([\vec{a}, \vec{b}, \vec{c}]\)
The scalar triple product can be calculated using the determinant:
\[ [\vec{a}, \vec{b}, \vec{c}] = \begin{vmatrix} 0 & 2 & 2 \\ 1 & -1 & 1 \\ 1 & 2 & 0 \end{vmatrix} \]

Calculating the determinant:
\[ = 0 \cdot \begin{vmatrix} -1 & 1 \\ 2 & 0 \end{vmatrix} - 2 \cdot \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} + 2 \cdot \begin{vmatrix} 1 & -1 \\ 1 & 2 \end{vmatrix} \]

Calculating the 2x2 determinants:
\[ = 0 - 2(1 \cdot 0 - 1 \cdot 1) + 2(1 \cdot 2 - (-1) \cdot 1) \]
\[ = 0 - 2(-1) + 2(2 + 1) \]
\[ = 0 + 2 + 2 \cdot 3 = 2 + 6 = 8 \]

### Final Answer:
The value of \([\vec{a}, \vec{b}, \vec{c}] = 8\).