Let `vec a= hat j+hat j,vec b=hat j+hat k` and `vec c=alpha vec a+beta vec b`. If the vectors, `hat i-2hat j+hat k,3 hat i+2 hat j-hat k` and `vec c` are coplanar then `alpha/beta` is

Options

A. `1`
B. `2`
C. `3`
D. `-3`

Answer

To solve the problem, we need to determine the ratio \(\frac{\alpha}{\beta}\) given that the vectors \(\vec{a} = \hat{i} + \hat{j}\), \(\vec{b} = \hat{j} + \hat{k}\), and \(\vec{c} = \alpha \vec{a} + \beta \vec{b}\) are coplanar with the vectors \(\hat{i} - 2\hat{j} + \hat{k}\) and \(3\hat{i} + 2\hat{j} - \hat{k}\).

### Step 1: Write the expressions for the vectors
We have:
\[ \vec{a} = \hat{i} + \hat{j} \]
\[ \vec{b} = \hat{j} + \hat{k} \]
Thus, we can express \(\vec{c}\) as:
\[ \vec{c} = \alpha \vec{a} + \beta \vec{b} = \alpha(\hat{i} + \hat{j}) + \beta(\hat{j} + \hat{k}) = \alpha \hat{i} + (\alpha + \beta) \hat{j} + \beta \hat{k} \]

### Step 2: Set up the vectors for coplanarity
The three vectors we are considering for coplanarity are:
1. \(\vec{u} = \hat{i} - 2\hat{j} + \hat{k}\)
2. \(\vec{v} = 3\hat{i} + 2\hat{j} - \hat{k}\)
3. \(\vec{c} = \alpha \hat{i} + (\alpha + \beta) \hat{j} + \beta \hat{k}\)

### Step 3: Use the scalar triple product condition
For the vectors to be coplanar, the scalar triple product must equal zero:
\[ \vec{u} \cdot (\vec{v} \times \vec{c}) = 0 \]

### Step 4: Calculate the cross product \(\vec{v} \times \vec{c}\)
We can calculate the cross product using the determinant:
\[ \vec{v} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -1 \\ \alpha & \alpha + \beta & \beta \end{vmatrix} \]
Calculating this determinant, we expand it:
\[ = \hat{i} \begin{vmatrix} 2 & -1 \\ \alpha + \beta & \beta \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & -1 \\ \alpha & \beta \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 2 \\ \alpha & \alpha + \beta \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. \(\begin{vmatrix} 2 & -1 \\ \alpha + \beta & \beta \end{vmatrix} = 2\beta - (-1)(\alpha + \beta) = 2\beta + \alpha + \beta = \alpha + 3\beta\)
2. \(\begin{vmatrix} 3 & -1 \\ \alpha & \beta \end{vmatrix} = 3\beta - (-1)(\alpha) = 3\beta + \alpha\)
3. \(\begin{vmatrix} 3 & 2 \\ \alpha & \alpha + \beta \end{vmatrix} = 3(\alpha + \beta) - 2\alpha = 3\alpha + 3\beta - 2\alpha = \alpha + 3\beta\)

So, we have:
\[ \vec{v} \times \vec{c} = (\alpha + 3\beta)\hat{i} - (3\beta + \alpha)\hat{j} + (\alpha + 3\beta)\hat{k} \]

### Step 5: Calculate the dot product \(\vec{u} \cdot (\vec{v} \times \vec{c})\)
Now we compute:
\[ \vec{u} \cdot (\vec{v} \times \vec{c}) = (1)(\alpha + 3\beta) + (-2)(- (3\beta + \alpha)) + (1)(\alpha + 3\beta) \]
This simplifies to:
\[ = \alpha + 3\beta + 6\beta + 2\alpha + \alpha + 3\beta = 4\alpha + 12\beta \]

### Step 6: Set the equation to zero
Setting the scalar triple product to zero for coplanarity:
\[ 4\alpha + 12\beta = 0 \]
This gives:
\[ 4\alpha = -12\beta \implies \frac{\alpha}{\beta} = -\frac{12}{4} = -3 \]

### Final Answer
Thus, the value of \(\frac{\alpha}{\beta}\) is:
\[ \frac{\alpha}{\beta} = -3 \]