If `veca =hati + hatj - hatk, vecb = 2hati + 3hatj + hatk` and `vec c = hati + alpha hatj` are coplanar vector , then the value of `alpha` is :

Options

A. `-4/3`
B. `3/4`
C. `4/3`
D. `2`

Answer

To solve the problem, we need to find the value of \( \alpha \) such that the vectors \( \vec{a} = \hat{i} + \hat{j} - \hat{k} \), \( \vec{b} = 2\hat{i} + 3\hat{j} + \hat{k} \), and \( \vec{c} = \hat{i} + \alpha \hat{j} \) are coplanar.

The condition for three vectors to be coplanar is that their scalar triple product must be zero. The scalar triple product of vectors \( \vec{a}, \vec{b}, \vec{c} \) can be computed using the determinant of a matrix formed by these vectors.

### Step 1: Write the vectors in matrix form
We can express the vectors in a matrix as follows:

\[ \begin{vmatrix} 1 & 1 & -1 \\ 2 & 3 & 1 \\ 1 & \alpha & 0 \end{vmatrix} \]

### Step 2: Calculate the determinant
We will calculate the determinant of the matrix:

\[ D = \begin{vmatrix} 1 & 1 & -1 \\ 2 & 3 & 1 \\ 1 & \alpha & 0 \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix, we have:

\[ D = 1 \cdot \begin{vmatrix} 3 & 1 \\ \alpha & 0 \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & 1 \\ 1 & 0 \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & 3 \\ 1 & \alpha \end{vmatrix} \]

### Step 3: Calculate the 2x2 determinants
Now, we calculate each of the 2x2 determinants:

1. \( \begin{vmatrix} 3 & 1 \\ \alpha & 0 \end{vmatrix} = (3)(0) - (1)(\alpha) = -\alpha \)
2. \( \begin{vmatrix} 2 & 1 \\ 1 & 0 \end{vmatrix} = (2)(0) - (1)(1) = -1 \)
3. \( \begin{vmatrix} 2 & 3 \\ 1 & \alpha \end{vmatrix} = (2)(\alpha) - (3)(1) = 2\alpha - 3 \)

### Step 4: Substitute back into the determinant
Substituting these values back into the determinant expression:

\[ D = 1(-\alpha) - 1(-1) - 1(2\alpha - 3) \]

This simplifies to:

\[ D = -\alpha + 1 - (2\alpha - 3) \]

### Step 5: Simplify the expression
Now, simplify the expression:

\[ D = -\alpha + 1 - 2\alpha + 3 = -3\alpha + 4 \]

### Step 6: Set the determinant to zero
For the vectors to be coplanar, we set the determinant equal to zero:

\[ -3\alpha + 4 = 0 \]

### Step 7: Solve for \( \alpha \)
Now, solve for \( \alpha \):

\[ -3\alpha = -4 \implies \alpha = \frac{4}{3} \]

Thus, the value of \( \alpha \) is \( \frac{4}{3} \).

### Final Answer
The value of \( \alpha \) is \( \frac{4}{3} \).

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