The value of `alpha` for which point `M(alpha hati+2hatj+hatk)`, lie in the plane containing three points `A(hati+hatj+hatk) and C(3hati-hatk)` is :

Options

A. 1
B. 5
C. `(1)/(2)`
D. `-(1)/(2)`

Answer

To find the value of `alpha` for which the point \( M(\alpha \hat{i} + 2 \hat{j} + \hat{k}) \) lies in the plane containing the points \( A(\hat{i} + \hat{j} + \hat{k}) \) and \( C(3 \hat{i} - \hat{k}) \), we will use the condition of coplanarity of the vectors.

### Step 1: Define the vectors
Let:
- \( \vec{M} = \alpha \hat{i} + 2 \hat{j} + \hat{k} \)
- \( \vec{A} = \hat{i} + \hat{j} + \hat{k} \)
- \( \vec{C} = 3 \hat{i} + 0 \hat{j} - \hat{k} \)

### Step 2: Find the vectors \( \vec{A} \) and \( \vec{C} \)
We can express the vectors \( \vec{A} \) and \( \vec{C} \) in component form:
- \( \vec{A} = (1, 1, 1) \)
- \( \vec{C} = (3, 0, -1) \)

### Step 3: Calculate \( \vec{A} \times \vec{C} \)
To find the cross product \( \vec{A} \times \vec{C} \), we set up the determinant:

\[ \vec{A} \times \vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 3 & 0 & -1 \end{vmatrix} \]

Calculating the determinant:

\[ \vec{A} \times \vec{C} = \hat{i} \begin{vmatrix} 1 & 1 \\ 0 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 3 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 3 & 0 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} 1 & 1 \\ 0 & -1 \end{vmatrix} = (1)(-1) - (1)(0) = -1 \)
2. \( \begin{vmatrix} 1 & 1 \\ 3 & -1 \end{vmatrix} = (1)(-1) - (1)(3) = -1 - 3 = -4 \)
3. \( \begin{vmatrix} 1 & 1 \\ 3 & 0 \end{vmatrix} = (1)(0) - (1)(3) = -3 \)

Substituting back, we get:

\[ \vec{A} \times \vec{C} = -1 \hat{i} + 4 \hat{j} - 3 \hat{k} = -\hat{i} + 4\hat{j} - 3\hat{k} \]

### Step 4: Find the dot product \( \vec{M} \cdot (\vec{A} \times \vec{C}) \)
Now, we compute the dot product \( \vec{M} \cdot (\vec{A} \times \vec{C}) \):

\[ \vec{M} = (\alpha, 2, 1) \]
\[ \vec{A} \times \vec{C} = (-1, 4, -3) \]

Calculating the dot product:

\[ \vec{M} \cdot (\vec{A} \times \vec{C}) = \alpha(-1) + 2(4) + 1(-3) \]
\[ = -\alpha + 8 - 3 \]
\[ = -\alpha + 5 \]

### Step 5: Set the dot product to zero for coplanarity
For the vectors to be coplanar, we set the dot product equal to zero:

\[ -\alpha + 5 = 0 \]

### Step 6: Solve for \( \alpha \)
Rearranging the equation gives:

\[ \alpha = 5 \]

Thus, the value of \( \alpha \) is \( 5 \).

### Final Answer
The value of \( \alpha \) is \( 5 \).

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