The value of `alpha` for which the vectors `2i-j+k, i+2j+alphak" and "3i-4j+5k` are coplanar is

Options

A. 3
B. -3
C. 2
D. none of these

Answer

To find the value of \( \alpha \) for which the vectors \( \mathbf{a} = 2\mathbf{i} - \mathbf{j} + \mathbf{k} \), \( \mathbf{b} = \mathbf{i} + 2\mathbf{j} + \alpha\mathbf{k} \), and \( \mathbf{c} = 3\mathbf{i} - 4\mathbf{j} + 5\mathbf{k} \) are coplanar, we can use the condition that the scalar triple product of the vectors must be equal to zero.

### Step 1: Set up the scalar triple product
The scalar triple product of vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c} \) is given by the determinant of the matrix formed by the components of these vectors:

\[ \begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & \alpha \\ 3 & -4 & 5 \end{vmatrix} \]

### Step 2: Calculate the determinant
We will compute the determinant using the formula for the determinant of a 3x3 matrix:

\[ \text{Det} = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Where:
- \( a = 2, b = -1, c = 1 \)
- \( d = 1, e = 2, f = \alpha \)
- \( g = 3, h = -4, i = 5 \)

Substituting the values into the determinant formula:

\[ \text{Det} = 2 \begin{vmatrix} 2 & \alpha \\ -4 & 5 \end{vmatrix} - (-1) \begin{vmatrix} 1 & \alpha \\ 3 & 5 \end{vmatrix} + 1 \begin{vmatrix} 1 & 2 \\ 3 & -4 \end{vmatrix} \]

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

1. \( \begin{vmatrix} 2 & \alpha \\ -4 & 5 \end{vmatrix} = (2)(5) - (-4)(\alpha) = 10 + 4\alpha \)
2. \( \begin{vmatrix} 1 & \alpha \\ 3 & 5 \end{vmatrix} = (1)(5) - (3)(\alpha) = 5 - 3\alpha \)
3. \( \begin{vmatrix} 1 & 2 \\ 3 & -4 \end{vmatrix} = (1)(-4) - (3)(2) = -4 - 6 = -10 \)

### Step 4: Substitute back into the determinant
Now substituting back into the determinant:

\[ \text{Det} = 2(10 + 4\alpha) + (5 - 3\alpha) - 10 \]

### Step 5: Simplify the expression
Expanding and simplifying:

\[ = 20 + 8\alpha + 5 - 3\alpha - 10 \]
\[ = 15 + 5\alpha \]

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

\[ 15 + 5\alpha = 0 \]

### Step 7: Solve for \( \alpha \)
Solving for \( \alpha \):

\[ 5\alpha = -15 \]
\[ \alpha = -3 \]

Thus, the value of \( \alpha \) for which the vectors are coplanar is \( \alpha = -3 \).

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