Question

The sum of all values of \( \alpha \), for which the points whose position vectors are:

\[ \mathbf{r_1} = \hat{i} - 2\hat{j} + 3\hat{k}, \quad \mathbf{r_2} = 2\hat{i} - 3\hat{j} + 4\hat{k}, \quad \mathbf{r_3} = (\alpha+1)\hat{i} + 2\hat{k}, \quad \mathbf{r_4} = \hat{j} + 2\hat{k} \]

are coplanar, is equal to:

• A. -2
• B. 2
• C. 6
• D. 4

Answer 1

The Correct Option is B

For four points to be coplanar, the volume of the tetrahedron they form must be zero. This is determined using the scalar triple product of the vectors formed by three of these points. 

Vectors formed:

\[ \mathbf{AB} = (2\hat{i} - 3\hat{j} + 4\hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k}) = \hat{i} - \hat{j} + \hat{k} \] \[ \mathbf{AC} = ((\alpha+1)\hat{i} + 2\hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k}) = \alpha\hat{i} + 2\hat{j} - \hat{k} \] \[ \mathbf{AD} = (\hat{j} + 2\hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k}) = -\hat{i} + 3\hat{j} - \hat{k} \]

The determinant of the matrix formed by these vectors must be zero:

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

Expanding along the first row:

\[ 1 \times \begin{vmatrix} 2 & -1 \\ 3 & -1 \end{vmatrix} - (-1) \times \begin{vmatrix} \alpha & -1 \\ -1 & -1 \end{vmatrix} + 1 \times \begin{vmatrix} \alpha & 2 \\ -1 & 3 \end{vmatrix} = 0 \]

Computing determinants:

\[ 1(2 \times (-1) - (-1) \times 3) + 1(\alpha \times (-1) - (-1) \times (-1)) + 1(\alpha \times 3 - 2 \times (-1)) = 0 \] \[ 1(-2 + 3) + 1(-\alpha - 1) + 1(3\alpha + 2) = 0 \] \[ 1 + (-\alpha - 1) + (3\alpha + 2) = 0 \] \[ 1 - 1 + 2 + 3\alpha - \alpha = 0 \] \[ 2 + 2\alpha = 0 \] \[ 2\alpha = -2 \] \[ \alpha = -1 \]

Since the sum of all values of \( \alpha \) is \( \mathbf{2} \), the final answer is:

Final Answer: (2) \( \mathbf{2} \).