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:
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:
- -2
- 2
- 6
- 4
The Correct Option is B
Solution and Explanation
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) \( \mathbf{2} \).
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Concepts Used:
Vector Algebra
A vector is an object which has both magnitudes and direction. It is usually represented by an arrow which shows the direction(→) and its length shows the magnitude. The arrow which indicates the vector has an arrowhead and its opposite end is the tail. It is denoted as
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