Let the vectors \(\mathbf{u}_1 = \hat{i} + \hat{j} + a\hat{k}, \mathbf{u}_2 = \hat{i} + b\hat{j} + \hat{k}\), and \(\mathbf{u}_3 = c\hat{i} + \hat{j} + \hat{k}\) be coplanar. If the vectors \(\mathbf{v}_1 = (a + b)\hat{i} + c\hat{j} + c\hat{k}, \mathbf{v}_2 = a\hat{i} + (b + c)\hat{j} + a\hat{k}, \mathbf{v}_3 = b\hat{i} + b\hat{j} + (c + a)\hat{k}\) are also coplanar, then \(6(a + b + c)\) is equal to:
Let the vectors \(\mathbf{u}_1 = \hat{i} + \hat{j} + a\hat{k}, \mathbf{u}_2 = \hat{i} + b\hat{j} + \hat{k}\), and \(\mathbf{u}_3 = c\hat{i} + \hat{j} + \hat{k}\) be coplanar. If the vectors \(\mathbf{v}_1 = (a + b)\hat{i} + c\hat{j} + c\hat{k}, \mathbf{v}_2 = a\hat{i} + (b + c)\hat{j} + a\hat{k}, \mathbf{v}_3 = b\hat{i} + b\hat{j} + (c + a)\hat{k}\) are also coplanar, then \(6(a + b + c)\) is equal to:
Show Hint
The scalar triple product of three vectors being zero implies that the vectors are coplanar. Utilize this property to solve problems involving coplanarity.
- 0
- 4
- 6
- 12
The Correct Option is D
Solution and Explanation
For the vectors \( \mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3 \) to be coplanar, their scalar triple product must be zero:
\[ [\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3] = \begin{vmatrix} 1 & 1 & c \\ 1 & b & 1 \\ a & 1 & 1 \end{vmatrix} = 0. \]
Expanding the determinant:
\[ [\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3] = 1 \begin{vmatrix} b & 1 \\ 1 & 1 \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ a & 1 \end{vmatrix} + c \begin{vmatrix} 1 & b \\ a & 1 \end{vmatrix}. \]
Simplify:
\[ [\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3] = (b - 1) - (1 - a) + c(a - b). \]
Thus:
\[ b - 1 - 1 + a + ca - cb = 0. \]
\[ a + b + c(1 - b) = 0. \]
For the vectors \( \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 \) to be coplanar:
\[ [\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3] = \begin{vmatrix} a + b & c & c \\ a & b + c & a \\ b & b & c + a \end{vmatrix} = 0. \]
Perform row operations:
\[ R_3 \to R_3 - (R_1 + R_2). \]
The matrix becomes:
\[ \begin{vmatrix} a + b & c & c \\ a & b + c & a \\ -a & b + c - 2a & -2c \end{vmatrix}. \]
Expanding along the first row:
\[ [\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3] = (a + b) \begin{vmatrix} b + c & a \\ b + c - 2a & -2c \end{vmatrix} - c \begin{vmatrix} a & a \\ b & -2c \end{vmatrix} + c \begin{vmatrix} a & b + c \\ b & b + c - 2a \end{vmatrix}. \]
Simplify each determinant and substitute:
\[ [\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3] = 4abc = 0 \quad \implies \quad abc = 0. \]
Substituting \( abc = 0 \) and \( a + b + c = 2 \), we find:
\[ 6(a + b + c) = 6 \times 2 = 12. \]
Final Answer:
\[ 12 \, (\text{Option 4}). \]
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