Question

If \( \vec{a} = \lambda\hat{i} + \hat{j} - 2\hat{k} \), \( \vec{b} = \hat{i} + \lambda\hat{j} - 2\hat{k} \), \( \vec{c} = \hat{i} + \hat{j} + \hat{k} \) and \( [\vec{a}\vec{b}\vec{c}] = 7 \), then the values of \( \lambda \) are

• A. 2, -6
• B. 6, -2
• C. 4, -2
• D. -4, 2

Hint:

The scalar triple product of three vectors is found by calculating the determinant of the 3x3 matrix where the rows (or columns) are the components of the vectors. Set this determinant equal to the given value and solve the resulting equation for the unknown variable.

Answer 1

The Correct Option is A

The scalar triple product \( [\vec{a}\vec{b}\vec{c}] \) is given by the determinant of the matrix formed by the components of the vectors. \[ [\vec{a}\vec{b}\vec{c}] = \begin{vmatrix} \lambda & 1 & -2 \\ 1 & \lambda & -2 \\ 1 & 1 & 1 \end{vmatrix} \] We are given that this determinant equals 7. Let's expand it along the first row: \[ \lambda(\lambda \cdot 1 - (-2) \cdot 1) - 1(1 \cdot 1 - (-2) \cdot 1) + (-2)(1 \cdot 1 - \lambda \cdot 1) = 7 \] \[ \lambda(\lambda + 2) - 1(1 + 2) - 2(1 - \lambda) = 7 \] \[ \lambda^2 + 2\lambda - 3 - 2 + 2\lambda = 7 \] \[ \lambda^2 + 4\lambda - 5 = 7 \] \[ \lambda^2 + 4\lambda - 12 = 0 \] This is a quadratic equation for \( \lambda \). We can solve it by factoring: \[ (\lambda + 6)(\lambda - 2) = 0 \] The solutions are \( \lambda = -6 \) and \( \lambda = 2 \). These values match option (A).