Question

The set of real values of \( \lambda \) for which the vectors \( \lambda\hat{i} - 3\hat{j} + 5\hat{k} \) and \( 2\lambda\hat{i} - \lambda\hat{j} + \hat{k} \) are perpendicular to each other is

• A. \( \{0, 1\} \)
• B. \( \{-2\} \)
• C. \( \{2, -1\} \)
• D. \( \phi \) (empty set)

Hint:

Two non-zero vectors are perpendicular if and only if their dot product is zero.
For \(\vec{A} = A_x\hat{i}+A_y\hat{j}+A_z\hat{k}\) and \(\vec{B} = B_x\hat{i}+B_y\hat{j}+B_z\hat{k}\), \(\vec{A}\cdot\vec{B} = A_x B_x + A_y B_y + A_z B_z\).
For a quadratic equation \(ax^2+bx+c=0\), real roots exist if the discriminant \(D = b^2-4ac \ge 0\). If \(D<0\), there are no real roots.

Answer 1

The Correct Option is D

Two vectors \( \vec{A} \) and \( \vec{B} \) are perpendicular if their dot product is zero: \( \vec{A} \cdot \vec{B} = 0 \). Let \( \vec{A} = \lambda\hat{i} - 3\hat{j} + 5\hat{k} \) Let \( \vec{B} = 2\lambda\hat{i} - \lambda\hat{j} + \hat{k} \) Calculate their dot product: \( \vec{A} \cdot \vec{B} = (\lambda)(2\lambda) + (-3)(-\lambda) + (5)(1) \) \( = 2\lambda^2 + 3\lambda + 5 \) For the vectors to be perpendicular, \( \vec{A} \cdot \vec{B} = 0 \). So, we need to solve the quadratic equation \( 2\lambda^2 + 3\lambda + 5 = 0 \) for real values of \(\lambda\). The discriminant of this quadratic equation \(ax^2+bx+c=0\) is \(D = b^2-4ac\). Here, \(a=2, b=3, c=5\). \( D = (3)^2 - 4(2)(5) = 9 - 40 = -31 \). Since the discriminant \(D = -31<0\), the quadratic equation \( 2\lambda^2 + 3\lambda + 5 = 0 \) has no real roots for \(\lambda\). The roots are complex. Therefore, there are no real values of \(\lambda\) for which the given vectors are perpendicular. The set of real values of \(\lambda\) is the empty set, denoted by \( \phi \). This matches option (d). \[ \boxed{\phi} \]