Question

If two vectors \( \vec{a} = 2\hat{i} + \lambda\hat{j} + \hat{k} \) and \( \vec{b} = 4\hat{i} - 2\hat{j} - 2\hat{k} \) are perpendicular to each other, determine the value of the scalar constant \( \lambda \).

• A. \( 3 \)
• B. \( -3 \)
• C. \( 6 \)
• D. \( 0 \)

Hint:

Keep this clear distinction in mind: use a zero dot product (\( \vec{a} \cdot \vec{b} = 0 \)) to test for perpendicular lines, and use proportional directional components (\( \frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} \)) to test for parallel lines.

Answer 1

The Correct Option is A

Concept: Two non-zero vector coordinates \( \vec{a} \) and \( \vec{b} \) are perpendicular (orthogonal) to each other if and only if their scalar dot product evaluates to exactly zero: \[ \vec{a} \cdot \vec{b} = 0 \] This follows from the definition \( \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta \), where a perpendicular alignment angle of \( \theta = 90^\circ \) makes \( \cos(90^\circ) = 0 \).

Step 1: Set up the algebraic expression for the scalar dot product.
Multiply corresponding components (\(x\), \(y\), and \(z\)) of both vectors and sum them up: \[ \vec{a} \cdot \vec{b} = (2)(4) + (\lambda)(-2) + (1)(-2) \] Simplify the arithmetic multiplication steps: \[ \vec{a} \cdot \vec{b} = 8 - 2\lambda - 2 = 6 - 2\lambda \]

Step 2: Equate the dot product to zero to solve for \( \lambda \).
Since the problem specifies that the lines run perpendicular to each other: \[ 6 - 2\lambda = 0 \quad \Rightarrow \quad 2\lambda = 6 \quad \Rightarrow \quad \lambda = 3 \]