Vector Algebra
Dot Product / Orthogonal Vectors
Grade None
Question:
<p>If the vectors \(\vec{a} = \hat{i} - \hat{j} + 2\hat{k}\), \(\vec{b} = 2\hat{i} + 4\hat{j} + \hat{k}\) and \(\vec{c} = \lambda\hat{i} + \hat{j} + \mu\hat{k}\) are mutually orthogonal, then \((\lambda, \mu) =\)</p>
<p>\((2, -3)\)</p>
<p>\((-2, 3)\)</p>
<p>\((3, 2)\)</p>
<p>\((-3, 2)\)</p>
Step-by-Step Solution
Key Concept: Two vectors are orthogonal when their dot product equals zero. Since all three vectors are mutually orthogonal, we need $\vec{a} \cdot \vec{c} = 0$ and $\vec{b} \cdot \vec{c} = 0$ to find two equations in two unknowns.
Step 1: Apply orthogonality condition $\vec{a} \cdot \vec{c} = 0$ $(\hat{i} - \hat{j} + 2\hat{k}) \cdot (\lambda\hat{i} + \hat{j} + \mu\hat{k}) = 0$ $\lambda - 1 + 2\mu = 0$ $\lambda + 2\mu = 1$ ... (equation 1) Step 2: Apply orthogonality condition $\vec{b} \cdot \vec{c} = 0$ $(2\hat{i} + 4\hat{j} + \hat{k}) \cdot (\lambda\hat{i} + \hat{j} + \mu\hat{k}) = 0$ $2\lambda + 4 + \mu = 0$ $2\lambda + \mu = -4$ ... (equation 2) Step 3: Solve the system of equations From equation 1: $\lambda = 1 - 2\mu$ Substitute into equation 2: $2(1 - 2\mu) + \mu = -4$ $2 - 4\mu + \mu = -4$ $-3\mu = -6$ $\mu = 2$ Therefore: $\lambda = 1 - 2(2) = 1 - 4 = -3$ Verification: $\vec{a} \cdot \vec{c} = -3 - 1 + 4 = 0$ ✓ and $\vec{b} \cdot \vec{c} = -6 + 4 + 2 = 0$ ✓ ∴ Answer: $(\lambda, \mu) = (-3, 2)$
Correct Answer: D