Question

Three honey bees were found flying along the vectors \( \vec{a} = 2\hat{i} - 3\hat{j} + \hat{k} \), \( \vec{b} = 4\hat{j} - 2\hat{k} \) and \( \vec{c} = 3\hat{i} + 2\hat{k} \) respectively. Find the value of \( \lambda \) such that the path for \( \vec{a} + \lambda \vec{b} \) is perpendicular to \( \vec{c} \).

Hint:

Always group components (\( \hat{i}, \hat{j}, \hat{k} \)) first before taking a dot product. Missing components (like \( \hat{j} \) in \( \vec{c} \)) have a coefficient of 0.

Answer 1

Concept:
• Two non-zero vectors \( \vec{u} \) and \( \vec{v} \) are perpendicular if their dot product is zero (\( \vec{u} \cdot \vec{v} = 0 \)).

Step 1: Determine the vector \( \vec{a} + \lambda \vec{b} \)
Substitute the components of \( \vec{a} \) and \( \vec{b} \):
\[ \vec{a} + \lambda \vec{b} = (2\hat{i} - 3\hat{j} + \hat{k}) + \lambda(0\hat{i} + 4\hat{j} - 2\hat{k}) \]
Combine corresponding components:
\[ \vec{a} + \lambda \vec{b} = 2\hat{i} + (4\lambda - 3)\hat{j} + (1 - 2\lambda)\hat{k} \]

Step 2: Apply the perpendicularity condition with \( \vec{c} \)
For \( (\vec{a} + \lambda \vec{b}) \perp \vec{c} \):
\[ (2\hat{i} + (4\lambda - 3)\hat{j} + (1 - 2\lambda)\hat{k}) \cdot (3\hat{i} + 0\hat{j} + 2\hat{k}) = 0 \]

Step 3: Calculate dot product and solve for \( \lambda \)
Multiply corresponding components and sum them:
\[ (2 \times 3) + (4\lambda - 3) \times 0 + (1 - 2\lambda) \times 2 = 0 \]
\[ 6 + 0 + 2 - 4\lambda = 0 \]
\[ 8 - 4\lambda = 0 \implies 4\lambda = 8 \implies \lambda = 2 \]