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:

Carefully watch the missing components (like \(\hat{i}\) in \(\vec{b}\) or \(\hat{j}\) in \(\vec{c}\)) and treat them as zero during calculations.
The dot product simplifies a vector equation into a simple linear scalar equation.

Answer 1

Concept:
• Vector addition: \((a_1\hat{i} + b_1\hat{j}) + (a_2\hat{i} + b_2\hat{j}) = (a_1+a_2)\hat{i} + (b_1+b_2)\hat{j}\).
• Perpendicularity: Two non-zero vectors are perpendicular if their dot product is zero.

Step 1: Find the combined vector \(\vec{a} + \lambda\vec{b}\)
\[ \vec{a} + \lambda\vec{b} = (2\hat{i} - 3\hat{j} + \hat{k}) + \lambda(0\hat{i} + 4\hat{j} - 2\hat{k}) \] \[ \vec{a} + \lambda\vec{b} = 2\hat{i} + (-3 + 4\lambda)\hat{j} + (1 - 2\lambda)\hat{k} \]

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

Step 3: Solve the resulting dot product equation
\[ (2)(3) + (-3 + 4\lambda)(0) + (1 - 2\lambda)(2) = 0 \] \[ 6 + 0 + 2 - 4\lambda = 0 \] \[ 8 - 4\lambda = 0 \] \[ 4\lambda = 8 \] \[ \lambda = 2 \]