Question

If \( \vec{a} = 2\hat{i} + 2\hat{j} + 3\hat{k} \), \( \vec{b} = -\hat{i} + 2\hat{j} + \hat{k} \), and \( \vec{c} = 3\hat{i} + \hat{j} \) are such that \( \vec{a} + \lambda \vec{b} \) is perpendicular to \( \vec{c} \), then find the value of \( \lambda \).

Answer 1

Given the vectors:

\[ \vec{a} = 2\hat{i} + 2\hat{j} + 3\hat{k}, \quad \vec{b} = -\hat{i} + 2\hat{j} + \hat{k}, \quad \vec{c} = 3\hat{i} + \hat{j} \] we are told that \( \vec{a} + \lambda \vec{b} \) is perpendicular to \( \vec{c} \).

For \( \vec{a} + \lambda \vec{b} \) to be perpendicular to \( \vec{c} \), their dot product must be zero:

\[ (\vec{a} + \lambda \vec{b}) \cdot \vec{c} = 0 \] Expanding the dot product: \[ (\vec{a} + \lambda \vec{b}) \cdot \vec{c} = \vec{a} \cdot \vec{c} + \lambda (\vec{b} \cdot \vec{c}) \] Now, calculate each dot product: \[ \vec{a} \cdot \vec{c} = (2\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (3\hat{i} + \hat{j}) = 2(3) + 2(1) + 3(0) = 6 + 2 = 8 \] \[ \vec{b} \cdot \vec{c} = (-\hat{i} + 2\hat{j} + \hat{k}) \cdot (3\hat{i} + \hat{j}) = (-1)(3) + 2(1) + 1(0) = -3 + 2 = -1 \] Now substitute these values into the equation: \[ 8 + \lambda (-1) = 0 \] Solving for \( \lambda \): \[ 8 - \lambda = 0 \quad \Rightarrow \quad \lambda = 8 \]

Final Answer:

\[ \lambda = 8 \]