Question

Let \( \vec{c} \) be the projection vector of \( \mathbf{b} = \lambda \hat{i} + 4 \hat{k}, \lambda > 0 \), on the vector \( \vec{a} = \hat{i} + 2 \hat{j} + 2 \hat{k} \). If \( |\vec{a} + \vec{c}| = 7 \), then the area of the parallelogram formed by the vectors \( \vec{b} \) and \( \vec{c} \) is:

Hint:

For the area of the parallelogram formed by two vectors, use the magnitude of their cross product.

Answer 1

Correct Answer: 16

The projection vector \( \vec{c} \) is given by: \[ \vec{c} = \frac{(\vec{b} \cdot \vec{a})}{|\vec{a}|} \vec{a} \] Substituting the values: \[ \vec{c} = \left( \frac{(\lambda \hat{i} + 4 \hat{k}) \cdot (\hat{i} + 2 \hat{j} + 2 \hat{k})}{|\hat{i} + 2 \hat{j} + 2 \hat{k}|} \right) (\hat{i} + 2 \hat{j} + 2 \hat{k}) \] \[ |\vec{a}| = \sqrt{1^2 + 2^2 + 2^2} = 3, \quad \vec{c} = \left( \frac{\lambda + 8}{9} \right) ( \hat{i} + 2 \hat{j} + 2 \hat{k}) \] Using \( |\vec{a} + \vec{c}| = 7 \), we get \( \lambda = 4 \). The area of the parallelogram is given by the magnitude of the cross product \( \vec{b} \times \vec{c} \): \[ \text{Area} = |\vec{b} \times \vec{c}| = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k}
\frac{4}{3} & \frac{8}{3} & \frac{8}{3}
4 & 8 & 8 \end{matrix} \right| = 16 \]