Question

Let $\vec{a} = \hat{i} + \alpha\hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} - \alpha\hat{j} + \hat{k}$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\vec{a}$ and $\vec{b}$ is 8$\sqrt{3}$ square units, then $\vec{a} \cdot \vec{b}$ is equal to ________ .

Hint:

There is a useful identity relating the dot product, cross product, and magnitudes of two vectors: $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$. This is known as Lagrange's identity and can sometimes provide an alternative solution path.

Answer 1

Correct Answer: 2

\[ \vec a=(1,\alpha,3),\quad \vec b=(3,-\alpha,1) \] \[ \vec a\times\vec b=(4\alpha,8,-4\alpha) \] \[ |\vec a\times\vec b|^2=32\alpha^2+64 \] Given area \(=8\sqrt3\): \[ 32\alpha^2+64=192 \Rightarrow \alpha^2=4 \] \[ \vec a\cdot\vec b=3-\alpha^2+3=2 \] \[ \boxed{2} \]