Question

If the area of a parallelogram, whose diagonals are $\hat{i} - \hat{j} + 2\hat{k}$ and $2\hat{i} + 3\hat{j} + \alpha \hat{k}$ is $\dfrac{\sqrt{93}}{2}$ sq. units, then find $\alpha$.

• A. -4, 2
• B. -3, -2
• C. 2, 1
• D. 4, 2

Hint:

Area of parallelogram $= |\vec{a} \times \vec{b}|$ (using sides) OR $\frac{1}{2} |\vec{d_1} \times \vec{d_2}|$ (using diagonals).

Answer 1

The Correct Option is B

Step 1: Formula for Area
Area $= \frac{1}{2} |\vec{d_1} \times \vec{d_2}|$.
Step 2: Compute Cross Product
$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 2 & 3 & \alpha \end{vmatrix} = (-\alpha-6)\hat{i} - (\alpha-4)\hat{j} + (3+2)\hat{k} = (-\alpha-6)\hat{i} + (4-\alpha)\hat{j} + 5\hat{k}$.
Step 3: Equate Area
$\frac{1}{2} \sqrt{(-\alpha-6)^2 + (4-\alpha)^2 + 25} = \frac{\sqrt{93}}{2}$.
$\alpha^2 + 12\alpha + 36 + 16 - 8\alpha + \alpha^2 + 25 = 93 \implies 2\alpha^2 + 4\alpha + 77 = 93 \implies 2\alpha^2 + 4\alpha - 16 = 0$.
$\alpha^2 + 2\alpha - 8 = 0 \implies (\alpha+4)(\alpha-2) = 0 \implies \alpha = -4, 2$.
Final Answer: (A)