MHT CET · Maths · Vector Algebra
If the area of parallelogram, whose diagonals are \(\hat{i}-\hat{j}+2 \hat{k}\) and \(2 \hat{i}+3 \hat{j}+\alpha \hat{k}\) is \(\frac{\sqrt{93}}{2}\) sq. unit, then \(\alpha=\)
- A \(-4,2\)
- B \(-3,-2\)
- C 2,1
- D 4,2
Answer & Solution
Correct Answer
(A) \(-4,2\)
Step-by-step Solution
Detailed explanation
Area \( = \frac{1}{2} | \vec{d_1} \times \vec{d_2} | \) \( \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 2 & 3 & \alpha \end{vmatrix} = \hat{i}(-\alpha-6) - \hat{j}(\alpha-4) + \hat{k}(3+2) = (-\alpha-6)\hat{i} + (4-\alpha)\hat{j} + 5\hat{k} \)
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