MHT CET · Maths · Vector Algebra
If the area of a parallelogram whose diagonals are represented by vectors \(3 \hat{i}+\lambda \hat{j}+2 \hat{k}\) and \(\hat{i}-2 \hat{j}+3 \hat{k}\) is \(\frac{\sqrt{117}}{2}\) sq. units, then \(\lambda=\)
- A \(-1\)
- B \(-2\)
- C \(-3\)
- D \(-4\)
Answer & Solution
Correct Answer
(D) \(-4\)
Step-by-step Solution
Detailed explanation
\(\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & \lambda & 2 \\ 1 & -2 & 3 \end{vmatrix} = (3\lambda+4)\hat{i} - 7\hat{j} - (\lambda+6)\hat{k}\) \(|\vec{d_1} \times \vec{d_2}|^2 = (3\lambda+4)^2 + (-7)^2 + (-\lambda-6)^2 = 9\lambda^2+24\lambda+16+49+\lambda^2+12\lambda+36 = 10\lambda^2+36\lambda+101\)
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