JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(A(2,3,5)\) and \(C(-3,4,-2)\) be opposite vertices of a parallelogram \(A B C D\) if the diagonal \(\overrightarrow{B D}=\hat{i}+2 \hat{j}+3 \hat{k}\) then the area of the parallelogram is equal to
- A \(\frac{1}{2} \sqrt{410}\)
- B \(\frac{1}{2} \sqrt{474}\)
- C \(\frac{1}{2} \sqrt{586}\)
- D \(\frac{1}{2} \sqrt{306}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{2} \sqrt{474}\)
Step-by-step Solution
Detailed explanation
Area \(=|\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{BD}}|\) \(=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 5 & -1 & 7 \\ 1 & 2 & 3\end{array}\right|\) \(=\frac{1}{2}|-17 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}+11 \hat{\mathrm{k}}|=\frac{1}{2} \sqrt{474}\)
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