Question:medium
Let \( \vec{a} = \hat{i}+\hat{j}+\hat{k}, \vec{b} = \hat{i}+3\hat{j}+5\hat{k}, \vec{c} = 7\hat{i}+9\hat{j}+11\hat{k} \). Then the area of parallelogram with diagonals \( \vec{a}+\vec{b} \) and \( \vec{b}+\vec{c} \) is
Let \( \vec{a} = \hat{i}+\hat{j}+\hat{k}, \vec{b} = \hat{i}+3\hat{j}+5\hat{k}, \vec{c} = 7\hat{i}+9\hat{j}+11\hat{k} \). Then the area of parallelogram with diagonals \( \vec{a}+\vec{b} \) and \( \vec{b}+\vec{c} \) is
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Area from diagonals uses half of cross product magnitude.
Updated On: May 1, 2026
- \( 4\sqrt{6} \)
- \( \frac{1}{2}\sqrt{21} \)
- \( \frac{\sqrt{6}}{2} \)
- \( \sqrt{6} \)
- \( \frac{1}{\sqrt{6}} \)
Show Solution
The Correct Option is D
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