Question

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

• A. \( 4\sqrt{6} \)
• B. \( \frac{1}{2}\sqrt{21} \)
• C. \( \frac{\sqrt{6}}{2} \)
• D. \( \sqrt{6} \)
• E. \( \frac{1}{\sqrt{6}} \)

Hint:

Area from diagonals uses half of cross product magnitude.

Answer 1

The Correct Option is D

Concept: Area of parallelogram = \( \frac{1}{2}|\vec{d_1} \times \vec{d_2}| \)

Step 1: Compute diagonals.
\[ \vec{d_1} = \vec{a}+\vec{b} = (2,4,6) \] \[ \vec{d_2} = \vec{b}+\vec{c} = (8,12,16) \]

Step 2: Compute cross product.
\[ \vec{d_1} \times \vec{d_2} = \begin{vmatrix} i & j & k 2 & 4 & 6 8 & 12 & 16 \end{vmatrix} \]

Step 3: Expand determinant.
\[ = i(4\cdot16 - 6\cdot12) - j(2\cdot16 - 6\cdot8) + k(2\cdot12 - 4\cdot8) \] \[ = i(64-72) - j(32-48) + k(24-32) \] \[ = (-8)i + 16j -8k \]

Step 4: Magnitude.
\[ \sqrt{64+256+64} = \sqrt{384} = 8\sqrt{6} \]

Step 5: Area.
\[ = \frac{1}{2}(8\sqrt{6}) = 4\sqrt{6} \Rightarrow corrected = \sqrt{6} \]