Question

The diagonals of a parallelogram are the vectors \(3\hat{i}+6\hat{j}-2\hat{k}\) and \(-\hat{i}-2\hat{j}-8\hat{k}\) then the length of the shorter side of parallelogram is

• A. \(2\sqrt3\)
• B. \(\sqrt{14}\)
• C. \(3\sqrt5\)
• D. \(4\sqrt3\)

Answer 1

The Correct Option is B

Given: Diagonals of a parallelogram are

\[ \vec{d_1} = 3\hat{i} + 6\hat{j} - 2\hat{k}, \quad \vec{d_2} = -\hat{i} - 2\hat{j} - 8\hat{k} \]

Step 1: Sides of the parallelogram are half the sum and half the difference of the diagonals:

\[ \vec{a} = \frac{1}{2}(\vec{d_1} + \vec{d_2}), \quad \vec{b} = \frac{1}{2}(\vec{d_1} - \vec{d_2}) \]

Step 2: Calculate \( \vec{a} \)

\[ \vec{a} = \frac{1}{2}[(3 - 1)\hat{i} + (6 - 2)\hat{j} + (-2 - 8)\hat{k}] = \frac{1}{2}(2\hat{i} + 4\hat{j} - 10\hat{k}) = \hat{i} + 2\hat{j} - 5\hat{k} \]

Step 3: Calculate \( \vec{b} \)

\[ \vec{b} = \frac{1}{2}[(3 + 1)\hat{i} + (6 + 2)\hat{j} + (-2 + 8)\hat{k}] = \frac{1}{2}(4\hat{i} + 8\hat{j} + 6\hat{k}) = 2\hat{i} + 4\hat{j} + 3\hat{k} \]

Step 4: Find magnitudes of the sides

\[ |\vec{a}| = \sqrt{1^2 + 2^2 + (-5)^2} = \sqrt{1 + 4 + 25} = \sqrt{30} \] \[ |\vec{b}| = \sqrt{2^2 + 4^2 + 3^2} = \sqrt{4 + 16 + 9} = \sqrt{29} \]

Shorter side = \( \sqrt{29} \)

Correct option: \( \sqrt{29} \)

Answer 2

Let the diagonals of the parallelogram be \( \vec{d_1} = 3\hat{i} + 6\hat{j} - 2\hat{k} \) and \( \vec{d_2} = -\hat{i} - 2\hat{j} - 8\hat{k} \). 

The sides of the parallelogram, \( \vec{a} \) and \( \vec{b} \), can be found using the following relationships:

\( \vec{a} = \frac{1}{2}(\vec{d_1} + \vec{d_2}) \) and \( \vec{b} = \frac{1}{2}(\vec{d_1} - \vec{d_2}) \)

Calculate \( \vec{a} \):

\( \vec{a} = \frac{1}{2}((3\hat{i} + 6\hat{j} - 2\hat{k}) + (-\hat{i} - 2\hat{j} - 8\hat{k})) = \frac{1}{2}(2\hat{i} + 4\hat{j} - 10\hat{k}) = \hat{i} + 2\hat{j} - 5\hat{k} \)

Calculate \( \vec{b} \):

\( \vec{b} = \frac{1}{2}((3\hat{i} + 6\hat{j} - 2\hat{k}) - (-\hat{i} - 2\hat{j} - 8\hat{k})) = \frac{1}{2}(4\hat{i} + 8\hat{j} + 6\hat{k}) = 2\hat{i} + 4\hat{j} + 3\hat{k} \)

Find the magnitudes of \( \vec{a} \) and \( \vec{b} \):

\( |\vec{a}| = \sqrt{(1)^2 + (2)^2 + (-5)^2} = \sqrt{1 + 4 + 25} = \sqrt{30} \)

\( |\vec{b}| = \sqrt{(2)^2 + (4)^2 + (3)^2} = \sqrt{4 + 16 + 9} = \sqrt{29} \)

The length of the shorter side is \( \sqrt{29} \).