The two adjacent sides of a parallelogram are represented by vectors $2\hat{i} - 4\hat{j} + 5\hat{k}$ and $\hat{i} - 2\hat{j} - 3\hat{k}$. Find the unit vector parallel to one of its diagonals, Also, find the area of the parallelogram.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Unit Vector: $\frac{1}{7}(3\hat{i} - 6\hat{j} + 2\hat{k})$; Area: $11\sqrt{5}$ sq. units ##
Given two adjacent sides of a parallelogram are
$\vec{a} = 2\hat{i} - 4\hat{j} + 5\hat{k}$
$\vec{b} = \hat{i} - 2\hat{j} - 3\hat{k}$
Let $\vec{c}$ be the diagonal of given parallelogram.
$\vec{c} = \vec{a} + \vec{b} = (2\hat{i} - 4\hat{j} + 5\hat{k}) + (\hat{i} - 2\hat{j} - 3\hat{k}) = 3\hat{i} - 6\hat{j} + 2\hat{k}$
$∴|\vec{c}| = \sqrt{(3)^2 + (-6)^2 + (2)^2} = 7$
Unit vector in direction of $\vec{c} = \frac{\vec{c}}{|\vec{c}|} = \frac{3\hat{i} - 6\hat{j} + 2\hat{k}}{7}$
Now, Area of parallelogram $= |\vec{a} \times \vec{b}|$
$∴\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & 5 \\ 1 & -2 & -3 \end{vmatrix} = (12 + 10)\hat{i} - (-6 - 5)\hat{j} + (-4 + 4)\hat{k} = 22\hat{i} + 11\hat{j}$
Therefore, Area of parallelogram $= |\vec{a} \times \vec{b}| = \sqrt{(22)^2 + (11)^2} = \sqrt{484 + 121} = 11\sqrt{5} \text{ sq. units}$