Question

The area of the parallelogram, whose adjacent sides are given by the vectors $\vec{a} = 2\hat{i} - \hat{j} + 5\hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} + 2\hat{k}$, is:

• A. $\sqrt{105}$
• B. $\sqrt{101}$
• C. $\sqrt{103}$
• D. $\sqrt{102}$

Answer 1

Answer: (B)

$\sqrt{101}$

Answer 2

The area of a parallelogram formed by two vectors is given by the magnitude of their cross product:

$\text{Area} = \| \vec{a} \times \vec{b} \|.$

Compute $\vec{a} \times \vec{b}$:

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\\ 2 & -1 & 5 \\\ 2 & 1 & 2 \end{vmatrix}.$

Expand the determinant:

$\vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} -1 & 5 \\\ 1 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 5 \\\ 2 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -1 \\\ 2 & 1 \end{vmatrix}.$

Calculate each minor:

$\begin{vmatrix} -1 & 5 \\\ 1 & 2 \end{vmatrix} = (-1)(2) - (5)(1) = -2 - 5 = -7,$

$\begin{vmatrix} 2 & 5 \\\ 2 & 2 \end{vmatrix} = (2)(2) - (5)(2) = 4 - 10 = -6,$

$\begin{vmatrix} 2 & -1 \\\ 2 & 1 \end{vmatrix} = (2)(1) - (-1)(2) = 2 + 2 = 4.$

Substitute back into the determinant:

$\vec{a} \times \vec{b} = -7 \hat{i} + 6 \hat{j} + 4 \hat{k}.$

Find the magnitude of $\vec{a} \times \vec{b}$:

$\| \vec{a} \times \vec{b} \| = \sqrt{(-7)^2 + (6)^2 + (4)^2} = \sqrt{49 + 36 + 16} = \sqrt{101}.$

Thus, the area of the parallelogram is:

$\text{Area} = \sqrt{101}.$