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

The Correct Option is B

The area of a parallelogram formed by vectors can be found using the cross product of the vectors. Given vectors \(\vec{a} = 2\hat{i} - \hat{j} + 5\hat{k}\) and \(\vec{b} = 2\hat{i} + \hat{j} + 2\hat{k}\), the first step is to find the cross product \(\vec{a} \times \vec{b}\). 

The formula for the cross product is: \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 5 \\ 2 & 1 & 2 \end{vmatrix} \]

Compute the determinant:

\[ \vec{a} \times \vec{b} = \hat{i}((-1)\cdot 2 - 5\cdot 1) - \hat{j}(2\cdot 2 - 5\cdot 2) + \hat{k}(2\cdot 1 - (-1)\cdot 2) \]

Simplify each component:

\[ = \hat{i}(-2 - 5) - \hat{j}(4 - 10) + \hat{k}(2 + 2) \]

\[ = -7\hat{i} + 6\hat{j} + 4\hat{k} \]

Thus, \(\vec{a} \times \vec{b} = -7\hat{i} + 6\hat{j} + 4\hat{k}\).

The magnitude of this vector gives the area of the parallelogram:

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

\[ = \sqrt{101} \]

Therefore, the area of the parallelogram is \(\sqrt{101}\).

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:

\[ \text{Area} = \sqrt{101}. \]