Question

The sides of a parallelogram are represented by vectors $$ \vec{p} = 5\hat{i} - 4\hat{j} + 3\hat{k} \quad \text{and} \quad \vec{q} = 3\hat{i} + 2\hat{j} - \hat{k}. $$ Then, the area of the parallelogram is

• A. \( \sqrt{684} \) sq units
• B. \( \sqrt{72} \) sq units
• C. 171 sq units
• D. 72 sq units

Hint:

The area of the parallelogram formed by two vectors is the magnitude of their cross product. Remember to use the determinant method for calculating cross products of vectors.

Answer 1

The Correct Option is A

The area of a parallelogram formed by two vectors \( \vec{p} \) and \( \vec{q} \) is given by the magnitude of the cross product of the vectors: \[ \text{Area} = |\vec{p} \times \vec{q}| \] To calculate the cross product \( \vec{p} \times \vec{q} \), we use the determinant method: \[ \vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 5 & -4 & 3 3 & 2 & -1 \end{vmatrix} \] Expanding the determinant: \[ \vec{p} \times \vec{q} = \hat{i} \begin{vmatrix} -4 & 3 2 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 5 & 3 \\ 3 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 5 & -4 \\ 3 & 2 \end{vmatrix} \] Solving each 2x2 determinant: \[ = \hat{i} [(-4)(-1) - (3)(2)] - \hat{j} [(5)(-1) - (3)(3)] + \hat{k} [(5)(2) - (-4)(3)] \] \[ = \hat{i} [4 - 6] - \hat{j} [-5 - 9] + \hat{k} [10 + 12] \] \[ = \hat{i} (-2) - \hat{j} (-14) + \hat{k} (22) \] \[ = -2\hat{i} + 14\hat{j} + 22\hat{k} \] Now, calculate the magnitude of the cross product: \[ |\vec{p} \times \vec{q}| = \sqrt{(-2)^2 + 14^2 + 22^2} \] \[ = \sqrt{4 + 196 + 484} = \sqrt{684} \]
Thus, the area of the parallelogram is \( \sqrt{684} \) square units.

Answer 2

To determine the area of a parallelogram formed by vectors \( \vec{p} \) and \( \vec{q} \), we need to calculate the magnitude of their cross product, \( \vec{p} \times \vec{q} \).

Given vectors are:

\( \vec{p} = 5\hat{i} - 4\hat{j} + 3\hat{k} \)

\( \vec{q} = 3\hat{i} + 2\hat{j} - \hat{k} \)

The cross product \( \vec{p} \times \vec{q} \) is computed using the determinant:

\(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & -4 & 3 \\ 3 & 2 & -1 \end{vmatrix}\)

Expanding the determinant, we have:

\(\vec{p} \times \vec{q} = \hat{i}((-4)(-1) - 3(2)) - \hat{j}(5(-1) - 3(3)) + \hat{k}(5(2) - (-4)(3))\)

Calculate step-by-step:

• For \( \hat{i} \): \( 4 - 6 = -2 \)
• For \( \hat{j} \): \( -5 - 9 = -14 \) (remember the negative sign from the determinant expansion, giving us \( 14 \hat{j} \))
• For \( \hat{k} \): \( 10 + 12 = 22 \)

This yields:

\(\vec{p} \times \vec{q} = -2\hat{i} + 14\hat{j} + 22\hat{k} \)

The magnitude of this cross product vector is:

\(|\vec{p} \times \vec{q}| = \sqrt{(-2)^2 + (14)^2 + (22)^2}\)

Calculating further:

\(|\vec{p} \times \vec{q}| = \sqrt{4 + 196 + 484} = \sqrt{684}\)

Thus, the area of the parallelogram is \( \sqrt{684} \) square units.