Q167 / 180 · COMEDK 2023 (Morning Shift) · Physics
The sides of a parallelogram are represented by vectors $\vec{p}=5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ and $\vec{q}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}$. Then, the area of the parallelogram is
- A$\sqrt{72}$ sq units
- B72 sq units
- Correct answerC$\sqrt{684}$ sq units
- D171 sq units
Solution
The area of a parallelogram with adjacent sides represented by vectors $\vec{p}$ and $\vec{q}$ is given by the magnitude of their cross product, $|\vec{p} \times \vec{q}|$.
First, calculate the cross product $\vec{p} \times \vec{q}$:
$\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & -4 & 3 \\ 3 & 2 & -1 \end{vmatrix}$
Expanding the determinant:
$\vec{p} \times \vec{q} = \hat{i}((-4)(-1) - (3)(2)) - \hat{j}((5)(-1) - (3)(3)) + \hat{k}((5)(2) - (-4)(3))$
$\vec{p} \times \vec{q} = \hat{i}(4 - 6) - \hat{j}(-5 - 9) + \hat{k}(10 + 12)$
$\vec{p} \times \vec{q} = -2\hat{i} + 14\hat{j} + 22\hat{k}$
Now, calculate the magnitude of the resulting vector:
$|\vec{p} \times \vec{q}| = \sqrt{(-2)^2 + (14)^2 + (22)^2}$
$|\vec{p} \times \vec{q}| = \sqrt{4 + 196 + 484}$
$|\vec{p} \times \vec{q}| = \sqrt{684}$
The area of the parallelogram is $\sqrt{684}$ sq units.
Answer: $\sqrt{684}$ sq units
First, calculate the cross product $\vec{p} \times \vec{q}$:
$\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & -4 & 3 \\ 3 & 2 & -1 \end{vmatrix}$
Expanding the determinant:
$\vec{p} \times \vec{q} = \hat{i}((-4)(-1) - (3)(2)) - \hat{j}((5)(-1) - (3)(3)) + \hat{k}((5)(2) - (-4)(3))$
$\vec{p} \times \vec{q} = \hat{i}(4 - 6) - \hat{j}(-5 - 9) + \hat{k}(10 + 12)$
$\vec{p} \times \vec{q} = -2\hat{i} + 14\hat{j} + 22\hat{k}$
Now, calculate the magnitude of the resulting vector:
$|\vec{p} \times \vec{q}| = \sqrt{(-2)^2 + (14)^2 + (22)^2}$
$|\vec{p} \times \vec{q}| = \sqrt{4 + 196 + 484}$
$|\vec{p} \times \vec{q}| = \sqrt{684}$
The area of the parallelogram is $\sqrt{684}$ sq units.
Answer: $\sqrt{684}$ sq units
Hence, the correct answer is option: 'C'.