The area of a parallelogram with diagonals $\mathbf{a} = 3\mathbf{i} + \mathbf{j} - 2\mathbf{k}$ and $\mathbf{b} = \mathbf{i} - 3\mathbf{j} + 4\mathbf{k}$ is

Question

The lines $ \frac{1 - x}{2} = \frac{y - 1}{3} = \frac{z}{1} $ and $ \frac{2x - 3}{2p} = \frac{y}{-1} = \frac{z - 4}{7} $ are perpendicular to each other for $ p $ equal to:

Answer 1

To determine the area of a parallelogram given vectors representing its diagonals, we utilize the formula for the area $\text{Area} = \frac{1}{2} |\mathbf{a} \times \mathbf{b}|$, where $\mathbf{a}$ and $\mathbf{b}$ are the diagonal vectors.

The vectors provided are:

$\mathbf{a} = 3\mathbf{i} + \mathbf{j} - 2\mathbf{k}$
$\mathbf{b} = \mathbf{i} - 3\mathbf{j} + 4\mathbf{k}$

To find the cross product $\mathbf{a} \times \mathbf{b}$, we use the determinant method:

\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 1 & -2 \\ 1 & -3 & 4 \end{vmatrix} \]

Calculate the determinant to find $\mathbf{a} \times \mathbf{b}$:

Along $\mathbf{i}$: $=(1)(4) - (-3)(-2) = 4 - 6 = -2$
Along $\mathbf{j}$: $=(4)(3) - (-2)(1) = 12 + 2 = 14$ (Note that the sign before $\mathbf{j}$ must be negative)
Along $\mathbf{k}$: $=(3)(-3) - (1)(1) = -9 - 1 = -10$

So, $\mathbf{a} \times \mathbf{b} = -2\mathbf{i} - 14\mathbf{j} - 10\mathbf{k}$

Next, find the magnitude of this vector:

\[ |\mathbf{a} \times \mathbf{b}| = \sqrt{(-2)^2 + (-14)^2 + (-10)^2} = \sqrt{4 + 196 + 100} = \sqrt{300} = 10\sqrt{3} \]

The area of parallelogram is:

\[ \text{Area} = \frac{1}{2} \times 10\sqrt{3} = 5\sqrt{3} \]

Therefore, the area of the parallelogram is $5\sqrt{3}$, thus the correct answer is:

Option: $5\sqrt{3}$

Answer 2