Three vertices A, B, C and D of a parallelogram ABCD are given by, $A(0, -3, 3), B(-5, m - 3, 0)$ and $D(1, -3, 4)$. The area of the parallelogram ABCD is 6 sq. units. Using vector method, find the value(s) of $m$.

$\pm 4$

The correct answer is Option (1) → $\pm 4$ ##

Area of $||$gm ABCD} $= |\vec{AB} \times \vec{AD}|$

Now, $\vec{AB} = (-5-0)\hat{i} + (m-3-(-3))\hat{j} + (0-3)\hat{k}$

or, $\vec{AB} = -5\hat{i} + m\hat{j} - 3\hat{k}$

and $\vec{AD} = (1-0)\hat{i} + (-3-(-3))\hat{j} + (4-3)\hat{k}$

or, $\vec{AD} = \hat{i} + \hat{k}$

$∴\vec{AB} \times \vec{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & m & -3 \\ 1 & 0 & 1 \end{vmatrix}$

or, $\vec{AB} \times \vec{AD} = \hat{i}(m - 0) - \hat{j}(-5 + 3) + \hat{k}(0 - m)$

or, $\vec{AB} \times \vec{AD} = m\hat{i} + 2\hat{j} - m\hat{k}$

Now, $|\vec{AB} \times \vec{AD}| = \sqrt{(m)^2 + (2)^2 + (-m)^2}$

or, $|\vec{AB} \times \vec{AD}| = \sqrt{m^2 + 4 + m^2} = \sqrt{2m^2 + 4}$

Given, area of parallelogram ABCD $= 6$ sq. units

$∴|\sqrt{2m^2 + 4}| = 6$

$\Rightarrow 2m^2 + 4 = 36$

$\Rightarrow 2m^2 = 32$

$\Rightarrow m^2 = 16$

$\Rightarrow m = \pm 4$

Thus, value of $m$ is $\pm 4$.