The area (in sq. units) of the parallelogram | vedclass

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(C) Let $\vec{d_1} = 8 \hat{i} - 6 \hat{j}$ and $\vec{d_2} = 3 \hat{i} + 4 \hat{j} - 12 \hat{k}$ be the diagonals of the parallelogram. The area of a

Vedclass · Accepted Answer

$65$

Vedclass · Answer

(C) Let $\vec{d_1} = 8 \hat{i} - 6 \hat{j}$ and $\vec{d_2} = 3 \hat{i} + 4 \hat{j} - 12 \hat{k}$ be the diagonals of the parallelogram. The area of a parallelogram with diagonals $\vec{d_1}$ and $\vec{d_2}$ is given by $	ext{Area} = \frac{1}{2} |\vec{d_1} 	imes \vec{d_2}|$. First,calculate the cross product $\vec{d_1} 	imes \vec{d_2}$: $\vec{d_1} 	imes \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 8 & -6 & 0 \ 3 & 4 & -12 \end{vmatrix}$ $= \hat{i}((-6)(-12) - (0)(4)) - \hat{j}((8)(-12) - (0)(3)) + \hat{k}((8)(4) - (-6)(3))$ $= \hat{i}(72 - 0) - \hat{j}(-96 - 0) + \hat{k}(32 + 18)$ $= 72 \hat{i} + 96 \hat{j} + 50 \hat{k}$. Now,find the magnitude $|\vec{d_1} 	imes \vec{d_2}|$: $|\vec{d_1} 	imes \vec{d_2}| = \sqrt{72^2 + 96^2 + 50^2} = \sqrt{5184 + 9216 + 2500} = \sqrt{16900} = 130$. Finally,the area is $\frac{1}{2} 	imes 130 = 65$ sq. units.

The area (in sq. units) of the parallelogram whose diagonals are along the vectors $8 \hat{i}-6 \hat{j}$ and $3 \hat{i}+4 \hat{j}-12 \hat{k}$ is:

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