Let overrightarrow{a}=hat{i}+2 hat{j}+3 hat{k | vedclass

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(A) Given $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}$ and $\vec{b}=\hat{i}-2 \hat{j}-3 \hat{k}$.$A_1$ is the area of the quadrilateral with diagonals $\vec{

Vedclass · Accepted Answer

$26$

Vedclass · Answer

(A) Given $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}$ and $\vec{b}=\hat{i}-2 \hat{j}-3 \hat{k}$.$A_1$ is the area of the quadrilateral with diagonals $\vec{a}$ and $\vec{b}$,given by $A_1 = \frac{1}{2} |\vec{a} 	imes \vec{b}|$.First,calculate the cross product $\vec{a} 	imes \vec{b}$:$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ 1 & -2 & -3 \end{vmatrix} = \hat{i}(-6 - (-6)) - \hat{j}(-3 - 3) + \hat{k}(-2 - 2) = 0\hat{i} + 6\hat{j} - 4\hat{k}$.The magnitude is $|\vec{a} 	imes \vec{b}| = \sqrt{0^2 + 6^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52}$.Thus,$A_1 = \frac{1}{2} \sqrt{52}$.$A_2$ is the area of the parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$,given by $A_2 = |\vec{a} 	imes \vec{b}| = \sqrt{52}$.Finally,$A_1 \cdot A_2 = (\frac{1}{2} \sqrt{52}) \cdot (\sqrt{52}) = \frac{52}{2} = 26$.

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