The vectors from the origin to the points A a | vedclass

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(A) Given $\overrightarrow{OA} = \overrightarrow{a} = 3\hat{i} - 6\hat{j} + 2\hat{k}$ and $\overrightarrow{OB} = \overrightarrow{b} = 2\hat{i} + \hat{

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$\frac{5}{2}\sqrt{17}$ sq. unit

Vedclass · Answer

(A) Given $\overrightarrow{OA} = \overrightarrow{a} = 3\hat{i} - 6\hat{j} + 2\hat{k}$ and $\overrightarrow{OB} = \overrightarrow{b} = 2\hat{i} + \hat{j} - 2\hat{k}$.The area of triangle $OAB$ is given by $	ext{Area} = \frac{1}{2} |\overrightarrow{a} 	imes \overrightarrow{b}|$.First,calculate the cross product $\overrightarrow{a} 	imes \overrightarrow{b}$:$\overrightarrow{a} 	imes \overrightarrow{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -6 & 2 \ 2 & 1 & -2 \end{vmatrix}$$= \hat{i}((-6)(-2) - (2)(1)) - \hat{j}((3)(-2) - (2)(2)) + \hat{k}((3)(1) - (-6)(2))$$= \hat{i}(12 - 2) - \hat{j}(-6 - 4) + \hat{k}(3 + 12)$$= 10\hat{i} + 10\hat{j} + 15\hat{k}$.Now,calculate the magnitude $|\overrightarrow{a} 	imes \overrightarrow{b}|$:$|\overrightarrow{a} 	imes \overrightarrow{b}| = \sqrt{10^2 + 10^2 + 15^2} = \sqrt{100 + 100 + 225} = \sqrt{425} = \sqrt{25 	imes 17} = 5\sqrt{17}$.Therefore,the area of $\Delta OAB = \frac{1}{2} 	imes 5\sqrt{17} = \frac{5}{2}\sqrt{17}$ sq. unit.

The vectors from the origin to the points $A$ and $B$ are $\overrightarrow A = 3\hat i - 6\hat j + 2\hat k$ and $\overrightarrow B = 2\hat i + \hat j - 2\hat k$ respectively. The area of the triangle $OAB$ is:

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