If P=3 hat{i}+5 hat{j}-hat{k} and Q=hat{i}+2 | vedclass

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(C) The area of a triangle with sides represented by vectors $P$ and $Q$ is given by the formula: $	ext{Area} = \frac{1}{2} |P 	imes Q|$.First,we ca

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$\frac{\sqrt{390}}{2}$

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(C) The area of a triangle with sides represented by vectors $P$ and $Q$ is given by the formula: $	ext{Area} = \frac{1}{2} |P 	imes Q|$.First,we calculate the cross product $P 	imes Q$:$P 	imes Q = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 5 & -1 \ 1 & 2 & 3 \end{vmatrix}$$= \hat{i}(5 	imes 3 - (-1) 	imes 2) - \hat{j}(3 	imes 3 - (-1) 	imes 1) + \hat{k}(3 	imes 2 - 5 	imes 1)$$= \hat{i}(15 + 2) - \hat{j}(9 + 1) + \hat{k}(6 - 5)$$= 17 \hat{i} - 10 \hat{j} + \hat{k}$.Now,find the magnitude of the cross product:$|P 	imes Q| = \sqrt{17^2 + (-10)^2 + 1^2} = \sqrt{289 + 100 + 1} = \sqrt{390}$.Finally,the area of the triangle is:$	ext{Area} = \frac{1}{2} |P 	imes Q| = \frac{\sqrt{390}}{2} 	ext{ sq units}$.Thus,option $(C)$ is correct.

If $P=3 \hat{i}+5 \hat{j}-\hat{k}$ and $Q=\hat{i}+2 \hat{j}+3 \hat{k}$ are two sides of a triangle,then its area is equal to . . . . . . sq units.

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