If the area of the triangle with vertices hat | vedclass

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(A) Let the vertices be $A = (1, y, 0)$,$B = (1, 0, 2)$,and $C = (0, 3, 1)$.$\overrightarrow{AB} = (1-1)\hat{i} + (0-y)\hat{j} + (2-0)\hat{k} = -y\hat

Vedclass · Accepted Answer

$2, 4$

Vedclass · Answer

(A) Let the vertices be $A = (1, y, 0)$,$B = (1, 0, 2)$,and $C = (0, 3, 1)$.$\overrightarrow{AB} = (1-1)\hat{i} + (0-y)\hat{j} + (2-0)\hat{k} = -y\hat{j} + 2\hat{k}$.$\overrightarrow{AC} = (0-1)\hat{i} + (3-y)\hat{j} + (1-0)\hat{k} = -\hat{i} + (3-y)\hat{j} + \hat{k}$.The area of the triangle is given by $\frac{1}{2} |\overrightarrow{AB} 	imes \overrightarrow{AC}| = \sqrt{6}$.$\overrightarrow{AB} 	imes \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & -y & 2 \ -1 & 3-y & 1 \end{vmatrix} = \hat{i}(-y - 2(3-y)) - \hat{j}(0 - (-2)) + \hat{k}(0 - y) = \hat{i}(y-6) - 2\hat{j} - y\hat{k}$.$|\overrightarrow{AB} 	imes \overrightarrow{AC}| = \sqrt{(y-6)^2 + (-2)^2 + (-y)^2} = \sqrt{y^2 - 12y + 36 + 4 + y^2} = \sqrt{2y^2 - 12y + 40}$.Since $\frac{1}{2} \sqrt{2y^2 - 12y + 40} = \sqrt{6}$,we have $\sqrt{2y^2 - 12y + 40} = 2\sqrt{6} = \sqrt{24}$.Squaring both sides: $2y^2 - 12y + 40 = 24 \Rightarrow 2y^2 - 12y + 16 = 0 \Rightarrow y^2 - 6y + 8 = 0$.Factoring: $(y-2)(y-4) = 0$,so $y = 2, 4$.

If the area of the triangle with vertices $\hat{i}+y \hat{j}$,$\hat{i}+2 \hat{k}$,and $3 \hat{j}+\hat{k}$ is $\sqrt{6}$ sq. units,then the values of $y$ are

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