Three lines are given by overrightarrow{r} = | vedclass

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(A) The lines are given by $\overrightarrow{r} = \lambda \hat{i}$,$\overrightarrow{r} = \mu(\hat{i} + \hat{j})$,and $\overrightarrow{r} = v(\hat{i} +

Vedclass · Accepted Answer

$0.75$

Vedclass · Answer

(A) The lines are given by $\overrightarrow{r} = \lambda \hat{i}$,$\overrightarrow{r} = \mu(\hat{i} + \hat{j})$,and $\overrightarrow{r} = v(\hat{i} + \hat{j} + \hat{k})$.Substituting these into the plane equation $x + y + z = 1$:For line $A$: $\lambda + 0 + 0 = 1 \Rightarrow \lambda = 1$,so $A = (1, 0, 0)$.For line $B$: $\mu + \mu + 0 = 1 \Rightarrow 2\mu = 1 \Rightarrow \mu = 1/2$,so $B = (1/2, 1/2, 0)$.For line $C$: $v + v + v = 1 \Rightarrow 3v = 1 \Rightarrow v = 1/3$,so $C = (1/3, 1/3, 1/3)$.Vectors are $\overrightarrow{AB} = B - A = (-1/2, 1/2, 0)$ and $\overrightarrow{AC} = C - A = (-2/3, 1/3, 1/3)$.The cross product $\overrightarrow{AB} 	imes \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1/2 & 1/2 & 0 \ -2/3 & 1/3 & 1/3 \end{vmatrix} = \hat{i}(1/6) - \hat{j}(-1/6) + \hat{k}(-1/6 + 1/3) = (1/6, 1/6, 1/6)$.The area $\Delta = \frac{1}{2} |\overrightarrow{AB} 	imes \overrightarrow{AC}| = \frac{1}{2} \sqrt{(1/6)^2 + (1/6)^2 + (1/6)^2} = \frac{1}{2} \sqrt{3/36} = \frac{\sqrt{3}}{12}$.Then $(6 \Delta)^2 = (6 	imes \frac{\sqrt{3}}{12})^2 = (\frac{\sqrt{3}}{2})^2 = 3/4 = 0.75$.

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