If overline{AB} parallel overline{CD} and the | vedclass

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(B) The position vectors are $\vec{A} = 2i + j$,$\vec{B} = i - 3j$,$\vec{C} = 3i + 2j$,and $\vec{D} = i + \lambda j$.Calculate the vectors $\overline{

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$-6$

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(B) The position vectors are $\vec{A} = 2i + j$,$\vec{B} = i - 3j$,$\vec{C} = 3i + 2j$,and $\vec{D} = i + \lambda j$.Calculate the vectors $\overline{AB}$ and $\overline{CD}$:$\overline{AB} = \vec{B} - \vec{A} = (i - 3j) - (2i + j) = -i - 4j$.$\overline{CD} = \vec{D} - \vec{C} = (i + \lambda j) - (3i + 2j) = -2i + (\lambda - 2)j$.Since $\overline{AB} \parallel \overline{CD}$,their components must be proportional:$\frac{-1}{-2} = \frac{-4}{\lambda - 2}$.Simplify the equation:$\frac{1}{2} = \frac{-4}{\lambda - 2}$.$\lambda - 2 = 2 	imes (-4)$.$\lambda - 2 = -8$.$\lambda = -6$.

If $\overline{AB} \parallel \overline{CD}$ and the position vectors of the four points $A, B, C,$ and $D$ are $2i + j, i - 3j, 3i + 2j,$ and $i + \lambda j$ respectively,then the value of $\lambda$ is .....

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