If the position vectors of A, B, C, and D are | vedclass

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

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(B) Given position vectors are $\vec{A} = 2i + j,$ $\vec{B} = i - 3j,$ $\vec{C} = 3i + 2j,$ and $\vec{D} = i + \lambda j.$First,calculate the vectors $\overrightarrow{AB}$ and $\overrightarrow{CD}$:$\overrightarrow{AB} = \vec{B} - \vec{A} = (i - 3j) - (2i + j) = -i - 4j.$$\overrightarrow{CD} = \vec{D} - \vec{C} = (i + \lambda j) - (3i + 2j) = -2i + (\lambda - 2)j.$Since $\overrightarrow{AB} \parallel \overrightarrow{CD},$ their components must be proportional:$\frac{-1}{-2} = \frac{-4}{\lambda - 2}.$Solving for $\lambda$:$\frac{1}{2} = \frac{-4}{\lambda - 2}$$\lambda - 2 = 2 	imes (-4)$$\lambda - 2 = -8$$\lambda = -6.$

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

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