The position vectors of three points A, B and | vedclass

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(C) Given points are $A(1, 3, x)$,$B(3, 5, 8)$,and $C(y, -1, -6)$.The vectors $\vec{AB}$ and $\vec{AC}$ are given by:$\vec{AB} = (3-1)\hat{i} + (5-3)\

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$\left(\frac{10}{3}, -3\right)$

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(C) Given points are $A(1, 3, x)$,$B(3, 5, 8)$,and $C(y, -1, -6)$.The vectors $\vec{AB}$ and $\vec{AC}$ are given by:$\vec{AB} = (3-1)\hat{i} + (5-3)\hat{j} + (8-x)\hat{k} = 2\hat{i} + 2\hat{j} + (8-x)\hat{k}$$\vec{AC} = (y-1)\hat{i} + (-1-3)\hat{j} + (-6-x)\hat{k} = (y-1)\hat{i} - 4\hat{j} - (6+x)\hat{k}$Since $A, B, C$ are collinear,$\vec{AB} = \lambda \vec{AC}$ for some scalar $\lambda$.Comparing the components:$\frac{2}{y-1} = \frac{2}{-4} = \frac{8-x}{-(6+x)}$From $\frac{2}{y-1} = \frac{2}{-4}$,we get $y-1 = -4$,so $y = -3$.From $\frac{2}{-4} = \frac{8-x}{-(6+x)}$,we have $-\frac{1}{2} = \frac{8-x}{-6-x}$.$6+x = 2(8-x) \Rightarrow 6+x = 16-2x \Rightarrow 3x = 10 \Rightarrow x = \frac{10}{3}$.Thus,$(x, y) = \left(\frac{10}{3}, -3\right)$.

The position vectors of three points $A, B$ and $C$ are $(1, 3, x), (3, 5, 8)$ and $(y, -1, -6)$ respectively. If $A, B$ and $C$ are collinear,then $(x, y) =$

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