If three points A, B and C have position vect | vedclass

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Question

(A) Given points are $A(1, x, 3)$,$B(3, 4, 7)$,and $C(y, -2, -5)$.Since the points are collinear,the vectors $\overrightarrow{AB}$ and $\overrightarro

Vedclass · Accepted Answer

$(2, -3)$

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(A) Given points are $A(1, x, 3)$,$B(3, 4, 7)$,and $C(y, -2, -5)$.Since the points are collinear,the vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$ must be parallel,i.e.,$\overrightarrow{AB} = \lambda \overrightarrow{BC}$ for some scalar $\lambda$.$\overrightarrow{AB} = (3-1)\hat{i} + (4-x)\hat{j} + (7-3)\hat{k} = 2\hat{i} + (4-x)\hat{j} + 4\hat{k}$.$\overrightarrow{BC} = (y-3)\hat{i} + (-2-4)\hat{j} + (-5-7)\hat{k} = (y-3)\hat{i} - 6\hat{j} - 12\hat{k}$.Equating the components of $\overrightarrow{AB} = \lambda \overrightarrow{BC}$:$2 = \lambda(y-3)$ ... $(i)$$4-x = -6\lambda$ ... $(ii)$$4 = -12\lambda$ ... $(iii)$From $(iii)$,$\lambda = \frac{4}{-12} = -\frac{1}{3}$.Substituting $\lambda = -\frac{1}{3}$ into $(i)$: $2 = -\frac{1}{3}(y-3) \Rightarrow -6 = y-3 \Rightarrow y = -3$.Substituting $\lambda = -\frac{1}{3}$ into $(ii)$: $4-x = -6(-\frac{1}{3}) \Rightarrow 4-x = 2 \Rightarrow x = 2$.Thus,$(x, y) = (2, -3)$.

If three points $A, B$ and $C$ have position vectors $(1, x, 3)$,$(3, 4, 7)$ and $(y, -2, -5)$ respectively and if they are collinear,then $(x, y) = $

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