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

(2, - 3)

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(- 2, 3)

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(- 2, - 3)

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(2, 3)

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Solution

The correct option is A $(2, - 3)$
$(1, x, 3) = λ (3, 4, 7) + μ (y, - 2, - 5)$
$1 = 3 λ + μ y$
$x = 4 λ + (- 2 μ)$
$3 = 7 λ - 5 μ$
$2 - x = (- 3 - y) μ$
So only $x = 2$ , $y = - 3$

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