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Equal Vectors

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Question

If $\to x$ and $\to y$ are two non-collinear vectors such that $(9 a - 3 b) \to x + (8 b - 6 c) \to y + (8 c - 16 a) (\to x \times \to y) = \to 0,$ then which of the following is/are correct ?

A

a = 3 b

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B

4 b = 3 c

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C

c = 2 a

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D

b = 3 a

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Solution

The correct option is D $b = 3 a$
Given, $(9 a - 3 b) \to x + (8 b - 6 c) \to y + (8 c - 16 a) (\to x \times \to y) = \to 0,$
As, $\to x, \to y, \to x \times \to y$ are non-coplanar, i.e. linearly independent
So, all scalars should be zero
$\Rightarrow 9 a - 3 b = 8 b - 6 c = 8 c - 16 a = 0 \Rightarrow b = 3 a, 4 b = 3 c, c = 2 a$

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