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Condition for Coplanarity of Four Points

If a , b ...

Question

If $a$ , $b$ and $c$ are non-co-planar vectors such that $x_{1} a + y_{1} b + z_{1} c = 3 x_{2} a + 4 y_{2} b + 5 z_{2} c$ then value of $\frac{x_{1}}{x_{2}} + \frac{y_{1}}{y_{2}} + \frac{z_{1}}{z_{2}}$ is:

10

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12

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22

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Solution

The correct option is B $12$
Given $x_{1} a + y_{1} b + z_{1} c = 3 x_{2} a + 4 y_{2} b + 5 z^{2} c$
$\Rightarrow (x_{1} - 3 x_{2}) a + (y_{1} - 4 y_{2}) b + (z_{1} - 5 z_{2}) c = 0$
Given that $a, b, c$ are non coplanar vectors
So we get $x_{1} - 3 x_{2} = 0$ , $y_{1} - 4 y_{2} = 0$ and $z_{1} - 5 z_{2} = 0$
$\Rightarrow \frac{x_{1}}{x_{2}} + \frac{y_{1}}{y_{2}} + \frac{z_{1}}{z_{2}} = 3 + 4 + 5 = 12$
Therefore the correct option is $B$

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