If a-b-c-d- b-c-d-a and a- b- c are non-coplanar- then the sum of a-b-c-d-

Given, nbsp; a+b+c=αd, nbsp; b+c+d=βa Eliminating d ⇒a+b+c=αd ⇒a+b+c=α(βa b c) ⇒(1 αβ)a+(1+α)b+(1+α)c=0 Since nbsp;a, b, c are non coplanar ⇒(1 αβ)=0, 1+α ...

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Standard XII

Mathematics

Equation of a Plane Passing through Three Points

If a+b+c=αd, ...

Question

If $\to a + \to b + \to c = α \to d,$ $\to b + \to c + \to d = β \to a$ and $\to a, \to b, \to c$ are non-coplanar, then the sum of $\to a + \to b + \to c + \to d =$

\to 0

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(β - 1) \cdot \to d + (α - 1) \cdot \to a

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(α - 1) \cdot \to d - (β - 1) \cdot \to a

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(α - 1) \cdot \to d + (β - 1) \cdot \to a

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Solution

The correct option is A $\to 0$
Given,
$\to a + \to b + \to c = α \to d,$ $\to b + \to c + \to d = β \to a$
Eliminating $\to d$
$\Rightarrow$ $\to a + \to b + \to c = α \to d$
$\Rightarrow \to a + \to b + \to c = α (β \to a - \to b - \to c)$
$\Rightarrow (1 - α β) \to a + (1 + α) \to b + (1 + α) \to c = 0$
Since $\to a, \to b, \to c$ are non-coplanar
$\Rightarrow (1 - α β) = 0, 1 + α = 0$
$\Rightarrow α = - 1$
$∴ β = - 1$
$\to a + \to b + \to c = α \to d = - \to d$
$∴$ $\to a + \to b + \to c + \to d = 0$

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