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Orthogonal System of Vectors

In a four dim...

Question

In a four dimensional space, where unit vectors along the axes are $^i,^j,^k$ and $^l,$ $_{1},_{2},_{3},_{4}$ are four non-zero vectors such that no vector can be expressed as a linear combination of others and $(λ - 1) ({\to a}_{1} - {\to a}_{2}) + α ({\to a}_{2} + {\to a}_{3}) + γ ({\to a}_{3} + {\to a}_{4} - 2 {\to a}_{2}) + {\to a}_{3} + δ {\to a}_{4} = \to 0 .$
Then which of the following is/are correct?

A

λ = 1

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B

α = 2

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C

γ = 1

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D

δ = \frac{1}{3}

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Solution

The correct option is D $δ = \frac{1}{3}$
Here, $(λ - 1) ({\to a}_{1} - {\to a}_{2}) + α ({\to a}_{2} + {\to a}_{3}) + γ ({\to a}_{3} + {\to a}_{4} - 2 {\to a}_{2}) + {\to a}_{3} + δ {\to a}_{4} = \to 0$

$\Rightarrow (λ - 1) {\to a}_{1} + (1 - λ + α - 2 γ) {\to a}_{2} + (α + γ + 1) {\to a}_{3} + (γ + δ) {\to a}_{4} = \to 0$
Since $a_{1}, a_{2}, a_{3}, a_{4}$ are linearly independent, therefore
$λ - 1 = 0,$
$1 - λ + α - 2 γ = 0,$
$α + γ + 1 = 0,$
and $γ + δ = 0$

On solving above equations, we get
$λ = 1$
$α = \frac{- 2}{3}$
$γ = \frac{- 1}{3}$
$δ = \frac{1}{3}$

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