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?

λ = 1

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α = 2

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γ = 1

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δ = \frac{1}{3}

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Solution

The correct options are
A $λ = 1$
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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Similar questions

Q. 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?

Q. In a four-dimensional space where unit vectors along the axes are

ˆ i, ˆ j, ˆ k

and

ˆ l

, and

_{1},_{2},_{3},_{4}

are four non-zero vectors such that no vector can be expressed as linear combination of others and

(λ - 1) (_{1} -_{2}) + μ (_{2} +_{3}) + γ (_{3} +_{4} - 2_{2}) +_{3} + δ_{4} = \to 0

, then

If $a_{1}, a_{2}, a_{3}$ and $a_{4}$ are the coefficients of four consecutive terms in the expansion of $(1 + x)^{n},$ prove that $\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{a_{3} + a_{4}} = \frac{2 a_{2}}{a_{2} + a_{3}} .$

If $a_{1}, a_{2}, a_{3}, a_{4}$ are the coefficients of any four

consecutive terms in the expansion of ( $1 + x)^{n}$ , then

$\frac{a_{1}}{a_{1} + a_{2}}$ + $\frac{a_{3}}{a_{3} + a_{4}}$ =

If $a_{1}, a_{2}, a_{3}, a_{4}$ be the coefficeints of four consecutive terms in the expansion of $(1 + x)^{n}$ then prove that
$\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{(a_{3} + a_{4})} = \frac{2 a_{2}}{(a_{2} + a_{3})} .$

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