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Linear Dependence and Independence of Vectors

If a⃗=î+ĵ+k...

Question

If $\to a =^i +^j +^k, \to b = 4^i + 3^j + 4^k$ and $\to c =^i + α^j + β^k$ are linearly dependent vectors and $| \to c | = \sqrt{3}$ , then

A

α = 1, β = - 1

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B

α = 1, β = \pm 1

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C

α = - 1, β = \pm 1

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D

α = \pm 1, β = 1

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Solution

The correct option is A $α = 1, β = - 1$
According to question...............
$\begin{matrix} \to a =^i +^j +^k, \to b = 4^i + 3^j + 4^k \to c =^i + α^j + β^k N o w, a, b, c i s l i n e a r l y . . . . . \Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} 111434 1 α β \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow 1 (3 β - 4 α) - 1 (4 β - 4) + 1 (4 α - 3) = 0 \Rightarrow 3 β - 4 α - 4 β + 4 + 4 α - 3 = 0 \Rightarrow - β + 1 = 0 ∴ β = 1 N o w, v a l u e o f v e c t o r : \to c =^i + α^j + β^k | \to c | = \sqrt{3} \Rightarrow | \to c | = \sqrt{1 + α^{2} + β^{2}} \Rightarrow \sqrt{1 + α^{2} + β^{2}} = \sqrt{3} \Rightarrow 1 + α^{2} + β^{2} = 3 \Rightarrow 1 + α^{2} + β^{2} = 3 \Rightarrow α^{2} + 1 + 1 = 3 \Rightarrow α^{2} = 1 ∴ α = \pm 1 S o, t h a t t h e v a l u e o f α = \pm 1, β = 1 a n d t h e c o r r e c t o p t i o n i s D . \end{matrix}$

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Similar questions

\to a =^i +^j +^k,^b = 4^i + 3^j + 4^k

\to c =^i + α^j + β^k

are linearly dependent vectors and

| \to c | = \sqrt{3},

A

\to a =^i +^i +^k, \to b = 4^i + 3^i + 4^k

\to c =^i + α^j + β^k

are linearly dependent and

| \to c | = \sqrt{3}

α = \pm 1, β = 1

R

: For coplanar vectors every vector can be expressed as linear combination of other.

\to a =^i +^j +^k, \to b = 4^i + 3^j + 4^k

\to c =^i + α^j + β^k

are linearly dependent vectors and

| \to c | = \sqrt{3}

Q. Consider three vectors

\to A =^i +^j - 2^k, \to B =^i -^j +^k

\to C = 2^i - 3^j + 4^k

\to X

α \to A + β \to B

α

β

are numbers) is perpendicular to

\to C

α

β

\to a =^i +^j +^k

\to b = 4^i + 3^j + 4^k

\to c =^i + α^j + β^k

are linearly dependent vectors and

| \to c | = \sqrt{3}

then the value of

α

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