If a→=i^+j^+k^,b→=4i^+3j^+4k^ and c→=i^+αj^+βk^ are linearly dependent vectors and |c→|=3, then

Given that a→=i^+j^+k^,b→=4i^+3j^+4k^ and c→=i^+αj^+βk^ are linearly dependent, 1 1 14 3 41 α β=01−β=0β=1 Also given that |c→|=3⇒1+α2+β2=3 Substituting the value of β, we get α2=1 or α=±1

If a→=i^+j^+k^,b→=4i^+3j^+4k^ and c→=i^+αj^+βk^ are linearly dependent vectors and |c→|=3, then

Given that a→=i^+j^+k^,b→=4i^+3j^+4k^ and c→=i^+αj^+βk^ are linearly dependent, 1 1 14 3 41 α β=01−β=0β=1 Also given that |c→|=3⇒1+α2+β2=3 Substituting the value of β, we get α2=1 or α=±1

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

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$α = - 1, β = \pm 1$

$α = 1, β = - 1$

$α = \pm 1, β = 1$

$α = 1, β = \pm 1$

answer is D.

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Detailed Solution

$Given that \vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = 4 \hat{i} + 3 \hat{j} + 4 \hat{k} and \vec{c} = \hat{i} + α \hat{j} + β \hat{k} are linearly dependent,$