If A→=i^+j^+k^,B→=4i^+3j^+4k^ and C→=i^+αj^+βk^ are linearly dependent vector and |C→|=3, then

Vector A→=i^+j^+k^,B→=4i^+3j^+4k^, And C→=i^+αj^+βk^ are linearly independent, if 1114341αβ=0 Operate C3→C3−C1;C2→C2−C1 we get, 1004−101α−1β−1=0⇒−(β−1)=0⇒β=1 Again, |C→|=1+α2+β2=3⇒1+α2+β2=3⇒1+α2+1=3∴α2=1⇒α=±1Hence α=±1,β=1

If A→=i^+j^+k^,B→=4i^+3j^+4k^ and C→=i^+αj^+βk^ are linearly dependent vector and |C→|=3, then

Vector A→=i^+j^+k^,B→=4i^+3j^+4k^, And C→=i^+αj^+βk^ are linearly independent, if 1114341αβ=0 Operate C3→C3−C1;C2→C2−C1 we get, 1004−101α−1β−1=0⇒−(β−1)=0⇒β=1 Again, |C→|=1+α2+β2=3⇒1+α2+β2=3⇒1+α2+1=3∴α2=1⇒α=±1Hence α=±1,β=1

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$\begin{array}{l} If \vec{A} = \hat{i} + \hat{j} + \hat{k}, \vec{B} = 4 \hat{i} + 3 \hat{j} + 4 \hat{k} and \vec{C} = \hat{i} + α \hat{j} + β \hat{k} \\ are linearly dependent vector and | \vec{C} | = \sqrt{3}, then \end{array}$

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$α = 1, β = - 1$

$α = 1, β = \pm 1$

$α = - 1, β = \pm 1$

$α = \pm 1, β = 1$

answer is D.

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Detailed Solution

$\begin{array}{l} Vector \vec{A} = \hat{i} + \hat{j} + \hat{k}, \vec{B} = 4 \hat{i} + 3 \hat{j} + 4 \hat{k}, And \vec{C} = \hat{i} + α \hat{j} + β \hat{k} are linearly independent, if \\ |\begin{array}{l} 1 & 1 & 1 \\ 4 & 3 & 4 \\ 1 & α & β \end{array}| = 0 \\ {Operate C}_{3} \to C_{3} - C_{1}; C_{2} \to C_{2} - C_{1} \\ we get, |\begin{matrix} 1 & 0 & 0 \\ 4 & - 1 & 0 \\ 1 & α - 1 & β - 1 \end{matrix}| = 0 \\ \Rightarrow - (β - 1) = 0 \Rightarrow β = 1 \\ Again, | \vec{C} | = \sqrt{1 + α^{2} + β^{2}} = \sqrt{3} \\ \Rightarrow 1 + α^{2} + β^{2} = 3 \Rightarrow 1 + α^{2} + 1 = 3 \\ ∴ α^{2} = 1 \Rightarrow α = \pm 1 \\ H e n c e α = \pm 1, β = 1 \end{array}$