If a→ and c→ are unit vector and |b→|=4 and the angle between a→ and c→ is cos−1⁡(1/4) and b→−2c→=λa→ the value of λ is

b→−2c→=λa→ or b→=2c→+λa→ or |b→|2=|2c→+λa→|2⇒ 16=4|c→|2+λ2|a→|2+4λa→⋅c→ =4+λ2+4λ14 or λ2+λ−12=0 or λ=3,−4

If a→ and c→ are unit vector and |b→|=4 and the angle between a→ and c→ is cos−1⁡(1/4) and b→−2c→=λa→ the value of λ is

b→−2c→=λa→ or b→=2c→+λa→ or |b→|2=|2c→+λa→|2⇒ 16=4|c→|2+λ2|a→|2+4λa→⋅c→ =4+λ2+4λ14 or λ2+λ−12=0 or λ=3,−4

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If $\vec{a} and \vec{c}$ are unit vector and $| \vec{b} | = 4$ and the angle between $\vec{a} and \vec{c} is \cos^{- 1} (1 / 4) and \vec{b} - 2 \vec{c} = λ \vec{a}$ the value of $λ$ is

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3,- 4

$\frac{1}{4}, \frac{3}{4}$

-3, 4

$\frac{- 1}{4}, \frac{3}{4}$

answer is A.

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Detailed Solution

$\begin{array}{l} \vec{b} - 2 \vec{c} = λ \vec{a} \\ or \vec{b} = 2 \vec{c} + λ \vec{a} \\ or | \vec{b} |^{2} = | 2 \vec{c} + λ \vec{a} |^{2} \end{array}$

$\begin{array}{l} \Rightarrow 16 = 4 | \vec{c} |^{2} + λ^{2} | \vec{a} |^{2} + 4 λ \vec{a} \cdot \vec{c} \\ = 4 + λ^{2} + 4 λ \frac{1}{4} \\ or λ^{2} + λ - 12 = 0 \\ or λ = 3, - 4 \end{array}$