Let a→=2i^−j^+5k^ and b→=αi^+βj^+2k^. If ((a→×b→)×i^)⋅k^=232, then |b→×2j^| is equal to

a→=2i^−j^+5k^, b→=αi^+βj^+2k^((a→×b→)×i^)⋅k^=232, then |b→×2j^| is ((a→⋅i^)b→−(b→⋅i^)a→)⋅k^=232(a→⋅i^)(b→⋅k^)−(b→⋅i^)(a→⋅k^)=2322×2−α×5=232⇒5α=4−232⇒α=−32b→×2j^=i^j^k^αβ2020=−4i^+2αk^∴|b→×2j^|=16+4α2=16+4×94=5

Let a→=2i^−j^+5k^ and b→=αi^+βj^+2k^. If ((a→×b→)×i^)⋅k^=232, then |b→×2j^| is equal to

a→=2i^−j^+5k^, b→=αi^+βj^+2k^((a→×b→)×i^)⋅k^=232, then |b→×2j^| is ((a→⋅i^)b→−(b→⋅i^)a→)⋅k^=232(a→⋅i^)(b→⋅k^)−(b→⋅i^)(a→⋅k^)=2322×2−α×5=232⇒5α=4−232⇒α=−32b→×2j^=i^j^k^αβ2020=−4i^+2αk^∴|b→×2j^|=16+4α2=16+4×94=5

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Let $\vec{a} = 2 \hat{i} - \hat{j} + 5 \hat{k} and \vec{b} = α \hat{i} + β \hat{j} + 2 \hat{k}$ . If $((\vec{a} \times \vec{b}) \times \hat{i}) \cdot \hat{k} = \frac{23}{2}, then | \vec{b} \times 2 \hat{j} |$ is equal to

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$\sqrt{21}$

$\sqrt{17}$

answer is B.

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Detailed Solution

$\begin{array}{l} \vec{a} = 2 \hat{i} - \hat{j} + 5 \hat{k}, \vec{b} = α \hat{i} + β \hat{j} + 2 \hat{k} \\ ((\vec{a} \times \vec{b}) \times \hat{i}) \cdot \hat{k} = \frac{23}{2}, then | \vec{b} \times 2 \hat{j} | is \\ ((\vec{a} \cdot \hat{i}) \vec{b} - (\vec{b} \cdot \hat{i}) \vec{a}) \cdot \hat{k} = \frac{23}{2} \\ (\vec{a} \cdot \hat{i}) (\vec{b} \cdot \hat{k}) - (\vec{b} \cdot \hat{i}) (\vec{a} \cdot \hat{k}) = \frac{23}{2} \\ 2 \times 2 - α \times 5 = \frac{23}{2} \Rightarrow 5 α = 4 - \frac{23}{2} \Rightarrow α = \frac{- 3}{2} \\ \vec{b} \times 2 \hat{j} = |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ α & β & 2 \\ 0 & 2 & 0 \end{array}| = - 4 \hat{i} + 2 α \hat{k} \\ ∴ | \vec{b} \times 2 \hat{j} | = \sqrt{16 + 4 α^{2}} = \sqrt{16 + 4 \times \frac{9}{4}} = 5 \end{array}$