Let a→=2i^+j^−2k^ and b→=i^+j^. If c→ is a vector such that a→⋅c→=|c→|,|c→−a→|=22and the angle between (a→×b→) and c→ is 30∘, then |(a→×b→)×c→|=

|c→−a→|=22⇒|c→−a→|2=|c→|2−2c→⋅a→+|a→|2=8⇒|c→|2−2|c→|+9=8⇒(|c→|−1)2=0∴|c→|=1.∣j^k^21−2110=2i^−2j^+k^∴ |a→×b→|=4+4+1=3Now |(a→×b→)×c→|=|a→×b→||c→|sin⁡30∘=3⋅1⋅12=32

Let a→=2i^+j^−2k^ and b→=i^+j^. If c→ is a vector such that a→⋅c→=|c→|,|c→−a→|=22and the angle between (a→×b→) and c→ is 30∘, then |(a→×b→)×c→|=

|c→−a→|=22⇒|c→−a→|2=|c→|2−2c→⋅a→+|a→|2=8⇒|c→|2−2|c→|+9=8⇒(|c→|−1)2=0∴|c→|=1.∣j^k^21−2110=2i^−2j^+k^∴ |a→×b→|=4+4+1=3Now |(a→×b→)×c→|=|a→×b→||c→|sin⁡30∘=3⋅1⋅12=32

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Let $\vec{a} = 2 \hat{i} + \hat{j} - 2 \hat{k} and \vec{b} = \hat{i} + \hat{j} . If \vec{c}$ is a vector such that $\vec{a} \cdot \vec{c} = | \vec{c} |, | \vec{c} - \vec{a} | = 2 \sqrt{2}$
and the angle between $(\vec{a} \times \vec{b}) and \vec{c} is 30^{\circ}, then | (\vec{a} \times \vec{b}) \times \vec{c} | =$

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$\frac{2}{3}$

$\frac{3}{2}$

$2$

$3$

answer is B.

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Detailed Solution

$| \vec{c} - \vec{a} | = 2 \sqrt{2}$

$\begin{array}{l} \Rightarrow | \vec{c} - \vec{a} |^{2} = | \vec{c} |^{2} - 2 \vec{c} \cdot \vec{a} + | \vec{a} |^{2} = 8 \\ \Rightarrow | \vec{c} |^{2} - 2 | \vec{c} | + 9 = 8 \Rightarrow (| \vec{c} | - 1)^{2} = 0 \\ ∴ | \vec{c} | = 1 . |\begin{array}{ccc} ∣ & \hat{j} & \hat{k} \\ 2 & 1 & - 2 \\ 1 & 1 & 0 \end{array}| = 2 \hat{i} - 2 \hat{j} + \hat{k} \\ ∴ | \vec{a} \times \vec{b} | = \sqrt{4 + 4 + 1} = 3 \end{array}$

Now $| (\vec{a} \times \vec{b}) \times \vec{c} | = | \vec{a} \times \vec{b} | | \vec{c} | \sin 30^{\circ} = 3 \cdot 1 \cdot \frac{1}{2} = \frac{3}{2}$