Let a→=2i^+j^−2k^ and b→=i^+j^. If c→ is a vector such that a→.c→=|c→|,|c→−a→|=22 and the angle between (a→×b→) and c→ is π6 , then the value of |(a→×b→)×c→| is

|(a→×b→)×c→|=|a→×b→||c→|sinπ6Now, a→×b→=(2i^+j^−2k^)×(i^+j^)=2i^−j^+2k^From , |c→−a→|=22we get⇒c2−2c+1=0⇒c=1=|c→|Thus, from (1),|(a→×b→)×c→|=3×1×1/2=3/2

Let a→=2i^+j^−2k^ and b→=i^+j^. If c→ is a vector such that a→.c→=|c→|,|c→−a→|=22 and the angle between (a→×b→) and c→ is π6 , then the value of |(a→×b→)×c→| is

|(a→×b→)×c→|=|a→×b→||c→|sinπ6Now, a→×b→=(2i^+j^−2k^)×(i^+j^)=2i^−j^+2k^From , |c→−a→|=22we get⇒c2−2c+1=0⇒c=1=|c→|Thus, from (1),|(a→×b→)×c→|=3×1×1/2=3/2

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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 $\vec{a} . \vec{c} = | \vec{c} |, | \vec{c} - \vec{a} | = 2 \sqrt{2}$ and the angle between $(\vec{a} \times \vec{b})$ and $\vec{c} is \frac{π}{6}$ , then the value of $| (\vec{a} \times \vec{b}) \times \vec{c} |$ is

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$\frac{2}{3}$

$\frac{3}{2}$

answer is C.

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Detailed Solution

$| (\vec{a} \times \vec{b}) \times \vec{c} | = | \vec{a} \times \vec{b} | | \vec{c} | \sin \frac{π}{6}$
Now, $\begin{array}{l} \vec{a} \times \vec{b} = (2 \hat{i} + \hat{j} - 2 \hat{k}) \times (\hat{i} + \hat{j}) \\ = 2 \hat{i} - \hat{j} + 2 \hat{k} \end{array}$
From , $| \vec{c} - \vec{a} | = 2 \sqrt{2}$ we get
$\begin{array}{l} \Rightarrow c^{2} - 2 c + 1 = 0 \\ \Rightarrow c = 1 = | \vec{c} | \end{array}$
Thus, from (1),
$| (\vec{a} \times \vec{b}) \times \vec{c} | = 3 \times 1 \times 1 / 2 = 3 / 2$