Let a=2i¯+j¯−2k¯ and b=i¯+j¯ be two vectors ‘c’ is a vector such that a.c=|c| and |c−a|=22. If the angle between a×b and c is 30°, then |(a×b)×c| is equal to

a¯×b¯=2i¯−2j¯+k¯ |a¯×b¯|=4+4+1=3|c−a|=22⇒|c−a|2=8|c|2−2c⋅a+|a|2=8⇒|c|2−2|c|+9=8⇒(|c|−1)2=0⇒|c|=1|(a¯×b¯)×c¯|=|a¯×b¯||c|sin⁡(a×b,c)=3×1×sin⁡30∘=32

Let a=2i¯+j¯−2k¯ and b=i¯+j¯ be two vectors ‘c’ is a vector such that a.c=|c| and |c−a|=22. If the angle between a×b and c is 30°, then |(a×b)×c| is equal to

a¯×b¯=2i¯−2j¯+k¯ |a¯×b¯|=4+4+1=3|c−a|=22⇒|c−a|2=8|c|2−2c⋅a+|a|2=8⇒|c|2−2|c|+9=8⇒(|c|−1)2=0⇒|c|=1|(a¯×b¯)×c¯|=|a¯×b¯||c|sin⁡(a×b,c)=3×1×sin⁡30∘=32

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Let $a = 2 \bar{i} + \bar{j} - 2 \bar{k}$ and $b = \bar{i} + \bar{j}$ be two vectors ‘c’ is a vector such that $a . c = | c |$ and $| c - a | = 2 \sqrt{2}$ . If the angle between $a \times b$ and c is 30°, then $| (a \times b) \times c |$ is equal to

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$\frac{2}{3}$

$\frac{3}{2}$

$\frac{\sqrt{3}}{2}$

answer is A.

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Detailed Solution

$\bar{a} \times \bar{b} = 2 \bar{i} - 2 \bar{j} + \bar{k} \begin{array}{l} | \bar{a} \times \bar{b} | = \sqrt{4 + 4 + 1} = 3 \\ | c - a | = 2 \sqrt{2} \\ \Rightarrow | c - a |^{2} = 8 \\ | c |^{2} - 2 c \cdot a + | a |^{2} = 8 \\ \Rightarrow | c |^{2} - 2 | c | + 9 = 8 \\ \Rightarrow (| c | - 1)^{2} = 0 \Rightarrow | c | = 1 \\ | (\bar{a} \times \bar{b}) \times \bar{c} | = | \bar{a} \times \bar{b} | | c | \sin (a \times b, c) \\ = 3 \times 1 \times \sin 30^{\circ} = \frac{3}{2} \end{array}$