Let a¯=2i^+j^−2k^ and b¯=i^+j^. Let c→ be a vector such that |c→−a→|=3 , |(a→×b→)×c→|=3 and the angle between c→ and a→×b→ be 300 then a→ · c→ is equal to

|(a¯×b¯)×c¯|= 3⇒ |a¯×b¯| |c¯| sinθ=3 ⇒ |c¯| =33×12=2 |c¯−a¯|=3 ⇒ |c¯|2+|a¯|2−2|c¯||a¯| cos (c¯,a¯)=9 ⇒ a→. c→=2

Let a¯=2i^+j^−2k^ and b¯=i^+j^. Let c→ be a vector such that |c→−a→|=3 , |(a→×b→)×c→|=3 and the angle between c→ and a→×b→ be 300 then a→ · c→ is equal to

|(a¯×b¯)×c¯|= 3⇒ |a¯×b¯| |c¯| sinθ=3 ⇒ |c¯| =33×12=2 |c¯−a¯|=3 ⇒ |c¯|2+|a¯|2−2|c¯||a¯| cos (c¯,a¯)=9 ⇒ a→. c→=2

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Let $\bar{a} = 2 \hat{i} + \hat{j} - 2 \hat{k}$ and $\bar{b} = \hat{i} + \hat{j}$ . Let $\vec{c}$ be a vector such that $| \vec{c} - \vec{a} | = 3$ , $| (\vec{a} \times \vec{b}) \times \vec{c} | = 3$ and the angle between $\vec{c}$ and $\vec{a} \times \vec{b}$ be $30^{0}$ then $\vec{a} \cdot \vec{c}$ is equal to

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$\frac{1}{8}$

$\frac{25}{8}$

answer is C.

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Detailed Solution

$| (\bar{a} \times \bar{b}) \times \bar{c} | = 3 \Rightarrow | \bar{a} \times \bar{b} | | \bar{c} | \sin θ = 3 \Rightarrow | \bar{c} | = \frac{3}{3 \times \frac{1}{2}} = 2 | \bar{c} - \bar{a} | = 3 \Rightarrow {| \bar{c} |}^{2} + {| \bar{a} |}^{2} - 2 | \bar{c} | | \bar{a} | \cos (\bar{c}, \bar{a}) = 9 \Rightarrow \vec{a} . \vec{c} = 2$