Let a→, b→ and c→ be three non-zero vectors such that b→. c→=0 and a→×(b→×c→)=b→−c→2 . If d→ be a vector such that b→.d→=a→.b→, then (a→×b→).(c→×d→) is equal to

a→×(b→×c→)=b→−c→2(a¯. c¯)b¯−(a¯ .b¯)c¯=b¯−c¯2 ⇒a¯.c¯=12; a¯.b¯=12 (a¯×b¯).(c¯×d¯)=12a¯.d¯0b¯.d¯=b¯.d¯2=a¯.b¯2=14

Let a→, b→ and c→ be three non-zero vectors such that b→. c→=0 and a→×(b→×c→)=b→−c→2 . If d→ be a vector such that b→.d→=a→.b→, then (a→×b→).(c→×d→) is equal to

a→×(b→×c→)=b→−c→2(a¯. c¯)b¯−(a¯ .b¯)c¯=b¯−c¯2 ⇒a¯.c¯=12; a¯.b¯=12 (a¯×b¯).(c¯×d¯)=12a¯.d¯0b¯.d¯=b¯.d¯2=a¯.b¯2=14

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Let $\vec{a}, \vec{b} and \vec{c}$ be three non-zero vectors such that $\vec{b} . \vec{c} = 0 and \vec{a} \times (\vec{b} \times \vec{c}) = \frac{\vec{b} - \vec{c}}{2}$ . If $\vec{d}$ be a vector such that $\vec{b} . \vec{d} = \vec{a} . \vec{b}, then (\vec{a} \times \vec{b}) . (\vec{c} \times \vec{d})$ is equal to

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$\frac{3}{4}$

$\frac{1}{2}$

$- \frac{1}{4}$

$\frac{1}{4}$

answer is D.

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Detailed Solution

$\vec{a} \times (\vec{b} \times \vec{c}) = \frac{\vec{b} - \vec{c}}{2}$
$(\bar{a} . \bar{c}) \bar{b} - (\bar{a} . \bar{b}) \bar{c} = \frac{\bar{b} - \bar{c}}{2} \Rightarrow \bar{a} . \bar{c} = \frac{1}{2}; \bar{a} . \bar{b} = \frac{1}{2} (\bar{a} \times \bar{b}) . (\bar{c} \times \bar{d}) = |\begin{matrix} \frac{1}{2} & \bar{a} . \bar{d} \\ 0 & \bar{b} . \bar{d} \end{matrix}| = \frac{\bar{b} . \bar{d}}{2} = \frac{\bar{a} . \bar{b}}{2} = \frac{1}{4}$