Let a , b and c be three non-zero vectors such that b · c = 0 and a × ( b × c ) = b - c 2 . If d is a vector such that b · d = a · b , then ( a × b ) · ( c × d ) is equal to [2023]

Question

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero vectors such that $\vec{b} \cdot \vec{c} = 0$ and $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{\vec{b} - \vec{c}}{2} .$ If $\vec{d}$ is a vector such that $\vec{b} \cdot \vec{d} = \vec{a} \cdot \vec{b}$ , then $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d})$ is equal to [2023]

Smart Achievers · Accepted Answer

(4)

$\vec{a} \times (\vec{b} \times \vec{c}) = \frac{\vec{b} - \vec{c}}{2}$

$(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = \frac{\vec{b} - \vec{c}}{2} \Rightarrow \vec{a} \cdot \vec{c} = \frac{1}{2}$

$\vec{a} \cdot \vec{b} = \frac{1}{2};$ $\vec{b} \cdot \vec{d} = \vec{a} \cdot \vec{b} = \frac{1}{2}$

$(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = (\vec{c} \times \vec{d}) \cdot (\vec{a} \times \vec{b})$

$= [\vec{c} \times \vec{d} \vec{a} \vec{b}] = \vec{b} \cdot (\vec{c} \times \vec{d}) \times \vec{a} = - \vec{b} \cdot (\vec{a} \times (\vec{c} \times \vec{d}))$

$= - \vec{b} \cdot [(\vec{a} \cdot \vec{d}) \vec{c} - (\vec{a} \cdot \vec{c}) \vec{d}]$ $= - (\vec{a} \cdot \vec{d}) \cdot (\vec{b} \cdot \vec{c}) + (\vec{a} \cdot \vec{c}) (\vec{b} \cdot \vec{d})$

$= 0 + (\vec{a} \cdot \vec{c}) (\vec{b} \cdot \vec{d})$ $∵$ $(\vec{b} \cdot \vec{c}) = 0$

$= \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

Solution

1 $\frac{3}{4}$

2 $\frac{1}{2}$

3 $- \frac{1}{4}$

4 $\frac{1}{4}$

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Solution

Q. Let a→,b→ and c→ be three non-zero vectors such that b→·c→=0 and a→×(b→×c→)=b→-c→2. If d→ is a vector such that b→·d→=a→·b→, then (a→×b→)·(c→×d→) is equal to [2023]

1 34

2 12

3 -14

4 14

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1 $\frac{3}{4}$

2 $\frac{1}{2}$

3 $- \frac{1}{4}$

4 $\frac{1}{4}$