If a→,b→ and c→ arc unit vectors such that a→+2b→+2c→=0, thena¯×c¯= :

Given, a→+2b→+2c→=0⇒ a→+2c→=−2b→Squaring both sides, we get|a→|2+4|c→|2+4a→⋅c→=4|b→|2⇒1+4+4(a→⋅c→)=4⇒a→⋅c→=−14⇒|a→|⋅|c→|cos⁡θ=−14⇒cos⁡θ=−14Now, sin2⁡θ=1−cos2⁡θ=1−116=1516∴ |a→×c¯|=|a→||c→|sin⁡θ=154

If a→,b→ and c→ arc unit vectors such that a→+2b→+2c→=0, thena¯×c¯= :

Given, a→+2b→+2c→=0⇒ a→+2c→=−2b→Squaring both sides, we get|a→|2+4|c→|2+4a→⋅c→=4|b→|2⇒1+4+4(a→⋅c→)=4⇒a→⋅c→=−14⇒|a→|⋅|c→|cos⁡θ=−14⇒cos⁡θ=−14Now, sin2⁡θ=1−cos2⁡θ=1−116=1516∴ |a→×c¯|=|a→||c→|sin⁡θ=154

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If $\vec{a}, \vec{b}$ and $\vec{c}$ arc unit vectors such that $\vec{a} + 2 \vec{b} + 2 \vec{c} = 0$ , then $|\bar{a} \times \bar{c}| =$ :

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1/4

$\sqrt{15} / 16$

15/16

$\sqrt{15} / 4$

answer is D.

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Detailed Solution

Given, $\vec{a} + 2 \vec{b} + 2 \vec{c} = 0$

$\Rightarrow \vec{a} + 2 \vec{c} = - 2 \vec{b}$

Squaring both sides, we get

$\begin{array}{l} | \vec{a} |^{2} + 4 | \vec{c} |^{2} + 4 \vec{a} \cdot \vec{c} = 4 | \vec{b} |^{2} \Rightarrow 1 + 4 + 4 (\vec{a} \cdot \vec{c}) = 4 \\ \Rightarrow \vec{a} \cdot \vec{c} = \frac{- 1}{4} \Rightarrow | \vec{a} | \cdot | \vec{c} | \cos θ = \frac{- 1}{4} \Rightarrow \cos θ = \frac{- 1}{4} \end{array}$