Let a→ and b→ be the vectors along the diagonal of a parallelogram having area 22. Let the angle between a→ and b→ be acute. |a→|=1 and |a→.b→|=|a→×b→|. If c→=22(a→×b→)−2b→, then an angle between b→ and c→ is :

∴ 12|a→×b→|=22 |a→||b→|sin⁡θ=42⇒ |b→|sin⁡θ=42 and |a→⋅b→|=|a→×b→| |a→||b→|cos⁡θ=|a→||b→|sin⁡θ⇒tan⁡θ=1 ∴θ=π4By (i) |b→|sinπ4=42 ⇒|b→|=8 Now c→=22(a→×b→)−2b→ ⇒c→⋅b→=−2|b→|2=−128 and c→⋅c→=8|a→×b→|2+4|b→|2 ⇒|c→|2=8.32+4.64=512 ⇒|c→|=162From (ii) and (iii) |c→||b→|cos⁡α=−128⇒cos⁡α=−12 α=3π4

Let a→ and b→ be the vectors along the diagonal of a parallelogram having area 22. Let the angle between a→ and b→ be acute. |a→|=1 and |a→.b→|=|a→×b→|. If c→=22(a→×b→)−2b→, then an angle between b→ and c→ is :

∴ 12|a→×b→|=22 |a→||b→|sin⁡θ=42⇒ |b→|sin⁡θ=42 and |a→⋅b→|=|a→×b→| |a→||b→|cos⁡θ=|a→||b→|sin⁡θ⇒tan⁡θ=1 ∴θ=π4By (i) |b→|sinπ4=42 ⇒|b→|=8 Now c→=22(a→×b→)−2b→ ⇒c→⋅b→=−2|b→|2=−128 and c→⋅c→=8|a→×b→|2+4|b→|2 ⇒|c→|2=8.32+4.64=512 ⇒|c→|=162From (ii) and (iii) |c→||b→|cos⁡α=−128⇒cos⁡α=−12 α=3π4

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Let $\vec{a}$ and $\vec{b}$ be the vectors along the diagonal of a parallelogram having area $2 \sqrt{2}$ . Let the angle between $\vec{a}$ and $\vec{b}$ be acute. $| \vec{a} | = 1$ and $| \vec{a} . \vec{b} | = | \vec{a} \times \vec{b} |$ . If $\vec{c} = 2 \sqrt{2} (\vec{a} \times \vec{b}) - 2 \vec{b}$ , then an angle between $\vec{b}$ and $\vec{c}$ is :

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$\frac{3 π}{4}$

$\frac{5 π}{6}$

$\frac{π}{4}$

$- \frac{π}{4}$

answer is D.

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Detailed Solution

$\begin{aligned} ∴ & \frac{1}{2} | \vec{a} \times \vec{b} | = 2 \sqrt{2} \\ | \vec{a} | | \vec{b} | \sin θ = 4 \sqrt{2} \\ \Rightarrow & | \vec{b} | \sin θ = 4 \sqrt{2} \\ and & | \vec{a} \cdot \vec{b} | = | \vec{a} \times \vec{b} | \\ | \vec{a} | | \vec{b} | \cos θ = | \vec{a} | | \vec{b} | \sin θ \\ \Rightarrow & \tan θ = 1 ∴ θ = \frac{π}{4} \end{aligned}$

By (i) $| \vec{b} | \sin \frac{π}{4} = 4 \sqrt{2} \Rightarrow | \vec{b} | = 8$

$\begin{aligned} Now \vec{c} = 2 \sqrt{2} (\vec{a} \times \vec{b}) - 2 \vec{b} \\ \Rightarrow \vec{c} \cdot \vec{b} = - 2 | \vec{b} |^{2} = - 128 \\ and \vec{c} \cdot \vec{c} = 8 | \vec{a} \times \vec{b} |^{2} + 4 | \vec{b} |^{2} \\ \Rightarrow | \vec{c} |^{2} = 8.32 + 4.64 = 512 \\ \Rightarrow | \vec{c} | = 16 \sqrt{2} \end{aligned}$