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Let \mathrm{\underset{a}{\rightarrow}} and \mathrm{\underset{b}{\rightarrow}} be the vectors along the diagonals of a parallelogram having area \mathrm{2\sqrt{2}}. Let the angle between \mathrm{\underset{a}{\rightarrow}} and \mathrm{\underset{b}{\rightarrow}} be cute, |\vec{\mathrm{a}}|=1 and |\vec{\mathrm{a}} \cdot \vec{\mathrm{b}}|=|\vec{\mathrm{a}} \times \vec{\mathrm{b}}| . If  \vec{\mathrm{c}}=2 \sqrt{2}(\vec{\mathrm{a}} \times \vec{\mathrm{b}})-2 \vec{\mathrm{b}}, then an angle between \vec{\mathrm{b}}\: and\: \vec{\mathrm{c}} is

Option: 1

\mathrm{\frac{\pi}{4}}


Option: 2

-\mathrm{\frac{\pi}{4}}


Option: 3

\mathrm{\frac{5\pi}{6}}


Option: 4

\mathrm{\frac{3\pi}{4}}


Answers (1)

best_answer

Area of parallelogram     \mathrm{=\frac{1}{2}\left|\overrightarrow{d_{1}} \times \overrightarrow{d_{2}}\right|=2 \sqrt{2}} \\

                                    \mathrm{\Rightarrow|\vec{a} \times \vec{b}|=4 \sqrt{2}}

|\vec{\mathrm{a}} \cdot \vec{\mathrm{b}}|=|\vec{\mathrm{a}} \times \vec{\mathrm{b}}|

|\vec{\mathrm{a}}||\vec{\mathrm{b}}| \cos \theta =|\vec{\mathrm{a}}||\vec{\mathrm{b}}| \sin \theta \\

\mathrm{\Rightarrow \tan \theta=1 \Rightarrow \theta =\pi / 4} \\

|\vec{\mathrm{a}} \times \vec{\mathrm{b}}|=4 \sqrt{2} \Rightarrow|\vec{\mathrm{a}}||\vec{\mathrm{b}}| \sin \pi / 4=4 \sqrt{2} \\

=|\vec{\mathrm{b}}|=8

\vec{\mathrm{c}} =2 \sqrt{2}(\vec{\mathrm{a}} \times \vec{\mathrm{b}})-2 \mathrm{b} \\

|\vec{\mathrm{c}}|^{2} =(2 \sqrt{2}(\vec{\mathrm{a}} \times \vec{\mathrm{b}}))^{2}+4|\vec{\mathrm{b}}|^{2}+4 \sqrt{2}(\vec{\mathrm{a}} \times \vec{\mathrm{b}})-\vec{\mathrm{b}} \\

=256+256=512 \Rightarrow| \vec{\mathrm{c}} \mid=16 \sqrt{2} \\

\vec{\mathrm{c}}+2 \vec{\mathrm{b}} =2 \sqrt{2}(\vec{\mathrm{a}} \times \vec{\mathrm{b}})

\Rightarrow|\vec{\mathrm{c}}|^{2}+4|\vec{\mathrm{b}}|^{2}+4|\vec{\mathrm{b}}||\vec{\mathrm{c}}| \cos \phi=(2 \sqrt{2}|\vec{\mathrm{a}} \times \vec{\mathrm{b}}|)^{2} \\

\mathrm{\Rightarrow 51 2+256+4 \times 8 \times 16 \sqrt{2} \cos \phi=256} \\

\mathrm{\Rightarrow \cos \phi=-\frac{1}{\sqrt{2}} \Rightarrow \phi=\frac{3 \pi}{4}}

Hence the correct answer is option 4

Posted by

Suraj Bhandari

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