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Vector Algebra Part 2

Section: Mathematics

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Question : 56 of 100

Marks: +1, -0

Let

\overset{\rightarrow}{a}

\overset{\rightarrow}{b}

be the vectors along the diagonals of a parallelogram having area

2\sqrt{2}

. Let the angle between

\overset{\rightarrow}{a}

\overset{\rightarrow}{b}

\left|\overset{\rightarrow}{a}\right|=1

\left|\overset{\rightarrow}{a}\cdot\overset{\rightarrow}{b}\right|=\left|\overset{\rightarrow}{a}\times\overset{\rightarrow}{b}\right|

\overset{\rightarrow}{c}=2\sqrt{2}\left(\overset{\rightarrow}{a}\times\overset{\rightarrow}{b}\right)-2\overset{\rightarrow}{b}

, then an angle between

\overset{\rightarrow}{b}

\overset{\rightarrow}{c}

[27-Jun-2022-Shift-2]

$\frac{\pi}{4}$
$-\frac{\pi}{4}$
$\frac{5\pi}{6}$
$\frac{3\pi}{4}$

Solution:

\because \overset{\rightarrow}{a}

and

\overset{\rightarrow}{b}

be the vectors along the diagonals of a parallelogram having area

2\sqrt{2}

.

\therefore\;\frac{1}{2}\left|\overset{\rightarrow}{a}\times\overset{\rightarrow}{b}\right|=2\sqrt{2}

\left|\overset{\rightarrow}{a}\right|\left|\overset{\rightarrow}{b}\right|\sin\;\theta=4\sqrt{2}

\Rightarrow \left|\overset{\rightarrow}{b}\right|\sin\;\theta=4\sqrt{2} \ldots \;\text{(i)}\;

\;\text{and}\; \overset{\rightarrow}{a}\cdot\overset{\rightarrow}{b}|=|\overset{\rightarrow}{a}\times\overset{\rightarrow}{b}|

\left|\overset{\rightarrow}{a}\right|\left|\overset{\rightarrow}{b}\right|\cos\theta=\left|\overset{\rightarrow}{a}\right|\left|\overset{\rightarrow}{b}\right|\sin\;\theta

\Rightarrow \tan\theta=1

\therefore \theta=\frac{\pi}{4}

By (i)

\left|\overset{\rightarrow}{b}\right|=8

Now

\overset{\rightarrow}{c}=2\sqrt{2}\left(\overset{\rightarrow}{a}\times\overset{\rightarrow}{b}\right)-2\overset{\rightarrow}{b}

\Rightarrow \overset{\rightarrow}{c}\cdot\overset{\rightarrow}{b}=-2\left|\overset{\rightarrow}{b}\right|^2=-128

and

\overset{\rightarrow}{c}\cdot\overset{\rightarrow}{c}=8\left|\overset{\rightarrow}{a}\times\overset{\rightarrow}{b}\right|^2+4\left|\overset{\rightarrow}{b}\right|^2

\Rightarrow |c|^2=8.32+4.64

\Rightarrow \left|\overset{\rightarrow}{c}\right|=16\sqrt{2}

From (ii) and (iii)

\left|\overset{\rightarrow}{c}\right| \left|\overset{\rightarrow}{b}\right| \cos\alpha = -128

\Rightarrow \cos\alpha = \frac{-1}{\sqrt{2}}

\alpha = \frac{3\pi}{4}

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