If {{u}} = {{a}} - {{b}}, {{v}} = {{a}} + {{b}}, |{{a}}| = |{{b}}| = 2 then |{{u}} × {{v}}| is equal to

Correct Answer is : 216−(a⃗⋅b⃗)22 {16 - ({{a}} · {{b}})²216−(a⋅b)2

Correct Answer is : 216−(a⃗⋅b⃗)22 {16 - ({{a}} · {{b}})²216−(a⋅b)2

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EAMCET Engineering 2010 Solved Paper

Section: Mathematics

Question : 4 of 160

Marks: +1, -0

\vec{\mathbf{u}} = \vec{\mathbf{a}} - \vec{\mathbf{b}}, \vec{\mathbf{v}} = \vec{\mathbf{a}} + \vec{\mathbf{b}}, |\vec{\mathbf{a}}| = |\vec{\mathbf{b}}| = 2

|\vec{\mathbf{u}} \times \vec{\mathbf{v}}|

Solution:

We have,

\vec{\mathbf{u}} = \vec{\mathbf{a}} - \vec{\mathbf{b}}, \vec{\mathbf{v}} = \vec{\mathbf{a}} + \vec{\mathbf{b}}

\; \Rightarrow \vec{\mathbf{u}} \times \vec{\mathbf{v}} = (\vec{\mathbf{a}} - \vec{\mathbf{b}}) \times (\vec{\mathbf{a}} + \vec{\mathbf{b}})

\; = 0 - \vec{\mathbf{b}} \times \vec{\mathbf{a}} + \vec{\mathbf{a}} \times \vec{\mathbf{b}} - 0

\; = 2 \vec{\mathbf{a}} \times \vec{\mathbf{b}}

\; \Rightarrow \;\; |\vec{\mathbf{u}} \times \vec{\mathbf{v}}| = 2 |\vec{\mathbf{a}} \times \vec{\mathbf{b}}|

\; = 2 \sqrt{ |\vec{\mathbf{a}} \times \vec{\mathbf{b}}|^2 }

\; = 2 \sqrt{ |\vec{\mathbf{a}}|^2 |\vec{\mathbf{b}}|^2 \sin^2 \theta |\hat{\mathbf{n}}|^2 }

\; \overset{\hat{} }{ \because \mathbf{n} = \text{unit vector} \; |\hat{\mathbf{n}}| = 1 }

\; = 2 \sqrt{4 \cdot 4 \cdot \sin^2 \theta \cdot 1}

\; = 2 \sqrt{16 (1 - \cos^2 \theta)}

\; = 2 \sqrt{16 - 16 \cos^2 \theta}

\; = 2 \sqrt{16 - 16 \left( \frac{\mathbf{a} \cdot \vec{\mathbf{b}}}{|\mathbf{a}| |\mathbf{b}|} \right)^2}

\; \because \{ \vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = |\vec{\mathbf{a}}| |\vec{\mathbf{b}}| \cos \theta \}

\; = 2 \sqrt{16 - 16 \frac{( \mathbf{a} \cdot \vec{\mathbf{b}} )^2}{|\mathbf{a}|^2 |\mathbf{b}|^2}}

\; = 2 \sqrt{16 - 16 \frac{( \vec{\mathbf{a}} \cdot \vec{\mathbf{b}} )^2}{4 \cdot 4}}

\Rightarrow \;\; 2 \sqrt{16 - (\vec{\mathbf{a}} \cdot \vec{\mathbf{b}})^2}

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