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Question Number 44035 by rahul 19 last updated on 20/Sep/18

Commented by tanmay.chaudhury50@gmail.com last updated on 20/Sep/18

$(\vec{x} . \vec{z}) \vec{y} - (\vec{x} . \vec{y}) \vec{z} = \vec{a}$ $(\sqrt{2} \times \sqrt{2} c o s 60^{o}) \vec{y} - (\sqrt{2} \times \sqrt{2} c o s 60^{o}) \vec{z} = \vec{a}$ $\vec{y} - \vec{z} = \vec{a}$ $\vec{z} - \vec{x} = \vec{b}$ $\vec{x} \times \vec{y} = \vec{c}$ $(\vec{a} + \vec{b}) \times \vec{c} = (\vec{y} - \vec{x}) \times (\vec{x} \times \vec{y})$ $= \vec{y} \times (\vec{x} \times \vec{y}) - \vec{x} \times (\vec{x} \times \vec{y})$ $= (\vec{y} . \vec{y}) \vec{x} - (\vec{y} . \vec{x}) \vec{y} - (\vec{x} . \vec{y}) \vec{x} + (\vec{x} . \vec{x}) \vec{y}$ $= 2 \vec{x} - \vec{y} - \vec{x} + 2 \vec{y}$ $= \vec{x} + \vec{y}$ $\vec{x} + \vec{y} = (\vec{a} + \vec{b}) \times \vec{c}$ $\vec{y} - \vec{x} = \vec{a} + \vec{b}$ $2 \vec{x} = [(\vec{a} + \vec{b}) \times \vec{c} - (\vec{a} + \vec{b})]$ $\vec{x} = \frac{1}{2} [(\vec{a} + \vec{b}) \times \vec{c} - (\vec{a} + \vec{b})]$ $(\vec{a} + \vec{b}) \times \vec{c} + (\vec{a} + \vec{b}) = \vec{x} + \vec{y} + \vec{y} - \vec{x}$ $\vec{y} = \frac{1}{2} [(\vec{a} + \vec{b}) \times \vec{c} + (\vec{a} + \vec{b})]$ $(\vec{a} + \vec{b}) \times \vec{c} + \vec{b} - \vec{a}$ $= \vec{x} + \vec{y} + \vec{z} - \vec{x} - (\vec{y} - \vec{z})$ $= \vec{x} + \vec{y} + \vec{z} - \vec{x} - \vec{y} + \vec{z}$ $= 2 \vec{z}$ $\vec{z} = \frac{1}{2} [(\vec{a} + \vec{b}) \times \vec{c} + \vec{b} - \vec{a})$
Commented by rahul 19 last updated on 21/Sep/18
Thank you Sir ! ��
Commented by Micky24 last updated on 09/Dec/18

$p l s h o w$
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