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I was assuming all vectors that are parallel to a certain v must be the zero vector multiplied by a constant. But apparently this is false, since this would imply that if all vectors parallel to v are of the form vx + 0y + 0z, then all vectors parallel to v are zero, and that is false. What am I doing wrong? A: Suppose that $\vec v$ is parallel to $\vec a$ as well as to $\vec b$ and $\vec c$, and that $\vec v = k\vec a+l\vec b + m\vec c$. Then $k\vec a+l\vec b = \lambda \vec v, \ \lambda, m \in \mathbb{R}.$ But $\vec v = k\vec a+l\vec b+m\vec c = \lambda \vec v + (k\vec a-\lambda \vec v)+(m\vec c-\lambda \vec v)$ so we have $$k\vec a -\lambda \vec v = (k\vec a-\lambda \vec v)+(m\vec c-\lambda \vec v).$$ Now, does there exist some $t$ such that $\vec v=t(k\vec a-\lambda \vec v)?$ If yes, it follows that $k\vec a =\lambda \vec v + t\vec v = t(k\vec a - \lambda \vec v) + \lambda \vec v$ so $k=t$ and $0=\lambda,$ a contradiction. So this can't be. Therefore $k\vec a - \lambda \vec v =0.$ I think this argument is not going to get your desired proof. Design, synthesis, and biological evaluation of novel 1,2,4-triazol- System Requirements For Photoshop Gold Text Effects Psd Files Free Download: Windows Vista, Windows XP Processor: Intel Pentium 4 2.00 GHz or greater Memory: 2 GB RAM Screen resolution: 1024 x 768 DirectX 9.0c with Shader Model 3.0 or higher Recommended: Processor: Intel Core 2 Duo 2.4 GHz or greater Memory: 4 GB RAM
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