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lecture_17_slides
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| lecture_17_slides [2017/04/03 15:28] – [Example] rupert | lecture_17_slides [2017/04/03 15:34] (current) – [Corollary] rupert | ||
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| That is, multiplying a vector by a scalar $c$ scales its length by $|c|$, the absolute value of $c$. | That is, multiplying a vector by a scalar $c$ scales its length by $|c|$, the absolute value of $c$. | ||
| - | * Hints: $\sqrt{xy}=\sqrt{x}\sqrt{y}$, | + | * Hints: $\sqrt{xy}=\sqrt{x}\sqrt{y}$ whenever $x,y\ge0$, and $\sqrt{c^2}=|c|$ for any $c\in \mathbb R$. |
| ==== Distance between two points ==== | ==== Distance between two points ==== | ||
| Line 181: | Line 181: | ||
| ==== Example ==== | ==== Example ==== | ||
| - | * Let $\vec v=\m{1\\2}$ and $\vec w=\m{-2\\1}$ | + | What's the angle between |
| - | * Then $\dp vw=1(-2)+2(1)=-2+2=0$. | + | * $\dp vw=1(-2)+2(1)=-2+2=0$. |
| - | * On the other hand, $\|\vec v\|=\sqrt5=\|\vec w\|$ | + | * $\|\vec v\|=\sqrt5=\|\vec w\|$ |
| * So the angle $\theta$ between $\vec v$ and $\vec w$ has $ 0=\dp vw=\sqrt 5\times \sqrt 5 \times \cos\theta$ | * So the angle $\theta$ between $\vec v$ and $\vec w$ has $ 0=\dp vw=\sqrt 5\times \sqrt 5 \times \cos\theta$ | ||
| * So $5\cos\theta=0$, | * So $5\cos\theta=0$, | ||
| Line 225: | Line 225: | ||
| * So $\vv\cdot\ww=\|\vv\|\, | * So $\vv\cdot\ww=\|\vv\|\, | ||
| - | ==== Corollary ==== | + | ==== Corollary |
| If $\vv$ and $\ww$ are non-zero vectors and $\theta$ is the angle between them, then $\cos\theta=\displaystyle\frac{\vv\cdot\ww}{\|\vv\|\, | If $\vv$ and $\ww$ are non-zero vectors and $\theta$ is the angle between them, then $\cos\theta=\displaystyle\frac{\vv\cdot\ww}{\|\vv\|\, | ||
| - | ==== Corollary ==== | + | ==== Corollary |
| If $\vv$ and $\ww$ are non-zero vectors with $\vv\cdot\ww=0$, | If $\vv$ and $\ww$ are non-zero vectors with $\vv\cdot\ww=0$, | ||
lecture_17_slides.1491233287.txt.gz · Last modified: by rupert
