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3 years ago

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Recall that if a line is parallel to the vector v and passes through the point P0, which is the tip of the position vector r0, t

hen the vector equation of the line is given by r(t) = r0 + tv. For the given line, we have r0 = 7, −8, 3 and v = 1, 6, − 1 3 . So the vector equation for this line is r(t) = 7, −8, 3 + t 1, 6, − 1 3 = .

1 answer:

Elenna [48]3 years ago

8 0

Answer:

The vector equation of the line is

$\overrightarrow{r}=+t$

Parametric equations for given line are

$x=7+t\\y=-8+6t\\z=3-13t$

Explanation:

The vector equation of the line is given by

$r(t) = r_{o} + tv$

r₀ = (7, -8, 3)

v = (1, 6, -13)

At these points the vector equation for this line is:

$\overrightarrow{r}=\overrightarrow{r_{o}}+t\overrightarrow{v}\\\overrightarrow{r}=+t$

Parametric equations for given line are

$x=7+t\\y=-8+6t\\z=3-13t$

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$\lambda_{p} = \frac{h}{p_{p}}$

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$v_{p}$ = velocity of an proton

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Kinetic energy of a proton = thermal kinetic energy

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$}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}$

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$\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm$

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