Problem regarding vector equation of a plane in space

Vertciel

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13. a) The vectors \(\displaystyle \vec{a},\), \(\displaystyle \vec{b}\), \(\displaystyle \vec{c}\) are the position vectors of three points A, B, and C. Show that \(\displaystyle \vec{r} = p\vec{a} + s\vec{b} + t\vec{c}\), where \(\displaystyle p + s + t = 1\) is an equation of the plane containing these three points.

b) What region of the plane is determined by the equation, when the parameters s and t are restricted to the values \(\displaystyle 0 \leq s \leq 1\), and \(\displaystyle 0 \leq t \leq 1\)? (Hint: Replace p with (1 - s - t))

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Hello there,

I think that I have shown the formula in part a), but I did not get the right answer for part b)

Thank you for your help.

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a) Since p = 1 - s - t, then \(\displaystyle \vec{r} = (1 - s - t)\vec{a} + s\vec{b} + t\vec{c}\).

\(\displaystyle \vec{r} = 1 - s\vec{a} - t\vec{a} + \vec{b} + t\vec{c}\)

\(\displaystyle \vec{r} = [1,0,0] + s(\vec{b} - \vec{a}) + t(\vec{c} - \vec{a})\)

Since this matches the vector equation of a plane in space, then the equation given is the equation of the plane containing points A, B, and C.

b) I set s = t = 0 and then s = t = 1 to see the difference.

I) s = 0, t = 0: \(\displaystyle \vec{r} = [1,0,0] + (0)(\vec{b} - \vec{a}) + (0)(\vec{c} - \vec{a})\)

\(\displaystyle \vec{r} = [1,0,0])\)

II) s = 1, t = 1: \(\displaystyle \vec{r} = [1,0,0] + (1)(\vec{b} - \vec{a}) + (1)(\vec{c} - \vec{a})\)

\(\displaystyle \vec{r} = [1,0,0] + \vec{b} + \vec{c} - 2\vec{a}\)

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The answer given says that all points in and on the parallelogram whose vertices have positions vectors, \(\displaystyle \vec{a}, \vec{b}, -\vec{a} + \vec{b} + \vec{c}, \vec{c}\). However, I do not see how there is a parallelogram represented by \(\displaystyle 0 \leq s, t \leq 1\).
 
Vertciel said:
The answer given says that all points in and on the parallelogram whose vertices have positions vectors, \(\displaystyle \vec{a}, \vec{b}, -\vec{a} + \vec{b} + \vec{c}, \vec{c}\). However, I do not see how there is a parallelogram represented by \(\displaystyle 0 \leq s, t \leq 1\).

You know it will be a subset of the plane.

In fact, what you have is an affine map from R^2=(s,t) into R^3. The image is a plane.

When you say s,t\in[0,1]^2, you now want the image of a square under the map. Is it so surprising you get a parallelogram? Where do the 4 courners map to?
 
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