Show algebraically that a triangle cannot be drawn with vertices at the three points A(2, 5), B(6, 1), and C(8, -1). There are (at least) two distinct approaches to this problem.

Patricia Bean
2022-07-26
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grocbyntza

Answered 2022-07-27
Author has **25** answers

A(2,5), B(6,1), C(8,-1)

like equation of AB

$\Rightarrow y-5=\frac{4}{-4}(x-2)\phantom{\rule{0ex}{0ex}}\Rightarrow y-5=x+2\phantom{\rule{0ex}{0ex}}\Rightarrow x+y-7=0$

for point C (8,-1)

$\Rightarrow 8-1-7=0$

All three points are on line line $\mathrm{\u25b3}$ formation is not posible.

$\overrightarrow{AB}=4i-4i=4\phantom{\rule{0ex}{0ex}}\overrightarrow{AC}=6i-6i$

Both $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel to each other

like equation of AB

$\Rightarrow y-5=\frac{4}{-4}(x-2)\phantom{\rule{0ex}{0ex}}\Rightarrow y-5=x+2\phantom{\rule{0ex}{0ex}}\Rightarrow x+y-7=0$

for point C (8,-1)

$\Rightarrow 8-1-7=0$

All three points are on line line $\mathrm{\u25b3}$ formation is not posible.

$\overrightarrow{AB}=4i-4i=4\phantom{\rule{0ex}{0ex}}\overrightarrow{AC}=6i-6i$

Both $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel to each other

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Find the vector and parametric equations for the line segment connecting P to Q.

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Which of the following expressions are meaningful? Which are meaningless? Explain.

a)$(a\cdot b)\cdot c$

$(a\cdot b)\cdot c$ has ? because it is the dot product of ?.

b)$(a\cdot b)c$

$(a\cdot b)c$ has ? because it is a scalar multiple of ?.

c)$|a|(b\cdot c)$

$|a|(b\cdot c)$ has ? because it is the product of ?.

d)$a\cdot (b+c)$

$a\cdot (b+c)$ has ? because it is the dot product of ?.

e)$a\cdot b+c$

$a\cdot b+c$ has ? because it is the sum of ?.

f)$|a|\cdot (b+c)$

$|a|\cdot (b+c)$ has ? because it is the dot product of ?.

a)

b)

c)

d)

e)

f)

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Let $\mathbf{A}$ be a matrix, and let 𝐱 and 𝐲 be linearly independent vectors such that 𝐀𝐱=𝐲,𝐀𝐲=𝐱+2𝐲.Then we have that ${\mathbf{A}}^{5}\mathbf{x}=a\mathbf{x}+b\mathbf{y}$ for some scalars 𝑎 and 𝑏. Find the ordered pair (a, b).

So far I have ${\mathbf{A}}^{5}x={\mathbf{A}}^{4}y$

${\mathbf{A}}^{4}y={\mathbf{A}}^{3}x+2{\mathbf{A}}^{3}y$

So far I have ${\mathbf{A}}^{5}x={\mathbf{A}}^{4}y$

${\mathbf{A}}^{4}y={\mathbf{A}}^{3}x+2{\mathbf{A}}^{3}y$

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Geometrical Interpretation of Vectors Addition

I don't understand what it means by $({y}_{1}-{x}_{1},{y}_{2}-{x}_{2},{y}_{3}-{x}_{3})$

This is the part of the book:

The directed line segment PQ, from the point $P=({x}_{1},{x}_{2},{x}_{3})$ to the point $Q=({y}_{l},{y}_{2},{y}_{3})$, has the same length and direction as the directed line segment from the origin $0=(0,0,0)$ to the point $({y}_{l}-{x}_{1},{y}_{2}-{x}_{2},{y}_{3}-{x}_{3})$. Furthermore, this is the only segment emanating from the origin which has the same length and direction as PQ. Thus, if one agrees to treat only vectors which emanate from the origin, there is exactly one vector associated with each given length and direction.

I don't understand what it means by $({y}_{1}-{x}_{1},{y}_{2}-{x}_{2},{y}_{3}-{x}_{3})$

This is the part of the book:

The directed line segment PQ, from the point $P=({x}_{1},{x}_{2},{x}_{3})$ to the point $Q=({y}_{l},{y}_{2},{y}_{3})$, has the same length and direction as the directed line segment from the origin $0=(0,0,0)$ to the point $({y}_{l}-{x}_{1},{y}_{2}-{x}_{2},{y}_{3}-{x}_{3})$. Furthermore, this is the only segment emanating from the origin which has the same length and direction as PQ. Thus, if one agrees to treat only vectors which emanate from the origin, there is exactly one vector associated with each given length and direction.