3
Active contributors today

## Draw the image of VPQR. Confusing geometry graphing question?

Cesareo R.
Featured 5 months ago

See below.

#### Explanation:

The rotation matrix is

$R = \left(\begin{matrix}- 1 & 0 \\ 0 & - 1\end{matrix}\right)$

and the reflexion matrix about the $x$ axis is

$E = \left(\begin{matrix}1 & 0 \\ 0 & - 1\end{matrix}\right)$ so given

$T = \Delta P Q R$

then

${T}_{R} = R . T$

and

${T}_{E} = E . T$

Attached a plot showing

$T$ in black
${T}_{R}$ in red
${T}_{E}$ in blue

## explain and solve this question ?

sente
Featured 5 months ago

2) $80$

#### Explanation:

We will use several facts.

First, note that the volume of a parallelepiped with coterminous edges $\vec{a} , \vec{b} , \vec{c}$ can be found by the triple scalar product$\vec{a} \cdot \left(\vec{b} \times \vec{c}\right)$. As this may be a signed value, we will take the absolute value to get the volume. This allows us to interpret the given information as $| \vec{a} \cdot \left(\vec{b} \times \vec{c}\right) | = 40$.

Two more facts about the scalar triple product we will use:

• $\vec{a} \cdot \left(\vec{b} \times \vec{c}\right) = | \left({a}_{1} , {a}_{2} , {a}_{3}\right) , \left({b}_{1} , {b}_{2} , {b}_{3}\right) , \left({c}_{1} , {c}_{2} , {c}_{3}\right) |$

• $\vec{a} \cdot \left(\vec{b} \times \vec{c}\right) = - \vec{b} \cdot \left(\vec{a} \times \vec{c}\right) = - \vec{c} \cdot \left(\vec{b} \times \vec{a}\right)$

Note that the first fact means that any scalar triple product involving the same vector twice will clearly be $0$, as it is equivalent to taking a determinant of a matrix with repeated rows. Additionally, the cross product of any vector with itself can be immediately evaluated as $\vec{0}$.

Finally, we will use that both the dot product and the cross product distribute over addition. With all that, we will now calculate the volume of the new parallelepiped as the scalar triple product of its coterminous edges:

$| \left(\vec{a} + \vec{b}\right) \cdot \left[\left(\vec{a} + \vec{c}\right) \times \left(\vec{b} + \vec{c}\right)\right] |$

#=|(vec(a)+vec(b)) * [(vec(a) xx vec(b)) + (vec(a) xx vec(c)) + (vec(c) xx vec(b)) + cancel((vec(c) xx vec(c))]|#

$= | \cancel{\vec{a} \cdot \left(\vec{a} \times \vec{b}\right)} + \cancel{\vec{a} \cdot \left(\vec{a} \times \vec{c}\right)}$

$+ \vec{a} \cdot \left(\vec{c} \times \vec{b}\right) + \cancel{\vec{b} \cdot \left(\vec{a} \times \vec{b}\right)}$

$+ \vec{b} \cdot \left(\vec{a} \times \vec{c}\right) + \cancel{\vec{b} \cdot \left(\vec{c} \times \vec{b}\right)} |$

$= | - \vec{a} \cdot \left(\vec{b} \times \vec{c}\right) - \vec{a} \cdot \left(\vec{b} \times \vec{c}\right) |$

$= 2 | \vec{a} \cdot \left(\vec{b} \times \vec{c}\right) |$

$= 2 \left(40\right)$

$= 80$

## Two circles have the following equations: #(x -6 )^2+(y -4 )^2= 64 # and #(x -5 )^2+(y -9 )^2= 49 #. Does one circle contain the other? If not, what is the greatest possible distance between a point on one circle and another point on the other?

Douglas K.
Featured 4 months ago

The farthest distance that between to points on the circle is the sum of the distance between the centers and the two radii:

${d}_{c} + {r}_{1} + {r}_{2} = \sqrt{26} + 8 + 7 \approx 20.1$

#### Explanation:

Here is a graph of the two circles:

The circles overlap but the larger circle does not contain the other. This could have been deduced by the following process.

We assign ${r}^{2}$ to the constant terms to the right of the equal sign in the given equations and then solve for r:

For circle 1:

${r}_{1}^{2} = 64$

${r}_{1} = 8$

For circle 2:

${r}_{2}^{2} = 49$

${r}_{2} = 7$

Compute the distance, ${d}_{c}$, between the two centers:

${d}_{c} = \sqrt{{\left(6 - 5\right)}^{2} + {\left(4 - 9\right)}^{2}}$

${d}_{c} = \sqrt{{\left(1\right)}^{2} + {\left(- 5\right)}^{2}}$

${d}_{c} = \sqrt{26} \approx 5.1$

If the larger circle contained the smaller, then the distance between the centers would be less than difference between the two radii, 1. This is not the case.

The farthest distance that between two points on the circle is the sum of the distance between the centers and the two radii:

${d}_{c} + {r}_{1} + {r}_{2} = \sqrt{26} + 8 + 7 \approx 20.1$

## Let #l# be a line described by equation ax+by+c=0 and let #P(x,y)# be a point not on #l#. Express the distance, #d# between #l and P# in terms of the coefficients #a, b and c# of the equation of line?

Cesareo R.
Featured 4 months ago

$d = \frac{c + a {x}_{0} + b {y}_{0}}{\sqrt{{a}^{2} + {b}^{2}}}$

#### Explanation:

Let $l \to a x + b y + c = 0$ and ${p}_{0} = \left({x}_{0} , {y}_{0}\right)$ a point not on $l$.

Supposing that $b \ne 0$ and calling ${d}^{2} = {\left(x - {x}_{0}\right)}^{2} + {\left(y - {y}_{0}\right)}^{2}$ after substituting $y = - \frac{a x + c}{b}$ into ${d}^{2}$ we have

${d}^{2} = {\left(x - {x}_{0}\right)}^{2} + {\left(\frac{c + a x}{b} + {y}_{0}\right)}^{2}$. The next step is find the ${d}^{2}$ minimum regarding $x$ so we will find $x$ such that

$\frac{d}{\mathrm{dx}} \left({d}^{2}\right) = 2 \left(x - {x}_{0}\right) - \frac{2 a \left(\frac{c + a x}{b} + {y}_{0}\right)}{b} = 0$. This occours for

$x = \frac{{b}^{2} {x}_{0} - a b {y}_{0} - a c}{{a}^{2} + {b}^{2}}$ Now, substituting this value into ${d}^{2}$ we obtain

${d}^{2} = {\left(c + a {x}_{0} + b {y}_{0}\right)}^{2} / \left({a}^{2} + {b}^{2}\right)$ so

$d = \frac{c + a {x}_{0} + b {y}_{0}}{\sqrt{{a}^{2} + {b}^{2}}}$

## What is the ellipse which has vertices at #v_1 = (5,10)# and #v_2=(-2,-10)#, passing by point #p_1=(-5,-4)#?

Douglas K.
Featured 3 months ago

#### Explanation:

To go from ${v}_{1}$ to ${v}_{2}$ the x coordinate decreased by 7 and the y coordinate decreased by 20.

To go from ${v}_{1}$ to the center $\left(h , k\right)$, the x coordinate must decrease by 3.5 and the y coordinate must decrease by 10:

Therefore, $h = 1.5 , k = 0 ,$ and the center of the ellipse is $\left(1.5 , 0\right)$

The length of the semi-major axis, a, is the distance from the center to either vertex:

$a = \sqrt{{\left(5 - 1.5\right)}^{2} + {\left(10 - 0\right)}^{2}}$

$a = \sqrt{{\left(3.5\right)}^{2} + {10}^{2}}$

$a = \frac{\sqrt{449}}{2}$

Let A = the angle of rotation, then:

$\sin \left(A\right) = \frac{10}{\frac{\sqrt{449}}{2}} , \mathmr{and} \cos \left(A\right) = \frac{3.5}{\frac{\sqrt{449}}{2}}$

Rationalizing both denominators:

$\sin \left(A\right) = \frac{20 \sqrt{449}}{449} \mathmr{and} \cos \left(A\right) = \frac{7 \sqrt{449}}{449}$

Here is a reference for An Rotated Ellipse that I not at the origin

${\left(\left(x - h\right) \cos \left(A\right) + \left(y - k\right) \sin \left(A\right)\right)}^{2} / {a}^{2} + {\left(\left(x - h\right) \sin \left(A\right) - \left(y - k\right) \cos \left(A\right)\right)}^{2} / {b}^{2} = 1$

Solving for $b$

$1 - {\left(\left(x - h\right) \cos \left(A\right) + \left(y - k\right) \sin \left(A\right)\right)}^{2} / {a}^{2} = {\left(\left(x - h\right) \sin \left(A\right) - \left(y - k\right) \cos \left(A\right)\right)}^{2} / {b}^{2}$

${b}^{2} = \frac{{a}^{2} {\left(\left(x - h\right) \sin \left(A\right) - \left(y - k\right) \cos \left(A\right)\right)}^{2}}{{a}^{2} - {\left(\left(x - h\right) \cos \left(A\right) + \left(y - k\right) \sin \left(A\right)\right)}^{2}}$

$b = \sqrt{\frac{{a}^{2} {\left(\left(x - h\right) \sin \left(A\right) - \left(y - k\right) \cos \left(A\right)\right)}^{2}}{{a}^{2} - {\left(\left(x - h\right) \cos \left(A\right) + \left(y - k\right) \sin \left(A\right)\right)}^{2}}}$

Force this to contain the point $\left(- 5 , - 4\right)$:

$b = \sqrt{\frac{\left(\frac{449}{4}\right) {\left(\left(- 5 - 1.5\right) \frac{20 \sqrt{449}}{449} - \left(- 4 - 0\right) \frac{7 \sqrt{449}}{449}\right)}^{2}}{\left(\frac{449}{4}\right) - {\left(\left(- 5 - 1.5\right) \frac{7 \sqrt{449}}{449} + \left(- 4 - 0\right) \frac{20 \sqrt{449}}{449}\right)}^{2}}}$

I used WolframAlpha to do the evaluation:

$b = 5.80553$

Here is the final equation:

${\left(\left(x - 1.5\right) \frac{20 \sqrt{449}}{449} + \left(y - 0\right) \frac{7 \sqrt{449}}{449}\right)}^{2} / {\left(\frac{\sqrt{449}}{2}\right)}^{2} + {\left(\left(x - 1.5\right) \frac{7 \sqrt{449}}{449} - \left(y - 0\right) \frac{20 \sqrt{449}}{449}\right)}^{2} / {\left(5.80553\right)}^{2} = 1$

Here is a graph to prove it:

## How do you calculate the surface area of an icosahedron?

George C.
Featured 1 month ago

$\frac{20 \sqrt{3}}{\frac{2}{3} + \varphi} {r}^{2} \text{ }$ where $r$ is the inner radius

$\frac{20 \sqrt{3}}{2 + \varphi} {R}^{2} \text{ }$ where $R$ is the outer radius

#### Explanation:

Synopsis

• First, I will show that the corners of three intersecting golden rectangles with width $1$ and height $\varphi = \frac{1}{2} \left(1 + \sqrt{5}\right)$ lie at the vertices of an icosahedron with edges of length $1$.

• Second, I will determine the outer radius of such an icosahedron, that is the distance from the centre to each vertex.

• Third, I will determine the inner radius of the same icosahedron, that is the distance from the centre to the centre of each face.

• Fourth, I will determine the surface area of an icosahedron with edges of length $1$.

• Fifth, I will use that to write down formulae for the surface area in terms of the inner and outer radii.

$\textcolor{w h i t e}{}$
Proposition

The following $12$ points (all combinations of $\pm$) in ${\mathbb{R}}^{3}$ form the vertices of a regular icosahedron with edges of length $1$:

$\left(\pm \frac{1}{2} , \pm \frac{1}{2} \varphi , 0\right)$

$\left(0 , \pm \frac{1}{2} , \pm \frac{1}{2} \varphi\right)$

$\left(\pm \frac{1}{2} \varphi , 0 , \pm \frac{1}{2}\right)$

where $\varphi = \frac{1}{2} + \frac{\sqrt{5}}{2}$

Each of the three sets of $4$ points are the corners of a golden rectangle with width $1$ and length $\varphi$.

Proof

Note that $\varphi$ is the limit of the ratio between consecutive terms of the Fibonacci sequence, and satisfies:

${\varphi}^{2} = \varphi + 1$

Due to the symmetrical specification, we just need to check that the distance between a corner of one of the golden rectangles and a selected corner of another is $1$.

The distance between $\left({x}_{1} , {y}_{1} , {z}_{1}\right) = \left(\frac{1}{2} , \frac{1}{2} \varphi , 0\right)$ and $\left({x}_{2} , {y}_{2} , {z}_{2}\right) = \left(0 , \frac{1}{2} , \frac{1}{2} \varphi\right)$ is given by the distance formula:

$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}}$

$\textcolor{w h i t e}{d} = \sqrt{{\left(0 - \frac{1}{2}\right)}^{2} + {\left(\frac{1}{2} - \frac{1}{2} \varphi\right)}^{2} + {\left(\frac{1}{2} \varphi - 0\right)}^{2}}$

$\textcolor{w h i t e}{d} = \sqrt{\frac{1}{4} + \frac{1}{4} {\left(1 - \varphi\right)}^{2} + \frac{1}{4} {\varphi}^{2}}$

$\textcolor{w h i t e}{d} = \sqrt{\frac{1}{4} + \frac{1}{4} \left(2 - \varphi\right) + \frac{1}{4} \left(\varphi + 1\right)}$

$\textcolor{w h i t e}{d} = \sqrt{1}$

$\textcolor{w h i t e}{d} = 1$

$\textcolor{w h i t e}{}$

The outer radius of an icosahedron with edges of length $1$ is the distance between $\left(0 , 0 , 0\right)$ and $\left(\frac{1}{2} , \frac{1}{2} \varphi , 0\right)$, namely:

$\sqrt{{\left(\frac{1}{2}\right)}^{2} + {\left(\frac{1}{2} \varphi\right)}^{2} + {0}^{2}} = \sqrt{\frac{1}{4} + \frac{1}{4} {\varphi}^{2}}$

$\textcolor{w h i t e}{\sqrt{{\left(\frac{1}{2}\right)}^{2} + {\left(\frac{1}{2} \varphi\right)}^{2} + {0}^{2}}} = \sqrt{\frac{1}{4} \left(2 + \varphi\right)}$

$\textcolor{w h i t e}{\sqrt{{\left(\frac{1}{2}\right)}^{2} + {\left(\frac{1}{2} \varphi\right)}^{2} + {0}^{2}}} = \frac{1}{2} \sqrt{2 + \varphi}$

$\textcolor{w h i t e}{}$

One of the faces of the icoshedron described above has corners:

$\left(\frac{1}{2} , \frac{1}{2} \varphi , 0\right)$, $\left(0 , \frac{1}{2} , \frac{1}{2} \varphi\right)$, $\left(\frac{1}{2} \varphi , 0 , \frac{1}{2}\right)$

The centre of this face is therefore located at:

$\left(\frac{1}{6} \left(1 + \varphi\right) , \frac{1}{6} \left(1 + \varphi\right) , \frac{1}{6} \left(1 + \varphi\right)\right)$

The distance of this point from the origin is:

$\sqrt{3 {\left(\frac{1}{6} \left(1 + \varphi\right)\right)}^{2}} = \sqrt{\frac{1}{12} {\left(1 + \varphi\right)}^{2}} = \frac{1}{2} \sqrt{\frac{2}{3} + \varphi}$

This is the inner radius of an icosahedron with edges of length $1$.

$\textcolor{w h i t e}{}$
Area of an equilateral triangle with unit sides

An equilateral triangle with sides of length $1$ has height $\frac{\sqrt{3}}{2}$ and therefore area $\frac{\sqrt{3}}{4}$

$\textcolor{w h i t e}{}$
Surface area of icosahedron

The surface area of our icosahedron with edges of length $1$ is:

$20 \cdot \frac{\sqrt{3}}{4} = 5 \sqrt{3}$

If we had an icosahedron of inner radius $r$, then its surface area would be:

$\frac{5 \sqrt{3}}{\frac{1}{2} \sqrt{\frac{2}{3} + \varphi}} ^ 2 {r}^{2} = \frac{20 \sqrt{3}}{\frac{2}{3} + \varphi} {r}^{2}$

If we had an icosahedron of outer radius $R$, then its surface area would be:

$\frac{5 \sqrt{3}}{\frac{1}{2} \sqrt{2 + \varphi}} ^ 2 {R}^{2} = \frac{20 \sqrt{3}}{2 + \varphi} {R}^{2}$

##### Questions
• · 2 minutes ago
• · An hour ago · in Special Right Triangles
• An hour ago · in Volume of Solids
• · 2 hours ago · in Special Right Triangles
· 3 hours ago · in Volume of Solids
• · 6 hours ago · in Special Right Triangles
• · 8 hours ago · in Altitudes
• · 9 hours ago · in Circle Arcs and Sectors
• 9 hours ago · in Rotations
• · 9 hours ago · in Special Right Triangles
• · 10 hours ago · in Solving Similar Triangles
• 10 hours ago · in Volume of Solids
• · 10 hours ago · in Distance between Points
• · 14 hours ago · in Perpendicular Bisectors
• 14 hours ago · in Quadrilaterals
• · 15 hours ago · in Volume of Solids
• · 17 hours ago · in Quadrilaterals
• · 17 hours ago · in Volume of Solids
• 18 hours ago · in Triangle Similarity
• · 19 hours ago · in Circle Arcs and Sectors
• · 19 hours ago · in Rigid Transformations
• · 21 hours ago · in Quadrilaterals