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[![enter image description here][1]][1]I can figure out how to plot this for calc 3.

Question:

a) Plot the circle of radius 3 centered at the point $\{-1, 1, 1\}$ in the plane whose xyz-equation is $2(x + 1) + 3(y - 1) + (z - 1) = 0$. Include in your plot a big enough piece of the plane to accommodate the circle.

My answer:

We are given:

• Radius = $3$
• Center = $\{-1, 1, 1\}$

Then:

$2(x + 1) + 3(y - 1) + (z - 1) = 0$ so

$2x + 2 + 3y - 3 + z - 1 = 0$ so

$2x + 3y + z = -2 + 3 + 1$ so

$2x + 3y + z = 2$

Thus: Normal = $\{2, 3, 1\}$

Equation of line:

$r(t) = \{-1, 1, 1\} + t \{2, 3, 1\}$

Equation of plane:

$2x + 3y + z = 2$ so

$\left(x + \frac{3}{2}y + \frac{z}{2}\right) = 1$

So we can find that the:

• x intercept = $1$
• y intercept = $\frac{2}{3}$
• z intercept = $2$

How can I plot this?

b) Here's a plot of a spiral in the xy-plane:

 

Clear[spiral, t];
spiral[t_] = {t Cos[2 t], t Sin[2 t]};
ParametricPlot[spiral[t], {t, 0, 3 Pi}, AxesLabel -> {"x", "y"}]

 

Use your answer to part a) above to help plot a true scale duplicate copy of this spiral on the plane with xyz-equation $2(x + 1) + 3(y - 1) + (z - 1) = 0$. Center your spiral at $\{-1, 1, 1\}$ and include in your plot a big enough hunk of the plane to accommodate the spiral.

How do I plot this?

The image is what I have tried. [1]: https://i.sstatic.net/CzZE1.png

I can figure out how to plot this for calc 3.

Question:

a) Plot the circle of radius 3 centered at the point $\{-1, 1, 1\}$ in the plane whose xyz-equation is $2(x + 1) + 3(y - 1) + (z - 1) = 0$. Include in your plot a big enough piece of the plane to accommodate the circle.

My answer:

We are given:

• Radius = $3$
• Center = $\{-1, 1, 1\}$

Then:

$2(x + 1) + 3(y - 1) + (z - 1) = 0$ so

$2x + 2 + 3y - 3 + z - 1 = 0$ so

$2x + 3y + z = -2 + 3 + 1$ so

$2x + 3y + z = 2$

Thus: Normal = $\{2, 3, 1\}$

Equation of line:

$r(t) = \{-1, 1, 1\} + t \{2, 3, 1\}$

Equation of plane:

$2x + 3y + z = 2$ so

$\left(x + \frac{3}{2}y + \frac{z}{2}\right) = 1$

So we can find that the:

• x intercept = $1$
• y intercept = $\frac{2}{3}$
• z intercept = $2$

How can I plot this?

b) Here's a plot of a spiral in the xy-plane:

Clear[spiral, t];
spiral[t_] = {t Cos[2 t], t Sin[2 t]};
ParametricPlot[spiral[t], {t, 0, 3 Pi}, AxesLabel -> {"x", "y"}]

Use your answer to part a) above to help plot a true scale duplicate copy of this spiral on the plane with xyz-equation $2(x + 1) + 3(y - 1) + (z - 1) = 0$. Center your spiral at $\{-1, 1, 1\}$ and include in your plot a big enough hunk of the plane to accommodate the spiral.

How do I plot this?

The image is what I have tried.

I can figure out how to plot this for calc 3.

Question:

a) Plot the circle of radius 3 centered at the point $\{-1, 1, 1\}$ in the plane whose xyz-equation is $2(x + 1) + 3(y - 1) + (z - 1) = 0$. Include in your plot a big enough piece of the plane to accommodate the circle.

My answer:

We are given:

• Radius = $3$
• Center = $\{-1, 1, 1\}$

Then:

$2(x + 1) + 3(y - 1) + (z - 1) = 0$ so

$2x + 2 + 3y - 3 + z - 1 = 0$ so

$2x + 3y + z = -2 + 3 + 1$ so

$2x + 3y + z = 2$

Thus: Normal = $\{2, 3, 1\}$

Equation of line:

$r(t) = \{-1, 1, 1\} + t \{2, 3, 1\}$

Equation of plane:

$2x + 3y + z = 2$ so

$\left(x + \frac{3}{2}y + \frac{z}{2}\right) = 1$

So we can find that the:

• x intercept = $1$
• y intercept = $\frac{2}{3}$
• z intercept = $2$

How can I plot this?

b) Here's a plot of a spiral in the xy-plane:

Clear[spiral, t];
spiral[t_] = {t Cos[2 t], t Sin[2 t]};
ParametricPlot[spiral[t], {t, 0, 3 Pi}, AxesLabel -> {"x", "y"}]

Use your answer to part a) above to help plot a true scale duplicate copy of this spiral on the plane with xyz-equation $2(x + 1) + 3(y - 1) + (z - 1) = 0$. Center your spiral at $\{-1, 1, 1\}$ and include in your plot a big enough hunk of the plane to accommodate the spiral.

How do I plot this?

[![enter image description here][1]][1]I can figure out how to plot this for calc 3.

Question:

a) Plot the circle of radius 3 centered at the point $\{-1, 1, 1\}$ in the plane whose xyz-equation is $2(x + 1) + 3(y - 1) + (z - 1) = 0$. Include in your plot a big enough piece of the plane to accommodate the circle.

My answer:

We are given:

• Radius = $3$
• Center = $\{-1, 1, 1\}$

Then:

$2(x + 1) + 3(y - 1) + (z - 1) = 0$ so

$2x + 2 + 3y - 3 + z - 1 = 0$ so

$2x + 3y + z = -2 + 3 + 1$ so

$2x + 3y + z = 2$

Thus: Normal = $\{2, 3, 1\}$

Equation of line:

$r(t) = \{-1, 1, 1\} + t \{2, 3, 1\}$

Equation of plane:

$2x + 3y + z = 2$ so

$\left(x + \frac{3}{2}y + \frac{z}{2}\right) = 1$

So we can find that the:

• x intercept = $1$
• y intercept = $\frac{2}{3}$
• z intercept = $2$

How can I plot this?

b) Here's a plot of a spiral in the xy-plane:

Clear[spiral, t];
spiral[t_] = {t Cos[2 t], t Sin[2 t]};
ParametricPlot[spiral[t], {t, 0, 3 Pi}, AxesLabel -> {"x", "y"}]

Use your answer to part a) above to help plot a true scale duplicate copy of this spiral on the plane with xyz-equation $2(x + 1) + 3(y - 1) + (z - 1) = 0$. Center your spiral at $\{-1, 1, 1\}$ and include in your plot a big enough hunk of the plane to accommodate the spiral.

How do I plot this?

The image is what I have tried. [1]: https://i.sstatic.net/CzZE1.png