Question

1. write the equation for the plane with normal $\vec{n}$ which passes through the point $p$. do this in the form $(x,y,z) - p \cdot \vec{n} = 0$ and also in the form $ax + by + cz = d$. (a) $\vec{n} = \langle 1, 2, 3 \

angle$, $p = (4, 5, 6)$. (b) $\vec{n} = \langle 0, 2, 5 \
angle$, $p = (1, 3, 6)$.

Explanation:

Part (a)

Step 1: Write the vector form

We know that the vector form of the plane equation is \([(x,y,z)-P]\cdot\vec{n} = 0\). Given \(P=(4,5,6)\) and \(\vec{n}=\langle1,2,3
angle\), and \((x,y,z)-P=\langle x - 4,y - 5,z - 6
angle\)

The dot product \(\langle x - 4,y - 5,z - 6
angle\cdot\langle1,2,3
angle=0\)

So the vector form is \([(x,y,z)-(4,5,6)]\cdot\langle1,2,3
angle = 0\)

Step 2: Expand the dot product to get the standard form

We expand \(\langle x - 4,y - 5,z - 6
angle\cdot\langle1,2,3
angle\)

Using the formula for dot product \(\vec{a}\cdot\vec{b}=a_1b_1 + a_2b_2+a_3b_3\)

\((x - 4)\times1+(y - 5)\times2+(z - 6)\times3=0\)

\(x-4 + 2y-10+3z - 18=0\)

\(x + 2y+3z-(4 + 10+18)=0\)

\(x + 2y+3z=32\)

Part (b)

Step 1: Write the vector form

We know that the vector form of the plane equation is \([(x,y,z)-P]\cdot\vec{n}=0\). Given \(P=(1,3,6)\) and \(\vec{n}=\langle0,2,5
angle\), and \((x,y,z)-P=\langle x - 1,y - 3,z - 6
angle\)

The dot product \(\langle x - 1,y - 3,z - 6
angle\cdot\langle0,2,5
angle = 0\)

So the vector form is \([(x,y,z)-(1,3,6)]\cdot\langle0,2,5
angle=0\)

Step 2: Expand the dot product to get the standard form

We expand \(\langle x - 1,y - 3,z - 6
angle\cdot\langle0,2,5
angle\)

Using the formula for dot product \(\vec{a}\cdot\vec{b}=a_1b_1+a_2b_2 + a_3b_3\)

\((x - 1)\times0+(y - 3)\times2+(z - 6)\times5=0\)

\(0+2y-6 + 5z-30=0\)

\(2y+5z-(6 + 30)=0\)

\(2y + 5z=36\) (and since the coefficient of \(x\) is \(0\), the equation is \(0x+2y + 5z=36\))

Final Answers

Part (a)

• Vector form: \(\boldsymbol{[(x,y,z)-(4,5,6)]\cdot\langle1,2,3

angle = 0}\)

• Standard form: \(\boldsymbol{x + 2y+3z=32}\)

Part (b)

• Vector form: \(\boldsymbol{[(x,y,z)-(1,3,6)]\cdot\langle0,2,5

angle=0}\)

• Standard form: \(\boldsymbol{0x + 2y+5z=36}\) (or \(\boldsymbol{2y + 5z=36}\))

Answer:

Part (a)

Step 1: Write the vector form

We know that the vector form of the plane equation is \([(x,y,z)-P]\cdot\vec{n} = 0\). Given \(P=(4,5,6)\) and \(\vec{n}=\langle1,2,3
angle\), and \((x,y,z)-P=\langle x - 4,y - 5,z - 6
angle\)

The dot product \(\langle x - 4,y - 5,z - 6
angle\cdot\langle1,2,3
angle=0\)

So the vector form is \([(x,y,z)-(4,5,6)]\cdot\langle1,2,3
angle = 0\)

Step 2: Expand the dot product to get the standard form

We expand \(\langle x - 4,y - 5,z - 6
angle\cdot\langle1,2,3
angle\)

Using the formula for dot product \(\vec{a}\cdot\vec{b}=a_1b_1 + a_2b_2+a_3b_3\)

\((x - 4)\times1+(y - 5)\times2+(z - 6)\times3=0\)

\(x-4 + 2y-10+3z - 18=0\)

\(x + 2y+3z-(4 + 10+18)=0\)

\(x + 2y+3z=32\)

Part (b)

Step 1: Write the vector form

We know that the vector form of the plane equation is \([(x,y,z)-P]\cdot\vec{n}=0\). Given \(P=(1,3,6)\) and \(\vec{n}=\langle0,2,5
angle\), and \((x,y,z)-P=\langle x - 1,y - 3,z - 6
angle\)

The dot product \(\langle x - 1,y - 3,z - 6
angle\cdot\langle0,2,5
angle = 0\)

So the vector form is \([(x,y,z)-(1,3,6)]\cdot\langle0,2,5
angle=0\)

Step 2: Expand the dot product to get the standard form

We expand \(\langle x - 1,y - 3,z - 6
angle\cdot\langle0,2,5
angle\)

Using the formula for dot product \(\vec{a}\cdot\vec{b}=a_1b_1+a_2b_2 + a_3b_3\)

\((x - 1)\times0+(y - 3)\times2+(z - 6)\times5=0\)

\(0+2y-6 + 5z-30=0\)

\(2y+5z-(6 + 30)=0\)

\(2y + 5z=36\) (and since the coefficient of \(x\) is \(0\), the equation is \(0x+2y + 5z=36\))

Final Answers

Part (a)

• Vector form: \(\boldsymbol{[(x,y,z)-(4,5,6)]\cdot\langle1,2,3

angle = 0}\)

• Standard form: \(\boldsymbol{x + 2y+3z=32}\)

Part (b)

• Vector form: \(\boldsymbol{[(x,y,z)-(1,3,6)]\cdot\langle0,2,5

angle=0}\)

• Standard form: \(\boldsymbol{0x + 2y+5z=36}\) (or \(\boldsymbol{2y + 5z=36}\))