Scalar equation of a plane

**diet.pepsi** · April 12th 2008, 01:24 PM

Find the scalar equation of the plane that is perpendicular to the plane x+2y+4=0, contains the origin, and whose normal makes an angle of 30 with the z-axis.

attempt:

d1= (1,2,0)
P(0,0,0)
cos(theta)=z/|normal|

**Soroban** · April 13th 2008, 07:31 AM

Wow . . . what a stupid blunder!

Sorry, everyone . . .

**diet.pepsi** · April 13th 2008, 07:46 AM

"Its normal vector, (a,b,c), is perpendicular to (1,2,4)"

Wouldnt it be perpendicular to (1,2,0)? 4 is the D value. thnks.

**earboth** · April 13th 2008, 07:48 AM

Originally Posted by Soroban

Hello, diet.pepsi!

I thought I had a good approach . . . then I hit a wall.

[size=3]The general equation of a plane is: . $ax + by + cz + d \:=\:0$

Since it contains the origin: . $ax + by + cz \:=\:0$

Its normal vector, $\vec{n} \,=\,\langle a,b,c\rangle$ , is perpendicular to $\langle1,2,4\rangle$

The normal vector is $\langle1,2,{\color{red}0}\rangle$

...

I nearly finished this question ... but I couldn't find my mistake.

I've attached the sketches I made so far.

**Moo** · April 13th 2008, 07:52 AM

Hello earboth,

With what software did you sketch this ?

**earboth** · April 13th 2008, 09:55 PM

Originally Posted by diet.pepsi

Find the scalar equation of the plane that is perpendicular to the plane x+2y+4=0, contains the origin, and whose normal makes an angle of 30 with the z-axis.

attempt:

d1= (1,2,0)
P(0,0,0)
cos(theta)=z/|normal|

As I assume Soroban tried to attempt this problem algebraically. I've choosen a completely different way to do this question:

1. From the equation of the plane $p_1$ you know that the normal vector is $\overrightarrow{n_1} = (1,2,0)$ . That means $p_1$ is perpendicular to the x-y-plane.

2. I looked for a line which lies simultaneously in $p_1$ and the x-y-plane and took the line which connects the intersection points of $p_1$ and the x-axis and $p_1$ and the y-axis. The intersection points are A(-4,0,0) and B(0,-2,0).

3. I took the line $\vec r= t \cdot (1,2,0)$ as a kind of axis. I then "tilted" the x-y-plane (which is perpendicular to $p_1$ !) about this "axis" by 30° to get $p_2$ . If I tilt the x-y-plane by 30° then the normal vector of $p_2$ and the z-axis include an angle of 30°.

4. I took the vector $\overrightarrow{AB} =(-4,2,0)$ and rotated it by 30° to get the vector $\vec b = \left(-4,2,\sqrt{\frac{20}3}\right)$ (By the way: You probably have noticed that there must exist a second vector $\vec b '$ if you rotate the x-y-plane in the opposite direction)

5. Now I know 2 vectors which span the plane $p_2$ . Since $p_2$ has to contain the origin the parametric equation of $p_2$ is:

$p_2$ : $\vec r = s \cdot (1,2,0) + t \cdot \left(-4,2,\sqrt{\frac{20}3}\right)$ ....... or ....... $p_2: \left\{\begin{array}{lcr}x&=&s-4t \\ y&=&2s+2t \\ z&=&\sqrt{\frac{20}3} \cdot t \end{array}\right.$

6. The cross product of the 2 direction vectors which span $p_2$ is the normal vector of $p_2$ :

$(1,2,0) \times \left(-4,2,\sqrt{\frac{20}3}\right) = \left(\frac43 \cdot \sqrt{15}\ ,\ -\frac23 \cdot \sqrt{15}\ ,\ 10 \right)$

And therefore the scalar equation of $p_2$ is:

$\boxed{\frac43 \cdot \sqrt{15} \cdot x - \frac23 \cdot \sqrt{15} \cdot y + 10 \cdot z = 0}$

7. I finally used derive 6.0 to draw the 2 planes and the normal vector. This program is able to rotate the 3-D-graphs about an axis. I took 2 screenshots to show how the planes and the normal are placed.

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