Cross Product of Two Planes

1. Sep 20, 2011

andorrak

1. The problem statement, all variables and given/known data
Find a plane perpendicular to the two planes, X+Y=3 and X+2y-z=4

I know i take the cross product of both so i get

<1,1,0> and <1,2,-1>

But when i do the cross product i get x-y+z

book tells me x-y-z

what am i doing wrong? Not sure why the z is negative. The cross product formula tells me. that the only the J component is negative. (Using Rogawskis Multivariable book. Page. 698, 13.4.)

2. Sep 20, 2011

Dick

The book is right. Your cross product must be wrong. Can you show how you got it?

3. Sep 20, 2011

andorrak

This is what i did.

|i j k |
|1 1 0|
|1 2 -1| =

*simplified*

|-1-0|i-|-1-0|j+|2-1|k

which gives me

i - j +k

Last edited: Sep 20, 2011
4. Sep 20, 2011

Dick

There's some mistakes in there alright. Take the i component. I get the coefficient to be 1*(-1)-0*2=(-1). Check your examples again.

5. Sep 20, 2011

dynamicsolo

Are those really absolute value operations you're doing there? If so, they don't belong in the calculation.

6. Sep 20, 2011

andorrak

I thought that there were supposed to be absolute values in the equation.

If there are not, the way i would calculate it would be then:

-i + j + k

7. Sep 20, 2011

Dick

Now that's right! Thanks for picking up on the problem, dynamicsolo.

8. Sep 20, 2011

dynamicsolo

No, they shouldn't be there.

Much better. You now have the direction vector for a line mutually perpendicular to the two given planes and thus the normal vector to the plane you seek. You next need to find a point that lies in that plane in order to write its equation.

9. Sep 20, 2011

andorrak

Okay so i just used (2, 1, 0) final equation -x+y+z=-1

but the book is saying x-y-z=f

I guess they just wanted the normal but why -y-z???

10. Sep 20, 2011

andorrak

Is it something as simple as multiplying by -1? if so, i feel stupid. but thanks on the absolute values, that would have been a face palm if i hadn't known that on the test day.

11. Sep 20, 2011

dynamicsolo

There is no single way to write the normal vector for a plane. Any non-zero scalar (numerical) multiple of the vector will serve; multiplying by a negative number changes the signs of all the components, but that just gives you a vector which points in exactly the opposite direction along the same line.

So you may use < -1, 1, 1 > or < 1, -1, -1 > (or, heck, < 2011, -2011, -2011 > , were you so inclined) to represent a normal vector to the desired plane. Flipping the signs on the equation for the plane (or multiplying the equation by any non-zero real number) just gives another name for the same plane, since the set of points that satisfy the algebraic equation will remain unchanged.

12. Sep 20, 2011

andorrak

Yea i figured that. Thanks so much Dick and Dynamicsolo! Don't worry ill be back. lol