Perpendicular Plane to Two Given Planes

[andorrak]

Homework Statement

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.)

 

[Dick]

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

 

[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

 

[Dick]

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

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.

 

[dynamicsolo]

This is what i did.

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

*simplified*

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

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

 

[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

 

[Dick]

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

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

 

[dynamicsolo]

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

No, they shouldn't be there.

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

-i + j + k

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.

 

[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?

 

[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.

 

[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.

 

[andorrak]

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