What is the Correct Cross Product for Finding a Normal Vector?

  • Thread starter Thread starter char808
  • Start date Start date
  • Tags Tags
    Plane
Click For Summary
To find the equation of a plane through the point (2, 0, 2) that is perpendicular to the vectors <1, -1, 2> and <-1, 1, 0>, the correct normal vector is determined by their cross product. The attempted normal vector of <-2, -3, 0> was questioned, with the correct calculation yielding <-2, 3, 0>. It was emphasized that the resulting normal vector must be perpendicular to both original vectors, which can be verified through dot products. If the dot products yield zero, the calculations are confirmed to be correct. The discussion highlights the importance of accurate cross product calculations in determining the normal vector for the plane equation.
char808
Messages
27
Reaction score
0

Homework Statement



The whole problem is :
Find eqn of plane through (2, 0, 2) perpendicular to vectors <1, -1, 2> and <-1, 1, 0>


I am trying to figure out if I am making an error with cross products here...

Find the normal (perpendicular) vector of <1, -1, 2> and <-1, 1, 0>

Homework Equations



Cross products equation... <1, -1, 2> X <-1, 1, 0>


The Attempt at a Solution




Normal Vector: <-2, -3, 0>

So the plane equation would be
-2(x-2) -3(y-0)+0(z-2) =0
-2x-3y=4
 
Physics news on Phys.org
char808 said:

Homework Statement



The whole problem is :
Find eqn of plane through (2, 0, 2) perpendicular to vectors <1, -1, 2> and <-1, 1, 0>

I am trying to figure out if I am making an error with cross products here...

Find the normal (perpendicular) vector of <1, -1, 2> and <-1, 1, 0>

Homework Equations



Cross products equation... <1, -1, 2> X <-1, 1, 0>

The Attempt at a Solution



Normal Vector: <-2, -3, 0>

So the plane equation would be
-2(x-2) -3(y-0)+0(z-2) =0
-2x-3y=4
It's impossible for a plane to be perpendicular two two vectors which are not parallel to each other.

Do you mean that the normal to the plane is perpendicular to those two vectors?
 
I think I worded it incorrectly. Mostly I am interested in the cross products and if that particular cross product is correct. IE <1,-1,2> X <-1,1-0> = <-2, 3, 0>

ignore the rest of it...
 
char808 said:
I think I worded it incorrectly. Mostly I am interested in the cross products and if that particular cross product is correct. IE <1,-1,2> X <-1,1-0> = <-2, 3, 0>
ignore the rest of it...
No, it's not correct. You can check your work by dotting your result vector with each of the two vectors that make up the cross product. The result vector should be perpendicular to each of the other two, meaning that both dot products should be zero.

If you can't figure it out, show us your calculations.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
4
Views
2K
  • · Replies 13 ·
Replies
13
Views
4K
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
4
Views
5K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K