- #1
Ryoukomaru
- 55
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I don't have the question with me so I ll just describe it to you. It was a test question.
I was given 3 vectors [tex]\vec u[/tex], [tex]\vec v[/tex], [tex]\vec w[/tex] and [tex]\vec w[/tex] had an unknown parameter p in it.
We were supposed to calculate a value for p such that the three vectors are coplanar.
What I did was find the cross product of [tex]\vec u[/tex]and [tex]\vec v[/tex] and then take the scalar product of the product of [tex]\vec {uv}[/tex] and [tex]\vec w[/tex] and make it equal to 0 and solve for p, i.e. [tex] (\vec{u} \times \vec{v}).\vec w=0 [/tex]
Now apparently this was the correct answer but I do not understand how it proves that they are coplanar. Because scalar product just shows that they are perpendicular and thus the vector is parallel to the plane. But can't it just as well be below or above the plane ?
I was given 3 vectors [tex]\vec u[/tex], [tex]\vec v[/tex], [tex]\vec w[/tex] and [tex]\vec w[/tex] had an unknown parameter p in it.
We were supposed to calculate a value for p such that the three vectors are coplanar.
What I did was find the cross product of [tex]\vec u[/tex]and [tex]\vec v[/tex] and then take the scalar product of the product of [tex]\vec {uv}[/tex] and [tex]\vec w[/tex] and make it equal to 0 and solve for p, i.e. [tex] (\vec{u} \times \vec{v}).\vec w=0 [/tex]
Now apparently this was the correct answer but I do not understand how it proves that they are coplanar. Because scalar product just shows that they are perpendicular and thus the vector is parallel to the plane. But can't it just as well be below or above the plane ?
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