Express the following plane in vector form: P 1 ⊆ R 3 with equation 4...

Freddy Friedman

Freddy Friedman

Answered

2022-07-18

Express the following plane in vector form:
P 1 R 3 with equation 4 x z = 0
The answer is t ( 1 , 0 , 4 ) + s ( 0 , 1 , 0 ). I don't understand how they got (0,1,0) for the direction vector.

Answer & Explanation

Brienueentismvh

Brienueentismvh

Expert

2022-07-19Added 11 answers

The cartesian equation is z=4x
The parametrization is not unique. Anyway You need two linearly independent points on the plane.
One point can be A=(1,0,4). The second one cannot be (2,0,8) or other points you can get from A multiplying by a constant. Otherwise you get a line not a plane.
The second point cannot obviously be (0,0,0)
So we look for a point where the only nonzero coordinate is y, like (0,1,0) or (0,10,0)
Finally the parametric equation of the plane is
( 1 , 0 , 4 ) t + ( 0 , 1 , 0 ) s = 0
Note that ( 2 , 0 , 8 ) t + ( 0 , 10 , 0 ) s = 0 would be right.
Hope this helps
Lexi Mcneil

Lexi Mcneil

Expert

2022-07-20Added 2 answers

Let π 4 x z = 0 be your plane. We notice that the point ( 0 , 0 , 0 ) π and the orthogonal vector to the plane is u _ = ( 4 , 0 , 1 )
The vector space dir ( π ) has dimension 2 and the vectors that span this space are orthogonal to the vector u _
We can take for example the vectors v _ = ( 1 , 0 , 4 ) and w _ = ( 0 , 1 , 0 ), infact we have
u _ , v _ = 1 4 + 0 + 4 ( 1 ) = 0  they are orthogonal
u _ , w _ = 0 4 + 1 0 + 0 ( 1 ) = 0  they are orthogonal
Finally the equation of the plane π is
{ x = 0 + α y = 0 + β z = 0 + 4 α , α , β R .

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