Find an equation of the plane. The plane through the origin and perpendicular to

Reeves

Reeves

Answered question

2021-09-13

Find an equation of the plane. The plane through the origin and perpendicular to the vector <1, -2, 5>

Answer & Explanation

hesgidiauE

hesgidiauE

Skilled2021-09-14Added 106 answers

Perpendicular to plane and normal to the plane mean the same thing.
Theorem 7 states that:
The scalar equation of a plane that passes through the point (a,b,c) and has (I,m,n) as the normal vector, is given by
l(x-a)+m(y-b)+n(z-c) =0
Therefore, equation of the plane that has ⟨1,—2,5⟩ as the normal vector and passes through the point (0,0,0) is given by
1*(x-0) -2*(y-0)+5*(z-0) =0
Remove the brackets
x-2y+5z=0
Results:
x-2y+5z=0
nick1337

nick1337

Expert2023-05-29Added 777 answers

The general equation of a plane is given by: Ax+By+Cz=D, where 𝐧=⟨A,B,C⟩ is a normal vector to the plane, and D is a constant.
Since we want the plane to be perpendicular to 𝐯=⟨1,−2,5⟩, the normal vector 𝐧 of the plane will be parallel to 𝐯. Therefore, 𝐧 will have the same direction as 𝐯.
Since the plane passes through the origin, which is the point (0,0,0), we can substitute these values into the equation to find the constant D.
Plugging in the values, we have: A(0)+B(0)+C(0)=D. Simplifying this equation gives us D=0.
Therefore, the equation of the plane passing through the origin and perpendicular to the vector 𝐯=⟨1,−2,5⟩ is: Ax+By+Cz=0.
We can also express this equation in vector form as 𝐧·𝐫=0, where 𝐫=⟨x,y,z⟩ is a position vector on the plane.
In this case, the equation becomes ⟨A,B,C⟩·⟨x,y,z⟩=0.
Don Sumner

Don Sumner

Skilled2023-05-29Added 184 answers

Step 1. Let 𝐫=[xyz] be a general point on the plane.
Step 2. Since the plane passes through the origin, the position vector 𝐫 is parallel to the plane.
Step 3. The vector 𝐯 is perpendicular to the plane, so the dot product of 𝐯 and 𝐫 should be zero.
Step 4. Therefore, we have 𝐯·𝐫=0.
Step 5 Using the dot product formula, we can express the dot product as:
𝐯·𝐫=[1−25]·[xyz]=1x+(−2)y+5z=0.
Simplifying the equation, we obtain:
x−2y+5z=0.
Hence, the equation of the plane through the origin and perpendicular to the vector [1−25] is x−2y+5z=0.
Vasquez

Vasquez

Expert2023-05-29Added 669 answers

Result:
x−2y+5z=0
Solution:
To find an equation of the plane through the origin and perpendicular to the vector ⟨1,−2,5⟩, we can use the general form of a plane equation.
Let P be a point on the plane and N be the normal vector of the plane. Since the plane passes through the origin, a point on the plane can be chosen as P=⟨0,0,0⟩. The normal vector of the plane is given as N=⟨1,−2,5⟩.
Using the point-normal form of a plane equation, the equation of the plane is given by:
𝐍·(𝐫−𝐏)=0
where 𝐫 represents a general point on the plane.
Substituting the values of P and N into the equation, we have:
⟨1,−2,5⟩·(⟨x,y,z⟩−⟨0,0,0⟩)=0
Simplifying this equation further:
⟨1,−2,5⟩·⟨x,y,z⟩=0
Taking the dot product of the vectors:
x−2y+5z=0
Hence, the equation of the plane through the origin and perpendicular to the vector ⟨1,−2,5⟩ is x−2y+5z=0.

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