Proving the vector OA is normal to the plane OBC

[rainez]

Homework Statement

Given position vectors: OA ( i + 2j - k) ; OB ( -i + 2j +3k); OC ( 2i + j + 4k)

Given that OA is perpendicular to OB.

The Question : Show that OA is normal to the plane OBC.

Homework Equations

r . n = d

To find the normal of the plane OBC, I used n = BO x BC

d = OB . n

The Attempt at a Solution

Equation of plane OBC:

BO = -OB = i - 2j - 3k

BC = OC - OB = (2,1,4) - (-1,2,3) = (3,-1,1)

n = BO x BC = (5,10,-5)

d = OB . n = (-1,2,3) . (5,10,-5) = 0

∴ Equation of plane OBC = r. (5,10,-5) = 0


How do I show that OA is normal to the plane? I know that it is related to the statement - 'OA is perpendicular to OB'. However, I do not know how can it be applied into the question.
I hope someone can help me out.

Thank you for your time!

 

[CAF123]

I agree with the normal vector to the plane OBC, that is $$\vec{n} = 5\vec{i} +10\vec{j} -5\vec{k}.$$
You are given that OA is perpendicular to OB, which you can easily verify.
To show that OA is normal to the plane, show that the cross product of OA with $\vec{n}$ is 0.

 

[rainez]

I agree with the normal vector to the plane OBC, that is $$\vec{n} = 5\vec{i} +10\vec{j} -5\vec{k}.$$
You are given that OA is perpendicular to OB, which you can easily verify.
To show that OA is normal to the plane, show that the cross product of OA with $\vec{n}$ is 0.

Thank you very much for your help!

 

[LCKurtz]

I agree with the normal vector to the plane OBC, that is $$\vec{n} = 5\vec{i} +10\vec{j} -5\vec{k}.$$
You are given that OA is perpendicular to OB, which you can easily verify.
To show that OA is normal to the plane, show that the cross product of OA with $\vec{n}$ is 0.

Or even easier, observe that OA is a scalar multiple of $\vec n$.