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Find the distance from Q(3,1,-2) to the YOZ plane.
I don't really understand the concept of the plane, I know it is where x intersect the vector z which is coming straight out of the page but then, isn't the YOZ plane's coordinates equal to the origin's?
Please help, thank you!
Hey Tutu.
A plane equation is given by n . (r - r0) = 0 where r0 is a specific point on the plane and n is a plane normal.
Calculating n . (p - r0) will give the distance to the plane.
The concept of a plane is simply a flat object in so many dimensions that satisfy the above formula.
If the distance is less than zero its below the plane, if its positive its above and if its zero its on the plane.
The plane equation can be understood intuitively by considering that a vector is orthogonal to a normal when the inner product is zero.
Expanding n . (r - r0) = n . r - n . r0 shows that we adjust the plane by considering how its "shifted" from the origin and this term is in the n . r0 term: If the origin was on the plane then it would be zero and the plane equation would be n . r = 0.
I think I get what a plane is..but I still cannot apply this to the question..
Take a look at my response above: The distance is n . (p - p0) where n is the plane normal, p is the point you are checking (i.e. the Q vector) and p0 is an existing point on the plane.
I understand that, I don't know the value of the existing point on the plane and the plane normal? As for the Q vector, am I to sub all components into p?
If we take this question at face value, thenQuote:
Originally Posted by Tutu
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