# calculus

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What is the distance from the point P(x,y,z)to the z axis? Explain your answer by graphing.

can anyone explian it to me plezzz

• calculus - ,

Determine whether the plane passing through the points A(1,2,3) , B(0,1,0) and C(0,2,2) passes through the origin

• calculus - ,

What does the following equation represent in R^3 ? Justify your answer.
x=y=z

• calculus - ,

That's the distance from the point
p1 = (0, 0, z) to p2 = (x, y, z).

The square of the distance is the square of the norm of the difference of p1 and p2, which is the inner product of that difference vector with itself (also conventionally denoted as the square of the vector):

d^2 = (p1-p2)^2 = x^2 + y^2

• calculus - ,

Quastion:

Determine whether the plane passing through the points A(1,2,3) , B(0,1,0) and C(0,2,2) passes through the origin.

Let's perform a translation, so that B moves to the origin and the old origin moves to minus B:

A = (1,1,3)

C = (0,1,2)

(I prefer this notation instead of writing A(x,y,z), so I consider A, B, C etc as vectors).

The location of the old origin is denoted by X:

X = (0,-1,0)

If X can be written as a linear combination of A and C, then X is on the plane. So, what you need to do is to determine the rank of the matrix which has A, B and X as its rows (or columns).

I find that the rank is 3, so X is not a point on the plane.

• calculus - ,

>> Determine whether the plane passing through the points A(1,2,3) , B(0,1,0) and C(0,2,2) passes through the origin. <<

alternate method:
vector AB = (1,1,2)
vector AC = (1,0,1)

normal to these is (1,1,-1) , I took the cross-product
so the equation of the plane is
x + y - z = k
put in (1,2,3) , or any of the other two points,
1 + 2 - 3 = k = 0
plane equation is x + y - z = 0
and the point (0,0,0) satisfies this

• calculus - ,

>> What does the following equation represent in R^3 ? Justify your answer.
x=y=z <<

suppose I rewrote the statement this way

(x-0)/1 = (y-0)/1 = (z-0)/1

does that help?

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