The graph of the equation z=r*cos(theta) is a plane passing through the origin with normal vector <-1,0,1>.

The answer is true.

Why? Please teach me . Thank you very much.

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- Dec 20th 2006, 01:08 AMJenny20a true and false question
The graph of the equation z=r*cos(theta) is a plane passing through the origin with normal vector <-1,0,1>.

The answer is true.

Why? Please teach me . Thank you very much. - Dec 20th 2006, 06:49 AMSoroban
Hello, Jenny!

There are a number of approaches . . . here's one.

Quote:

The graph of the equation is a plane

passing through the origin with normal vector

The answer is True . . . why?

Converting from Cylindrical to Rectangular, we have: .

If you sketch (or visualize) this plane, its trace is the 45° line on the "left wall".

. . The plane extends infinitely to the left and right.

So it__does__pass through the origin.

Now we must find its normal vector.

Pick three points on the plane:

We have two vectors in the plane: .

The normal vector of the plane is perpendicular to both and .

Hence: .

Therefore: . . . . The statement is*true.*