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.
Hello, Jenny!
There are a number of approaches . . . here's one.
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.