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 $\displaystyle z = r\cos\theta$ is a plane
passing through the origin with normal vector $\displaystyle \langle-1,0,1\rangle$
The answer is True . . . why?
Converting from Cylindrical to Rectangular, we have: .$\displaystyle z \:=\:x$
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: $\displaystyle A(0,0,0),\;B(1,0,1),\;C(0,1,0)$
We have two vectors in the plane: .$\displaystyle \begin{array}{cc}\vec{u}\:=\:AB\:=\:\langle 1,0,1\rangle \\ \vec{v}\:=\:AC \:=\:\langle 0,1,0\rangle\end{array} $
The normal vector of the plane is perpendicular to both $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$.
Hence: .$\displaystyle \vec{n}\:=\:\vec{u} \times \vec{v} \:=\:\begin{vmatrix}i & j & k \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{vmatrix} \;= \;-i + 0j + k$
Therefore: .$\displaystyle \vec{n} \:=\:\langle -1,0,1\rangle$ . . . The statement is true.