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.

Printable View

- Dec 20th 2006, 12:08 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, 05:49 AMSoroban
Hello, Jenny!

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

Quote:

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.*