# Thread: Angle between 1 and x

1. ## Angle between 1 and x

In C[0,1], with inner product defined by (3), consider the vectors 1 and x.

Find the angle theta between 1 and x.

(3)$\displaystyle \int_{0}^{1}f(x)g(x)dx$

Find the angle theta between 1 and x

2. ## didnot understand

Originally Posted by dwsmith
In C[0,1], with inner product defined by (3), consider the vectors 1 and x.

Find the angle theta between 1 and x.

(3)$\displaystyle \int_{0}^{1}f(x)g(x)dx$

Find the angle theta between 1 and x
didnot get your question. could you please elaborate???

3. Originally Posted by dwsmith
In C[0,1], with inner product defined by (3), consider the vectors 1 and x.

Find the angle theta between 1 and x.

(3)$\displaystyle \int_{0}^{1}f(x)g(x)dx$

Find the angle theta between 1 and x
$\displaystyle \left\langle 1,x\right\rangle=\int_0^1 x dx=\frac{1}{2}$

4. Originally Posted by dwsmith
In C[0,1], with inner product defined by (3), consider the vectors 1 and x.

Find the angle theta between 1 and x.

(3)$\displaystyle \int_{0}^{1}f(x)g(x)dx$

Find the angle theta between 1 and x
You are thinking of C[0,1] as an inner product space with the standard inner product $\displaystyle <f, g>= \int_0^1 f(x)g(x)dx$.

Of course, in any inner product space, $\displaystyle <u, v>= |u||v|cos(\theta)$ where $\displaystyle \theta$ is the angle between u and v.

The particular calculation for the angle between u= 1 and v= x should be easy.

5. After taking the integral, I obtain 1/2. How does that help me obtain the angle of pi/6?

6. After taking what integral? The formula I gave you, $\displaystyle <u, v>= |u||v|cos(\theta)$ requires three integrals to determine <u, v>, |u|, and |v|.