Question

what is the angle \\( \theta \\) between the vectors \\( \vec{u}=\left\

$$\begin{array}{c}1 \\\\-1 \\\\0\\end{array}$$

\
ight \\) and \\( \vec{v}=\left\

$$\begin{array}{l}0 \\\\1 \\\\1\\end{array}$$

\
ight \\)?
(a) \\( \theta=\frac{\pi}{6} \\).
(b) \\( \theta=\frac{\pi}{4} \\).
(c) \\( \theta=\frac{\pi}{3} \\).
(d) \\( \theta=\frac{\pi}{2} \\).
(e) \\( \theta=\frac{2 \pi}{3} \\).

Explanation:

Step1: Calculate the dot - product

The dot - product formula is \(\vec{u}\cdot\vec{v}=u_1v_1 + u_2v_2+u_3v_3\).
Given \(\vec{u}=

$$\begin{bmatrix}1\\-1\\0\end{bmatrix}$$

\) and \(\vec{v}=

$$\begin{bmatrix}0\\1\\1\end{bmatrix}$$

\), then \(\vec{u}\cdot\vec{v}=(1)\times(0)+(-1)\times(1)+(0)\times(1)=- 1\).

Step2: Calculate the magnitudes

The magnitude formula is \(|\vec{a}|=\sqrt{a_1^{2}+a_2^{2}+a_3^{2}}\).
For \(\vec{u}\), \(|\vec{u}|=\sqrt{1^{2}+(-1)^{2}+0^{2}}=\sqrt{2}\).
For \(\vec{v}\), \(|\vec{v}|=\sqrt{0^{2}+1^{2}+1^{2}}=\sqrt{2}\).

Step3: Use the formula for the angle between two vectors

The formula is \(\cos\theta=\frac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|}\).
Substitute the values: \(\cos\theta=\frac{-1}{\sqrt{2}\times\sqrt{2}}=-\frac{1}{2}\).
Since \(0\leq\theta\leq\pi\), then \(\theta=\frac{2\pi}{3}\).

Answer:

E. \(\theta=\frac{2\pi}{3}\)