Questions: Consider the vectors u=(1,1,1), v=(2,1,-1). Answer the following questions, (a) Find a unit vector in the direction of v. (b) Find u · v. (c) Find u+v. (d) Find the angle between u and v.

Solution Steps

(a) To find a unit vector in the direction of \(\vec{v}\), we need to divide \(\vec{v}\) by its magnitude.

(b) To find the dot product \(\vec{u} \cdot \vec{v}\), we multiply corresponding components of \(\vec{u}\) and \(\vec{v}\) and sum the results.

(c) To find \(\|\vec{u} + \vec{v}\|\), we first find the vector sum \(\vec{u} + \vec{v}\) and then calculate its magnitude.

To find a unit vector in the direction of \( \vec{v} = (2, 1, -1) \), we first calculate its magnitude:

\[ \|\vec{v}\| = \sqrt{2^2 + 1^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \approx 2.4495 \]

The unit vector \( \hat{v} \) is given by:

\[ \hat{v} = \frac{\vec{v}}{\|\vec{v}\|} = \left( \frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{-1}{\sqrt{6}} \right) \approx (0.8165, 0.4082, -0.4082) \]

The dot product of \( \vec{u} = (1, 1, 1) \) and \( \vec{v} = (2, 1, -1) \) is calculated as follows:

\[ \vec{u} \cdot \vec{v} = 1 \cdot 2 + 1 \cdot 1 + 1 \cdot (-1) = 2 + 1 - 1 = 2 \]

First, we find the vector sum:

\[ \vec{u} + \vec{v} = (1 + 2, 1 + 1, 1 - 1) = (3, 2, 0) \]

Next, we calculate its magnitude:

\[ \|\vec{u} + \vec{v}\| = \sqrt{3^2 + 2^2 + 0^2} = \sqrt{9 + 4 + 0} = \sqrt{13} \approx 3.6056 \]

Final Answer