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.

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.
Transcript text: Consider the vectors $\vec{u}=(1,1,1), \vec{v}=(2,1,-1)$. Answer the following questions, (a) Find a unit vector in the direction of $\vec{v}$. (b) Find $\vec{u} \cdot \vec{v}$. (c) Find $\|\vec{u}+\vec{v}\|$. (d) Find the angle between $\vec{u}$ and $\vec{v}$.
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Solution

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Solution Steps

Solution Approach

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

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

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

Step 1: Unit Vector in the Direction of v \vec{v}

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

v=22+12+(1)2=4+1+1=62.4495 \|\vec{v}\| = \sqrt{2^2 + 1^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \approx 2.4495

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

v^=vv=(26,16,16)(0.8165,0.4082,0.4082) \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)

Step 2: Dot Product uv \vec{u} \cdot \vec{v}

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

uv=12+11+1(1)=2+11=2 \vec{u} \cdot \vec{v} = 1 \cdot 2 + 1 \cdot 1 + 1 \cdot (-1) = 2 + 1 - 1 = 2

Step 3: Magnitude of u+v \vec{u} + \vec{v}

First, we find the vector sum:

u+v=(1+2,1+1,11)=(3,2,0) \vec{u} + \vec{v} = (1 + 2, 1 + 1, 1 - 1) = (3, 2, 0)

Next, we calculate its magnitude:

u+v=32+22+02=9+4+0=133.6056 \|\vec{u} + \vec{v}\| = \sqrt{3^2 + 2^2 + 0^2} = \sqrt{9 + 4 + 0} = \sqrt{13} \approx 3.6056

Final Answer

  • Unit vector in the direction of v \vec{v} : v^(0.8165,0.4082,0.4082) \hat{v} \approx (0.8165, 0.4082, -0.4082)
  • Dot product uv=2 \vec{u} \cdot \vec{v} = 2
  • Magnitude of u+v3.6056 \vec{u} + \vec{v} \approx 3.6056

Thus, the final answers are: v^(0.8165,0.4082,0.4082),uv=2,u+v3.6056 \boxed{\hat{v} \approx (0.8165, 0.4082, -0.4082), \quad \vec{u} \cdot \vec{v} = 2, \quad \|\vec{u} + \vec{v}\| \approx 3.6056}

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