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Problem Statement

In the $Oxyz$ space, given two vectors $\vec{a} = (-2; 1; 2)$ and $\vec{b} = (1; 1; -1)$.

• a) Calculate the magnitude (length) of vector $\vec{a}$ and vector $\vec{b}$.
• b) Calculate the dot product of the two vectors: $\vec{a} \cdot \vec{b} = ?$
• c) Calculate the angle between the two vectors $(\vec{a}, \vec{b})$ (round the result to the first decimal place).

Solutions

a) Magnitude of the vectors

The magnitude of a vector $\vec{u} = (x; y; z)$ is calculated as $|\vec{u}| = \sqrt{x^2 + y^2 + z^2}$.

• Magnitude of $\vec{a}$:
• $|\vec{a}| = \sqrt{(-2)^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$
• Magnitude of $\vec{b}$:
• $|\vec{b}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$

b) Dot product of the two vectors

The dot product of $\vec{a} = (x_1; y_1; z_1)$ and $\vec{b} = (x_2; y_2; z_2)$ is $x_1x_2 + y_1y_2 + z_1z_2$.

$\vec{a} \cdot \vec{b} = (-2 \cdot 1) + (1 \cdot 1) + (2 \cdot -1)$

$\vec{a} \cdot \vec{b} = -2 + 1 - 2 = -3$

c) Angle between the two vectors

The cosine of the angle $\theta$ between two vectors is given by the formula:

$\cos(\theta) = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot |\vec{b}|}$

Substituting our values:

$\cos(\theta) = \frac{-3}{3 \cdot \sqrt{3}} = \frac{-1}{\sqrt{3}}$

To find the angle:

$\theta = \arccos\left(\frac{-1}{\sqrt{3}}\right) \approx 125.26^\circ$

Rounding to the first decimal place:

$\theta \approx 125.3^\circ$

Summary of Results

ParameterValue**Magnitude $\vec{a}**Magnitude $\vec{b}Dot Product $\vec{a} \cdot \vec{b}$$-3$Angle $(\vec{a}, \vec{b})$$125.3^\circ$