Questions: Problem A. 10 Find the magnitude of the vector u and a unit vector in the direction of the vector u. Do not approximate square roots and fractions as decimals. A.10.a u = [3,-2]

$Problem A. 10 Find the magnitude of the vector u and a unit vector in the direction of the vector u. Do not approximate square roots and fractions as decimals. A.10.a u = [3,-2]$

Transcript text: Problem A. 10 Find $\|\overrightarrow{\boldsymbol{u}}\|$ and a unit vector in the direction of $\overrightarrow{\boldsymbol{u}}$. Do not approximate square roots and fractions as decimals. \[ \text { A.10.a } \vec{u}=[3,-2] \]

Solution

Solution Steps

To find the magnitude of the vector $\vec{u}$, we use the formula $\|\vec{u}\| = \sqrt{u_1^2 + u_2^2}$. To find a unit vector in the direction of $\vec{u}$, we divide each component of $\vec{u}$ by its magnitude.

Solution Approach

Calculate the magnitude of $\vec{u}$ using the formula $\|\vec{u}\| = \sqrt{u_1^2 + u_2^2}$.
Find the unit vector by dividing each component of $\vec{u}$ by its magnitude.

Step 1: Calculate the Magnitude of $\vec{u}$

The magnitude of the vector $\vec{u} = [3, -2]$ is calculated using the formula: \[ \|\vec{u}\| = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \]

Step 2: Calculate the Unit Vector in the Direction of $\vec{u}$

To find the unit vector in the direction of $\vec{u}$, we divide each component of $\vec{u}$ by its magnitude: \[ \text{Unit vector} = \left[ \frac{3}{\sqrt{13}}, \frac{-2}{\sqrt{13}} \right] \]