Questions: Prove the following by showing the algebra. If vectors u and v are unit vectors and perpendicular and live in R^3, then u+v = sqrt(2).

Transcript text: Prove the following by showing the algebra. If vectors $\vec{u}$ and $\vec{v}$ are unit vectors and perpendicular and live in $\mathbb{R}^{3}$, then $\|\vec{u}+\vec{v}\|=\sqrt{2}$.

Solution

Solution Steps

To solve this problem, we need to use the properties of unit vectors and the fact that they are perpendicular. Since both vectors are unit vectors, their magnitudes are 1. The fact that they are perpendicular means their dot product is zero. We can use these properties to find the magnitude of their sum.

Use the formula for the magnitude of the sum of two vectors: $\|\vec{u} + \vec{v}\| = \sqrt{\|\vec{u}\|^2 + \|\vec{v}\|^2 + 2 \cdot \vec{u} \cdot \vec{v}}$.
Substitute the known values: $\|\vec{u}\| = 1$, $\|\vec{v}\| = 1$, and $\vec{u} \cdot \vec{v} = 0$.
Simplify the expression to find the magnitude of $\vec{u} + \vec{v}$.

Step 1: Define Unit Vectors and Their Properties

Given that $\vec{u}$ and $\vec{v}$ are unit vectors in $\mathbb{R}^3$, we have $\|\vec{u}\| = 1$ and $\|\vec{v}\| = 1$. Additionally, since they are perpendicular, their dot product is $\vec{u} \cdot \vec{v} = 0$.

Step 2: Use the Magnitude Formula for Vector Addition

The magnitude of the sum of two vectors $\vec{u}$ and $\vec{v}$ is given by: \[ \|\vec{u} + \vec{v}\| = \sqrt{\|\vec{u}\|^2 + \|\vec{v}\|^2 + 2 \cdot (\vec{u} \cdot \vec{v})} \]

Step 3: Substitute Known Values

Substitute the known values into the formula: \[ \|\vec{u} + \vec{v}\| = \sqrt{1^2 + 1^2 + 2 \cdot 0} = \sqrt{1 + 1} = \sqrt{2} \]