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(2.2) For any two subsets of $\mathcal{A}, \mathcal{B} \subseteq \mathbb{R}^{n}...

May 23, 2024

Solution by Steps

step 1

Given two orthogonal vectors

x, y \in \mathbb{R}^{n}

, we need to show that

\|x+y\|^{2}=\|x\|^{2}+\|y\|^{2}

step 2

Start by expanding the norm squared of the sum of the vectors:

\|x+y\|^{2} = (x+y) \cdot (x+y)

step 3

Use the distributive property of the dot product:

(x+y) \cdot (x+y) = x \cdot x + 2(x \cdot y) + y \cdot y

step 4

Since

x

and

y

are orthogonal,

x \cdot y = 0

. Thus, the equation simplifies to:

x \cdot x + y \cdot y

step 5

Recognize that

x \cdot x = \|x\|^{2}

and

y \cdot y = \|y\|^{2}

. Therefore,

\|x+y\|^{2} = \|x\|^{2} + \|y\|^{2}

Answer

\|x+y\|^{2} = \|x\|^{2} + \|y\|^{2}

Key Concept

Orthogonal Vectors and Norms

Explanation

The key concept here is that the dot product of orthogonal vectors is zero, which simplifies the expression for the norm of their sum.

Solution by Steps

step 1

Let

\mathcal{A}=\{A\} \in \mathbb{R}^{3}

be a single point and

\mathcal{B} \subseteq \mathbb{R}^{3}

a plane. We need to show that there is a unique point

B \in \mathcal{B}

such that

\overrightarrow{A B}

is orthogonal to

\mathcal{B}

step 2

A plane in

\mathbb{R}^{3}

can be defined by a point

P

and a normal vector

\mathbf{n}

step 3

The vector

\overrightarrow{A B}

from point

A

to point

B

on the plane must be parallel to the normal vector

\mathbf{n}

step 4

Therefore,

B

is the point on the plane such that

\overrightarrow{A B} = \lambda \mathbf{n}

for some scalar

\lambda

step 5

Since

\mathbf{n}

is unique for a given plane, the point

B

is uniquely determined by the intersection of the line through

A

in the direction of

\mathbf{n}

with the plane

\mathcal{B}

Answer

There is a unique point

B \in \mathcal{B}

such that

\overrightarrow{A B}

is orthogonal to

\mathcal{B}

Key Concept

Orthogonality and Planes

Explanation

The key concept is that the orthogonal projection of a point onto a plane is unique and lies along the direction of the plane's normal vector.

Solution by Steps

step 1

Given that

\overrightarrow{A B}

is orthogonal to

\mathcal{B}

, we need to conclude that the distance between

\mathcal{A}

and

\mathcal{B}

\|\overrightarrow{A B}\|

step 2

The distance between a point and a plane is defined as the length of the perpendicular (orthogonal) segment from the point to the plane

step 3

Since

\overrightarrow{A B}

is orthogonal to the plane

\mathcal{B}

, it represents the shortest distance from point

A

to the plane

step 4

Therefore, the distance between

\mathcal{A}

and

\mathcal{B}

is given by the magnitude of the vector

\overrightarrow{A B}

, which is

\|\overrightarrow{A B}\|

Answer

The distance between

\mathcal{A}

and

\mathcal{B}

\|\overrightarrow{A B}\|

Key Concept

Distance from Point to Plane

Explanation

The key concept is that the shortest distance from a point to a plane is along the line perpendicular to the plane, which is represented by the orthogonal vector.